Solutions
1. 1]
Solution:
Show "Ÿab Ÿ"a b from the field axioms.
You need to show that Ÿ"a b acts like the additive inverse of ab because then, it follows that it is
the additive inverse. However, ab Ÿ"a b Ÿa Ÿ"a b 0b 0 and so "Ÿab Ÿ"a b.
2. 2]
Solution:
Show the maximum (minimum) of ax 2 bx c, a p 0, always occurs when x "b/Ÿ2a .
Then you get this equals aŸx 2 ba x ac . It equals
2
2
a x 2 ba x b 2 " b 2 ac
4a
4a
2
2
a x b
4ac "2 b
2a
4a
b
Thus a minimum occurs if a 0 and it occurs at x " 2a
while if a 0, you get a maximum at
the same value of x.
3. 3]
Solution:
Let fŸt t 2 4t for t u "2. Find a formula for f "1 Ÿy .
Answer: Let y x 2 4x and solve for x f "1 Ÿy .
"4 o 4y 16
2
Here y u "4 and you also need x u "2. Therefore, you need to use the sign. Hence
x
f "1 Ÿy y4 "2
4. 4]
Solution:
Suppose
fŸx, y ax 2 by 2 2cxy
and suppose a b 0 and ab " c 2 0. Show that under these conditions, the above function is
nonnegative. This is a function of two variables. You can pick x, y to be any numbers and the
above expression under the given conditions is always nonnegative.
You just complete the square. Assume first that a 0. The function equals
2cxy
c2y2
c2y2
by 2
a a2 " a2
cy 2 c 2 y 2
a x a
" 2
by 2
a
2
cy 2
a x a
by 2 " ca y 2
cy 2 1 2
a x a
a y Ÿab " c 2 u 0
by assumption. If a 0, then b 0 since their sum is positive. Then a similar argument works.
a x2 2cxy
a
by 2 a x 2 You first complete the square on y and then check what is left over.
5. 5]
Solution:
Start with
"b o b 2 " 4ca
2a
and show directly that x given by either of the two numbers in the above formula satisfies the
quadratic equation ax 2 bx c 0.
x
a
"b b 2 "4ca
2a
a
1
2a 2
b2 "
2
1
a
b
c"
"b b 2 "4ca
2a
1
2a 2
c
b b 2 " 4ac
b
"b b 2 "4ca
2a
c 0
The other case is similar.
6. 6]
Solution:
Suppose there are two real solutions to ax 2 bx c 0, x 1 , x 2 . Show the maximum (minimum),
depending on sign of a, of fŸx ax 2 bx c occurs at Ÿx 1 x 2 /2.
When you complete the square, you obtain the polynomial equals
2
2
" b " 4ac
a x b
2a
4a 2
You are also given that x 1 and x 2 make this expression equal to 0. Therefore, the two solutions to
that equation are the two numbers given below.
"b o b 2 " 4ac
2a
It follows that if these are added and the result divided by 2 you obtain
"2b 1 " 1 b
2a 2
2a
which, from the above expression is the correct value to maximize or minimize the polynomial.
7. 7]
Solution:
For z equal to the following complex number, find z "1 .
a. 2 5i
2
5
" 29
i
29
b. 9 " 9i
1
1
18
i
18
c. "5 2i
5
2
" 29
" 29
i
d. 9 4i
9
4
" 97
i
97
8. 8]
Solution:
Let z 3 8i and let w 6 " 8i. Find zw, z w, z 2 , and w/z.
" 72
i
82 24i, 9, "55 48i, " 46
73
73
9. 9]
Solution:
Give the complete solution to x 4 16 0. Graph these points in the complex plane.
Ÿ1 i 2 , "Ÿ1 " i 2 , "Ÿ1 i 2 , Ÿ1 " i 2
y
4
2
-4
-2
2
4
x
-2
-4
10. 10]
Solution:
Give the complete solution to x 3 64 0. Graph these points in the complex plane.
"4, 2 " 2i 3 , 2i 3 2
y
4
2
-4
-2
2
4
x
-2
-4
11. 11]
Solution:
If z is a complex number, show there exists F a complex number with |F| 1 and Fz |z|.
If z 0, let F 1. If z p 0, let F |z|z
12. 12]
Solution:
De Moivre’s theorem says ¡rŸcos t i sin t ¢ n r n Ÿcos nt i sin nt for n a positive integer. Does
this formula continue to hold for all integers, n, even negative integers? Explain.
Yes, it holds for all integers. First of all, it clearly holds if n 0. Suppose now that n is a negative
integer. Then "n 0 and so
1
1
"n
¡rŸcos t i sin t ¢ n r ŸcosŸ"nt i sinŸ"nt ¡rŸcos t i sin t ¢ "n
r n ŸcosŸnt i sinŸnt rn
ŸcosŸnt " i sinŸnt ŸcosŸnt " i sinŸnt ŸcosŸnt i sinŸnt r n ŸcosŸnt i sinŸnt because ŸcosŸnt " i sinŸnt ŸcosŸnt i sinŸnt 1.
13. 13]
Solution:
You already know formulas for cosŸx y and sinŸx y and these were used to prove De
Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sinŸ4x and one for
cosŸ4x .
cos 4 x 4i cos 3 x sin x " 6 cos 2 x sin 2 x " 4i cos x sin 3 x sin 4 x
The real part is cosŸ4x and the imaginary part is sinŸ4x .
14. 14]
Solution:
If z and w are two complex numbers and the polar form of z involves the angle 2 while the polar
form of w involves the angle C, show that in the polar form for zw the angle involved is 2 C.
Also, show that in the polar form of a complex number, z, r |z|.
You have z |z|Ÿcos 2 i sin 2 and w |w|Ÿcos C i sin C . Then when you multiply these, you
get
|z||w|Ÿcos 2 i sin 2 Ÿcos C i sin C |z||w|Ÿcos 2 cos C " sin 2 sin C iŸcos 2 sin C cos C sin 2 |z||w|ŸcosŸ2 C i sinŸ2 C 15. 15]
Solution:
Factor x 3 64 as a product of linear factors.
Ÿx 4 x 2i 3 " 2 x " 2i 3 " 2
16. 16]
Solution:
Write x 3 27 in the form Ÿx 3 Ÿx 2 ax b where x 2 ax b cannot be factored any more
using only real numbers.
Ÿx 3 Ÿx 2 " 3x 9 17. 17]
Solution:
Completely factor x 4 256 as a product of linear factors.
x Ÿ2 2i 2 x Ÿ2 " 2i 2 x " Ÿ2 " 2i 2 x " Ÿ2 2i 2
18. 18]
Solution:
Completely factor x 4 81 as a product of of two quadratic polynomials each of which cannot be
factored further without using complex numbers.
x2 3 2 x 9 x2 " 3 2 x 9
19. 19]
Solution:
If z, w are complex numbers prove zw z w and then show by induction that z 1 Cz m z 1 Cz m .
m
m
Also verify that ! k1 z k ! k1 z k . In words this says the conjugate of a product equals the
product of the conjugates and the conjugate of a sum equals the sum of the conjugates.
Ÿa ib Ÿc id ac " bd iŸad bc Ÿac " bd " iŸad bc Ÿa " ib Ÿc " id ac " bd " iŸad bc which is the same thing. Thus it holds for a product of
two complex numbers. Now suppose you have that it is true for the product of n complex
numbers. Then
z 1 Cz n1 z 1 Cz n z n1
and now, by induction this equals
z 1 Cz n z n1
As to sums, this is even easier.
n
n
n
j1
j1
j1
!Ÿx j iy j ! x j i ! y j
n
n
n
n
j1
j1
j1
j1
! x j " i ! y j ! x j " iy j ! Ÿx j iy j .
20. 20]
Solution:
Suppose pŸx a n x n a n"1 x n"1 C a 1 x a 0 where all the a k are real numbers. Suppose also
that pŸz 0 for some z C. Show it follows that pŸ z 0 also.
You just use the above problem. If pŸz 0, then you have
pŸz 0 a n z n a n"1 z n"1 C a 1 z a 0
a n z n a n"1 z n"1 C a 1 z a 0
a n z n a n"1 z n"1 C a 1 z a 0
a n z n a n"1 z n"1 C a 1 z a 0
pŸ z 21. 21]
Solution:
Show that "2 5i and 9 " 10i are the only two zeros to the polynomial
pŸx x 2 Ÿ"7 5i x Ÿ32 65i Thus the zeros do not necessarily come in conjugate pairs if the coefficients of the polynomial are
not real.
Ÿx " Ÿ"2 5i Ÿx " Ÿ9 " 10i x 2 Ÿ"7 5i x Ÿ32 65i 22. 22]
Solution:
I claim that 1 "1. Here is why.
" 1 i2 "1 "1 Ÿ"1 2 1 1.
This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?
Something is wrong. There is no single "1 .
23. 23]
Solution:
De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents, not just
integers.
1 1 Ÿ1/4 Ÿcos 2= i sin 2= 1/4 cosŸ=/2 i sinŸ=/2 i.
Therefore, squaring both sides it follows 1 "1 as in the previous problem. What does this tell
you about De Moivre’s theorem? Is there a profound difference between raising numbers to
integer powers and raising numbers to non integer powers?
It doesn’t work. This is because there are four fourth roots of 1.
24. 24]
Solution:
If n is an integer, is it always true that Ÿcos 2 " i sin 2 n cosŸn2 " i sinŸn2 ? Explain.
Yes, this is true.
Ÿcos 2 " i sin 2 n ŸcosŸ"2 i sinŸ"2 n
cosŸ"n2 i sinŸ"n2 cosŸn2 " i sinŸn2 25. 25]
Solution:
Suppose you have any polynomial in cos 2 and sin 2. By this I mean an expression of the form
! m)0 ! n*0 a )* cos ) 2 sin * 2 where a )* C. Can this always be written in the form
nm
b cos +2 ! A"Ÿnm c A sin A2? Explain.
! mn
+"Ÿnm +
Yes it can. It follows from the identities for the sine and cosine of the sum and difference of
angles that
sin a sin b 1 ŸcosŸa " b " cosŸa b 2
1
cos a cos b ŸcosŸa b cosŸa " b 2
sin a cos b 1 ŸsinŸa b sinŸa " b 2
Now cos 2 1 cos 2 0 sin 2 and sin 2 0 cos 2 1 sin 2. Suppose that whenever k t n,
k
cos k Ÿ2 ! a j cosŸj2 b j sinŸj2 j"k
for some numbers a j , b j . Then
n
cos
n1
Ÿ2 ! a j cosŸ2 cosŸj2 b j cosŸ2 sinŸj2 j"n
Now use the above identities to write all products as sums of sines and cosines of
Ÿj " 1 2, j2, Ÿj 1 2. Then adjusting the constants, it follows
n1
cos
n1
Ÿ2 !
a Uj cosŸ2 cosŸj2 b Uj cosŸ2 sinŸj2 j"n1
Now use the above formulas to get the thing you need. You can do something similar with sin n Ÿ2 and with products of the form cos ) 2 sin * 2.
26. 26]
Solution:
Suppose pŸx a n x n a n"1 x n"1 C a 1 x a 0 is a polynomial and it has n zeros,
z 1 , z 2 , C, z n
listed according to multiplicity. (z is a root of multiplicity m if the polynomial fŸx Ÿx " z m
divides pŸx but Ÿx " z fŸx does not.) Show that
pŸx a n Ÿx " z 1 Ÿx " z 2 CŸx " z n .
pŸx Ÿx " z 1 qŸx rŸx where rŸx is a nonzero constant or equal to 0. However, rŸz 1 0
and so rŸx 0. Now do to qŸx what was done to pŸx and continue until the degree of the
resulting qŸx equals 0. Then you have the above factorization.
27. 27]
Solution:
Give the roots of the following quadratic polynomials having real coefficients.
a. x 2 " 6x 25, 3 4i, 3 " 4i
b. 3x 2 " 18x 75, 3 4i, 3 " 4i
c. 9x 2 " 24x 32, 43 43 i, 43 " 43 i
d. 16x 2 " 24x 25, 34 i, 34 " i
e. 16x 2 " 24x 25, 34 i, 34 " i
28. 28]
Solution:
Give the solutions to the following quadratic equations having possibly complex coefficients.
a. x 2 " Ÿ6 " i x 39 " 3i 0, 3 5i, 3 " 6i
b. x 2 " Ÿ6 " i x 39 " 3i 0, 3 5i, 3 " 6i
c. 9x 2 " Ÿ18 " 3i x 21 " 3i 0, 1 i, 1 " 43 i
d. 9x 2 21
ix 9 0, 34 i, " 43 i
4
Fn
1. 29]
Solution:
Find 8Ÿ9, 1, "6, "2 9Ÿ"1, "5, 6, 8 .
63 "37 6 56
2. 30]
Solution:
Find parametric equations for the line through Ÿ5, 7, 0 and Ÿ"4, "4, 3 . Recall that to find the
parametric line through two points Ÿa, b, c and Ÿd, e, f , you find a direction vector of the form
Ÿd " a, e " b, f " c and then the equation of the line is Ÿx, y, z Ÿa, b, c tŸd " a, e " b, f " c where t R. You should check that this works in the sense that when t 0, you are at Ÿa, b, c and when t 1, the value is Ÿd, e, f .
5 7 0
t
"9 "11 3
3. 31]
Solution:
Find parametric equations for the line through the point Ÿ"7, 7, 0 with a direction vector
Ÿ1, "5, "3 . Recall from calculus, the equation of a line through Ÿa, b, c having direction vector
Ÿf, g, h is given by Ÿx, y, z Ÿa, b, c tŸf, g, h .
t " 7 7 " 5t "3t
4. 32]
Solution:
Parametric equations of the line are Ÿx, y, z 2t " 6 3t 6 3t 3 . Find a direction vector
for the line and a point on the line. Recall that to find a direction vector, you could subtract one
point on the line from the other or else just recall that a parametric equation of a line is of the form
Point on the line t direction vector and use this in the above.
"6 6 3
is a point.
2 3 3
is a direction vector.
5. 33]
Solution:
"1 9 3 , 1 6 3 .
Recall that to find the parametric line through two points Ÿa, b, c and Ÿd, e, f , you find a direction
vector of the form Ÿd " a, e " b, f " c and then the equation of the line is
Ÿx, y, z Ÿa, b, c tŸd " a, e " b, f " c where t R. You should check that this works in the
sense that when t 0, you are at Ÿa, b, c and when t 1, the value is Ÿd, e, f .
Find parametric equations for the line through the two points
2t " 1 9 " 3t 3
6. 34]
Solution:
The equation of a line in two dimensions is written as y 2x " 2. Find parametric equations for
this line.
x t, y 2t " 2 and symmetric equations x 12 y 1.
7. 35]
Solution:
Find 5Ÿ3, "6, 0, 5 3Ÿ2, "6, 0, 9 .
21 "48 0 52
8. 36]
Solution:
Find parametric equations for the line through Ÿ2, 4, "1 and Ÿ1, 7, "6 . Recall that to find the
parametric line through two points Ÿa, b, c and Ÿd, e, f , you find a direction vector of the form
Ÿd " a, e " b, f " c and then the equation of the line is Ÿx, y, z Ÿa, b, c tŸd " a, e " b, f " c where t R. You should check that this works in the sense that when t 0, you are at Ÿa, b, c and when t 1, the value is Ÿd, e, f .
t 2 4 " 3t 5t " 1
9. 37]
Solution:
Find parametric equations for the line through the point Ÿ"7, 2, "4 with a direction vector
Ÿ1, 5, "6 .
t " 7 5t 2 "6t " 4
10. 38]
Solution:
Parametric equations of the line are Ÿx, y, z 2t 3 6t 8 3t 6
. Find a direction vector
for the line and a point on the line.
3 8 6
is a point.
2 6 3
is a direction vector.
11. 39]
Solution:
0 4 5 , 2 "5 "5 .
Recall that to find the parametric line through two points Ÿa, b, c and Ÿd, e, f , you find a direction
vector of the form Ÿd " a, e " b, f " c and then the equation of the line is
Ÿx, y, z Ÿa, b, c tŸd " a, e " b, f " c where t R. You should check that this works in the
sense that when t 0, you are at Ÿa, b, c and when t 1, the value is Ÿd, e, f .
Find parametric equations for the line through the two points
2t 4 " 9t 5 " 10t
12. 40]
Solution:
Find the point on the line segment from
from
"5 5 4
to
1 "6 "5
"5 5 4
to
1 "6 "5
which is
1
5
of the way
.
" 15 " 19
" 16
5
5
13. 41]
Solution:
P P
The point is P 1 P32 P 3 because this point is of the form P i 13 j 2 k 23 whenever i, j, k are distinct.
The last expression is a point on the line between the midpoint of P j and P k and the point P i . This
point is called the barycenter.
14. 42]
Solution:
The wind blows from the South at 50 kilometers per hour and an airplane which flies at 300
kilometers per hour in still air is heading East. What is the actual velocity of the airplane and
where will it be after two hours.
300 50
and after two hours it is at
.
600 100
15. 43]
Solution:
The wind blows from the West at 50 kilometers per hour and an airplane which flies at 500
kilometers per hour in still air is heading North East. What is the actual velocity of the airplane and
where will it be after two hours.
250 2 50 250 2
it will be at
500 2 100 500 2
16. 44]
Solution:
The wind blows from the North at 20 kilometers per hour and an airplane which flies at 300
kilometers per hour in still air is supposed to go to the point whose coordinates are at
200 200 kilometers. In what direction should the airplane head?
Let its direction be Ÿp, q . Then after much fuss, q need to solve the equations
449 1
30
1
30
,p 1
30
449 "
1
30
. You
300p 300q " 20
p2 q2 1
where also p, q u 0.
17. 45]
Solution:
Three forces act on an object. Two are
"7 "5 1
2 "1 "6
and
Newtons. Find the
third force if the object is not to move.
5 6 5
18. 46]
Solution:
Three forces act on an object. Two are
7 2 "1
third force if the total force on the object is to be
and
1 "6 "5
"6 "6 "5
Newtons. Find the
.
"14 "2 1
19. 47]
Solution:
A river flows West at the rate of b miles per hour. A boat can move at the rate of 1 miles per
hour. Find the smallest value of b such that it is not possible for the boat to proceed directly across
the river.
b 1.
20. 48]
Solution:
The wind blows from West to East at a speed of 20 miles per hour and an airplane which travels
at 300 miles per hour in still air is heading North West. What is the velocity of the airplane relative
to the ground? What is the component of this velocity in the direction North?
The velocity is the sum of two vectors.
20 " 150 2 150 2
In the direction of North, the component is 150 2 .
21. 49]
Solution:
The wind blows from West to East at a speed of 60 miles per hour and an airplane can travel
travels at 400 miles per hour in still air. How many degrees West of North should the airplane
head in order to travel exactly North?
Let Ÿa, b be a unit vector in the appropriate direction. Then you need
400Ÿa, b 60Ÿ1, 0 to have direction vector Ÿ0, 1 . Thus you need
a400 60 0
and so
a "60
400
and so
b
1"
"60
400
2
1 391
20
Thus the appropriate direction vector is
"60 , 1 391
400 20
To find the angle you would need to find the cosine of the angle between Ÿ0, 1 and this vector.
This is
1 391
20
and so the radian measure is
0. 150 57
Then this needs to be changed to degrees.
8. 626 9
22. 50]
Solution:
The wind blows from West to East at a speed of 60 miles per hour and an airplane which travels
at 300 miles per hour in still air heading somewhat West of North so that, with the wind, it is
flying due North. It uses 40. 0 gallons of gas every hour. It has to travel 200. 0 miles, how much
gas will it use in flying to its destination?
As above, the velocity of the plane is
"60. 0 293. 94
and then there is the wind which gives
a velocity vector of
60. 0 0
293. 94. How many hours?
. Therefore, the component in the direction North is just
200. 0 0. 680 41
293. 94
How much gas?
Ÿ40. 0 Ÿ0. 680 41 27. 217
23. 51]
Solution:
An airplane is flying due north at 100. 0 miles per hour but it is not actually going due North
because there is a wind which is pushing the airplane due east at 20. 0 miles per hour. After 1
hour, the plane starts flying 30 ( East of North. Assuming the plane starts at Ÿ0, 0 , where is it after
2 hours? Let North be the direction of the positive y axis and let East be the direction of the
positive x axis.
Where is it after one hour?
20. 0 100. 0
or in other words, in the direction of
1
2
,
3
2
. Now from here it heads 20 degrees east of North
. Thus the new velocity vector is
70. 0 86. 603
Then after another hour, it will have gone this vector added to the first. Thus its position then is
90. 0 186. 6
24. 52]
Solution:
City A is located at the origin while city B is located at Ÿ300, 500 where distances are in miles.
An airplane flies at 250 miles per hour in still air. This airplane wants to fly from city A to city B
but the wind is blowing in the direction of the positive y axis at a speed of 50 miles per hour. Find
a unit vector such that if the plane heads in this direction, it will end up at city B having flown the
shortest possible distance. How long will it take to get there?
Wind: Ÿ0, 50 . Direction it needs to travel: Ÿ3, 5 1 . Then you need 250Ÿa, b Ÿ0, 50 to have
34
this direction where Ÿa, b is an appropriate unit vector. Thus you need
a2 b2 1
250b 50 5
250a
3
3
4
Thus a 5 , b 5 . Then the velocity of the plane relative to the ground is
150 250
. The
speed of the plane relative to the ground is then
Ÿ150 2 Ÿ250 2 291. 55.
It has to go a distance of Ÿ300 2 Ÿ500 2 583. 10 miles. Therefore, it takes
583. 1 2 hours
291. 55
25. 53]
Solution:
A certain river is one half mile wide with a current flowing at 2. 0 miles per hour from East to
West. A man takes a boat directly toward the opposite shore from the South bank of the river at a
speed of 6. 0 miles per hour. How far down the river does he find himself when he has swam
across? How far does he end up travelling?
Water: Ÿ"2. 0, 0 Boat: Ÿ0, 6. 0 . Resultant velocity:
"2. 0 6. 0 Time taken to get across:
8. 333 3 • 10 "2 . How far down the river: "0. 166 67. Distance travelled: 0. 527 05.
26. 54]
Solution:
A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West.
A man can swim at 3 miles per hour in still water. In what direction should he swim in order to
travel directly across the river? What would the answer to this problem be if the river flowed at 3
miles per hour and the man could swim only at the rate of 2 miles per hour?
Man: 3Ÿa, b Water: Ÿ"2, 0 . Then you need 3a 2 and so a 2/3 and hence b 5 /3. The
5
vector is then 23 , 3 . In the second case, he could not do it. You would need to have a unit
vector Ÿa, b such that 2a 3. This is not possible, not even if you try real hard.
27. 55]
Solution:
Three forces are applied to a point which does not move. Two of the forces are
3. 0i 3. 0j 8. 0k Newtons and "7. 0i 8. 0j 3. 0k Newtons. Find the third force.
x " 4 y 11 z 11
needs to equal the zero vector. You need
4 "11 "11
28. 56]
Solution:
The total force acting on an object is to be Ÿ1 i Ÿ4 j Ÿ3 k Newtons. A force of
Ÿ"8 i Ÿ"2 j Ÿ8 k Newtons is being applied. What other force should be applied to achieve the
desired total force?
Need
1 4 3
x y z
"8 "2 8
x y z
Thus
9 6 "5
29. 57]
Solution:
A bird flies from its nest 8 km. in the direction 32 = north of east where it stops to rest on a tree. It
then flies 5 km. in the direction due southeast and lands atop a telephone pole. Place an xy
coordinate system so that the origin is the bird’s nest, and the positive x axis points east and the
positive y axis points north. Find the displacement vector from the nest to the telephone pole.
To the tree:
To pole:
0 "8
" 52 2 " 52 2
Total displacement:
" 52 2 " 52 2 " 8
30. 58]
Solution:
A car is stuck in the mud. There is a cable stretched tightly from this car to a tree which is 20 feet
long. A person grasps the cable in the middle and pulls with a force of 150 pounds perpendicular
to the stretched cable. The center of the cable moves 3 feet and remains still. What is the tension
in the cable? The tension in the cable is the force exerted on this point by the part of the cable
nearer the car as well as the force exerted on this point by the part of the cable nearer the tree.
Assume the cable cannot be lengthened so the car moves slightly in moving the center of the
cable.
3
2T cos ) 150 where cos ) 10
75
T cos)
250
31. 59]
Solution:
A car is stuck in the mud. There is a cable stretched tightly from this car to a tree which is 20 feet
long. A person grasps the cable in the middle and pulls with a force of 80 pounds perpendicular to
the stretched cable. The center of the cable moves 3 feet and remains still. What is the tension in
the cable? The tension in the cable is the force exerted on this point by the part of the cable nearer
the car as well as the force exerted on this point by the part of the cable nearer the tree. Assume
the cable does lengthen and the car does not move at all.
3
2T cos ) 80 where cos ) 109
109
40
40
T cos) 3 109
32. 60]
Solution:
A box is placed on a strong plank of wood and then one end of the plank of wood is gradually
raised. It is found that the box does not move till the angle of inclination of the wood equals 60
degrees at which angle the box begins to slide. What is the coefficient of static friction?
You need to have 6wcos 60 wsin 60 and so tan 60 6.
33. 61]
Solution:
Recall that the open ball having center at a and radius r is given by
BŸa, r q £x : |x " a| r¤
Show that if y BŸa, r , then there exists a positive number - such that BŸy, - – BŸa, r . The
symbol – means that every point in BŸy, - is also in BŸa, r . In words, it states that BŸy, - is
contained in BŸa, r . When you have done this, you will have shown that an open ball is open.
This is a fantastically important observation although its major implications will not be explored
very much in this book.
You let - r " |y " a|. Then if z BŸy, - , it follows from the triangle inequality that
|z " a| t |z " y| |y " a| - |y " a| r " |y " a| |y " a| r
Thus, z BŸa, r .
Dot Product
1. 62]
Solution:
Find the following dot products.
a. Ÿ1, 4, 3, 10 Ÿ2, "1, "1, 3 25
b. Ÿ2, 7, 5 Ÿ"1, "3, 2 "13
c. Ÿ4, 2, 5, 6, 7 Ÿ"1, 1, "1, 2, 0 5
2. 63]
Solution:
Use the geometric description of the dot product to verify the Cauchy Schwartz inequality and to
show that equality occurs if and only if one of the vectors is a scalar multiple of the other.
That formula said that a b |a||b|cos 2 where 2 is the included angle. Therefore,
|a b| ||a||b|cos 2| |a||b||cos 2| t |a||b|.
3. 64]
Solution:
For u, v vectors in R 3 , define the product, u ' v q u 1 v 1 2u 2 v 2 3u 3 v 3 . Show the axioms for a
dot product all hold for this funny product. Prove
|u ' v| t Ÿu ' u 1/2 Ÿv ' v 1/2 .
Hint: Do not try to do this with methods from trigonometry.
These axioms are all obvious. Therefore, since the proof of the Cauchy Schwarz inequality holds
for an arbitrary product satisfying those axioms, the Cauchy Schwarz inequality holds.
4. 65]
Solution:
Find the angle between the vectors 4i " j " k and i 8j 2k.
4 "1 "1
2 arccos
1 8 2
18 69
1
" 69
2
1
" 69
2 69
1. 741 9
69
5. 66]
Solution:
Find the angle between the vectors 4i " j " k and i 6j 2k.
4 "1 "1
2 arccos
18 41
2
" 123
1 6 2
2
" 123
2 41
1. 718 6
2 41
6. 67]
Solution:
Find proj u Ÿv where v Ÿ6. 0, 2. 0, "2 and u Ÿ1, 7. 0, 3 .
Ÿ1,7.0,3 Ÿ6. 0, 2. 0, "2 Ÿ1,7.0,3 14
59 Ÿ1,7.0,3 59
59
7. 68]
Solution:
59
59
Find proj u Ÿv where v Ÿ3. 0, 3. 0, "2 and u Ÿ1, 7. 0, 3 .
Ÿ1,7.0,3 Ÿ1,7.0,3 Ÿ1,7.0,3 Ÿ3. 0, 3. 0, "2 18
59
59
59
59
59
8. 69]
Solution:
Find proj u Ÿv where v Ÿ2. 0, "3. 0, "2 and u Ÿ1, 5. 0, 3 .
Ÿ1,5.0,3 Ÿ2. 0, "3. 0, "2 Ÿ1,5.0,3 " 19
35 Ÿ1,5.0,3 35
35
35
35
9. 70]
Solution:
A boy drags a sled for 40 feet along the ground by pulling on a rope which is 45 degrees from the
horizontal with a force of 10 pounds. How much work does this force do?
Ÿ40 Ÿ10 ŸcosŸ 14 = 17997.
10. 71]
Solution:
A dog drags a sled for 80 feet along the ground by pulling on a rope which is 30 degrees from the
horizontal with a force of 20 pounds. How much work does this force do?
Ÿ80 Ÿ20 ŸcosŸ 16 = 47993.
11. 72]
Solution:
A boy drags a sled for 70 feet along the ground by pulling on a rope which is 20 degrees from the
horizontal with a force of 60 pounds. How much work does this force do?
Ÿ70 Ÿ60 ŸcosŸ 19 = 83987.
12. 73]
Solution:
How much work in Newton meters does it take to slide a crate 80. 0 meters along a loading dock
by pulling on it with a 400. 0 Newton force at an angle of 20. 0 degrees from the horizontal?
80. 0Ÿ400. 0 cosŸ0. 349 07 6. 399 • 10 5
13. 74]
Solution:
How much work in Newton meters does it take to slide a crate 40. 0 meters along a loading dock
by pulling on it with a 100. 0 Newton force at an angle of 45. 0 degrees from the horizontal?
40. 0Ÿ100. 0 cosŸ0. 785 40 1. 799 7 • 10 5
14. 75]
Solution:
An object moves 11. 0 meters in the direction of j i. There are two forces acting on this object,
F 1 i j 1. 0k, and F 2 "5i Ÿ7. 0 j "6k. Find the total work done on the object by the two
forces. Hint: You can take the work done by the resultant of the two forces or you can add the
work done by each force. Why?
11.0
Ÿ1, 1, 0 Ÿ"4, 8. 0, "5. 0 12 2 11. 0Ÿ4. 0 31. 113
2
15. 76]
Solution:
An object moves 10. 0 meters in the direction of j i. There are two forces acting on this object,
F 1 i j 5. 0k, and F 2 "5i Ÿ1. 0 j "6k. Find the total work done on the object by the two
forces. Hint: You can take the work done by the resultant of the two forces or you can add the
work done by each force. Why?
10.0
Ÿ1, 1, 0 Ÿ"4, 2. 0, "1. 0 12 2 10. 0Ÿ"2. 0 "14. 142
2
16. 77]
Solution:
An object moves 7. 0 meters in the direction of j i. There are two forces acting on this object,
F 1 i j 5. 0k, and F 2 "5i Ÿ2. 0 j "6k. Find the total work done on the object by the two
forces. Hint: You can take the work done by the resultant of the two forces or you can add the
work done by each force. Why?
7.0
Ÿ1, 1, 0 Ÿ"4, 3. 0, "1. 0 12 2 7. 0Ÿ"1. 0 "4. 949 7
2
17. 78]
Solution:
An object moves 6. 0 meters in the direction of j i " k. There are two forces acting on this
object, F 1 i j 6. 0k, and F 2 "5i Ÿ6. 0 j "6k. Find the total work done on the object by
the two forces. Hint: You can take the work done by the resultant of the two forces or you can add
the work done by each force. Why?
6.0
Ÿ1, 1, "1 Ÿ"4, 7. 0, 0. 0 10. 392
3
18. 79]
Solution:
Does it make sense to speak of proj 0 Ÿv ?
No. The 0 vector has no direction and the formula doesn’t make sense either.
19. 80]
Solution:
If F is a force and D is a vector, show proj D ŸF Ÿ|F|cos 2 u where u is the unit vector in the
direction of D, u D/|D| and 2 is the included angle between the two vectors, F and D. |F|cos 2 is
sometimes called the component of the force, F in the direction, D.
D D
1
FD
u |F||D| |D|
proj D ŸF F |D|
Ÿcos 2 u Ÿ|F|cos 2 u
|D|
|D|
20. 81]
Solution:
If a, b, and c are vectors. Show that Ÿb c µ b µ c µ where b µ b " proj a Ÿb .
Ÿb c a
a
|a|2
b " b 2a a c " c 2a a
|a|
|a|
bµ cµ
Ÿb c µ b c "
21. 82]
Solution:
Show that Ÿa b 14 ¡|a b|2 " |a " b|2 ¢.
1
¡|a b|2 " |a " b|2 ¢ 14 ¡Ÿa b Ÿa b " Ÿa " b Ÿa " b ¢
4
14 ŸŸa a " Ÿa a Ÿb b " Ÿb b 4a b a b.
22. 83]
Solution:
Prove from the axioms of the dot product the parallelogram identity,
|a b|2 |a " b|2 2|a|2 2|b|2 .
Explain what this says geometrically in terms of the diagonals and sides of a parallelogram.
The left equals
|a|2 |b|2 2a b |a|2 |b|2 " 2a b 2|a|2 2|b|2
This follows right away from the properties of the dot product. It says that the sum of the squares
of the diagonals of a parallelogram equals the sum of the squares of the sides.
23. 84]
Solution:
Prove the Cauchy Schwarz inequality in R n as follows. For u, v vectors, note that
Ÿu " proj v u Ÿu " proj v u u 0
Now simplify using the axioms of the dot product and then put in the formula for the projection.
Show that you get equality in the Cauchy Schwarz inequality if and only if the above equals zero
which means that u proj v u. What is the geometric significance of this condition?
u " u 2v v
|v|
u " u 2v v
|v|
|u|2 " 2Ÿu v 2 1 2 Ÿu v 2 1 2 u 0
|v|
|v|
And so
|u|2 |v|2 u Ÿu v 2
You get equality exactly when u proj v u uv2 v, in other words, when u is a multiple of v.
|v|
24. 85]
Solution:
For C n , the dot product is often called the inner product. It was given by
n
x y Ÿx, y q
! xjyj
j1
Why place the conjugate on the y j ? Why is it not possible to simply define this the usual way?
It is because if you did, you would not necessarily have Ÿx, x u 0. For example, suppose
x Ÿi, 1 i . Then if you did the usual thing you would get "1 Ÿ1 i 2 " 1 2i which is not
nonnegative. Also, you could have Ÿx, x 0 even though x p 0. For example, Ÿi, 1 .
25. 86]
Solution:
For the complex inner product, prove the parallelogram identity.
|x y|2 |x " y|2 2|x|2 2|y|2
Start with the left. From axioms,
|x y|2 |x " y|2 |x|2 |y|2 2 ReŸx, y |x|2 |y|2 " 2 ReŸx, y 2|x|2 2|y|2
26. 87]
Solution:
For the complex inner product, show that if ReŸx, y 0 for all y C n , then x 0. In fact,
show that for all y, you have Ÿx, y 0.
Ÿx, y C and so there exists a complex number 2 such that |2| 1 and 2Ÿx, y |Ÿx, y |. You
Ÿx,y can take 2 |Ÿx,y |
if Ÿx, y p 0. Of course if this equals zero, there is nothing to show. Then from
the axioms
0 ReŸx, 2 y Re 2Ÿx, y Re|Ÿx, y | |Ÿx, y |.
Letting y x, shows that x 0.
27. 88]
Solution:
For the complex inner product, show that for Ÿx, y ReŸx, y i ImŸx, y , it follows that
ImŸx, y ReŸx, iy so that
Ÿx, y ReŸx, y i ReŸx, iy which shows that the inner product is determined by the real part.
ReŸx, iy i ImŸx, iy q Ÿx, iy "iŸx, y " i¡ReŸx, y i ImŸx, y ¢
and so, equating real parts,
ReŸx, iy ImŸx, y Therefore,
Ÿx, y ReŸx, y i ImŸx, y ReŸx, y i ReŸx, iy Cross Product
1. 89]
Solution:
Show that if a • u 0 for all unit vectors, u, then a 0.
If a p 0, then the condition says that |a • u||a|sin 2 0 for all angles 2. Hence a 0 after all.
2. 90]
Solution:
Find the area of the triangle determined by the three points, Ÿ0, 1, "3 , Ÿ2, 2, 0 and Ÿ"1, 1, "1 .
2 "7 1
is the cross product. Area is then
3
2
6
3. 91]
Solution:
Find the area of the triangle determined by the three points, Ÿ3, "2, 2 , Ÿ4, 2, 0 and Ÿ"1, 1, 2 .
6 8 19
is the cross product. Area is then
1
2
461
4. 92]
Solution:
Find the area of the triangle determined by the three points, Ÿ3, 2, 0 , Ÿ4, 3, 1 and Ÿ2, 1, "1 .
The area is 0.
5. 93]
Solution:
Find the area of the parallelogram determined by the two vectors, Ÿ3, 0, 2 , Ÿ0, "2, 0 .
4 0 "6
is the cross product. The area is then
2 13
6. 94]
Solution:
Find the area of the parallelogram determined by the two vectors, Ÿ"2, 2, "3 , Ÿ1, 5, 0 .
15 "3 "12
is the cross product. The area is then
3 3 14
7. 95]
Solution:
Find the volume of the parallelepiped determined by the vectors,
"2 "3 "3 , 4 2 1 , 1 1 "3 .
"2 "3 "3
•
4 2 1
1 1 "3
"31
8. 96]
Solution:
Find the volume of the parallelepiped determined by the vectors,
3 "1 3 , 1 "3 0 , 1 1 "1 .
3 "1 3
•
1 "3 0
1 1 "1
20
9. 97]
Solution:
Suppose a, b, and c are three vectors whose components are all integers. Can you conclude the
volume of the parallelepiped determined from these three vectors will always be an integer?
Yes. It will involve the sum of product of integers and so it will be an integer.
10. 98]
Solution:
What does it mean geometrically if the box product of three vectors gives zero?
It means that if you place them so that they all have their tails at the same point, the three will lie
in the same plane.
11. 99]
Solution:
Using the notion of the box product yielding either plus or minus the volume of the parallelepiped
determined by the given three vectors, show that
Ÿa • b c a Ÿb • c In other words, the dot and the cross can be switched as long as the order of the vectors remains
the same. Hint: There are two ways to do this, by the coordinate description of the dot and cross
product and by geometric reasoning. It is better if you use geometric reasoning.
In the picture, you have the same parallelepiped. The one on the left involves a • b c which is
equal to |a • b||c|cos 2 which is the area of the base determined by a, b times the altitude which
equals |c|cos 2. In the second picture, the altitude is measured from the base determined by c, b.
This altitude is |a|cos ) and the area of the new base is |c • d|. Hence the volume of the
parallelepiped, determined in this other way is |b • c||a|cos ) which equals a b • c. Thus both
a b • c and a • b c yield the same thing and it is the volume of the parallelepiped. If you switch
two of the vectors, then the sign would change but in this case, both expressions would give "1
times the volume of the parallelepiped.
12. 100]
Solution:
Is a •Ÿb • c Ÿa • b • c? What is the meaning of a • b • c? Explain. Hint: Try Ÿi • j •j.
Ÿi • j •j k • j "i. However, i •Ÿj • j 0 and so the cross product is not associative.
13. 101]
Solution:
Discover a vector identity for Ÿu • v •w and one for u •Ÿv • w .
ŸŸu • v • w i / ijk Ÿu • v j wk / ijk / jrs u r v s wk
"/ jik / jrs u r v s wk "Ÿ- ir - ks " - kr - is u r v s wk
u k v i wk " u i v k wk ŸŸu w v "Ÿv w u i
Hence
Ÿu • v •w Ÿu w v "Ÿv w u
You can do the same thing for the other one or you could do the following.
u •Ÿv • w "Ÿv • w • u "¡Ÿv u w "Ÿw u v¢
Ÿw u v "Ÿv u w
14. 102]
Solution:
Discover a vector identity for Ÿu • v • Ÿz • w .
It is easiest to use the identity just discovered, although you could do it directly with the
permutation symbol and reduction identity. The expression equals
u Ÿz • w v " v Ÿz • w u ¡u, z, w¢v "¡v, z, w¢u
Here ¡u, z, w¢ denotes the box product.
15. 103]
Solution:
Simplify Ÿu • v Ÿv • w • Ÿw • z .
Consider the cross product term. From the above,
Ÿv • w • Ÿw • z ¡v, w, z¢w "¡w, w, z¢v
¡v, w, z¢w
Thus it reduces to
Ÿu • v ¡v, w, z¢w ¡v, w, z¢¡u, v, w¢
16. 104]
Solution:
Simplify |u • v|2 Ÿu v 2 " |u|2 |v|2 .
|u • v|2 / ijk u j v k / irs u r v s Ÿ- jr - ks " - kr - js u r v s u j v k
u j v k u j v k " u k v j u j v k |u|2 |v|2 " Ÿu v 2
It follows that the expression reduces to 0. You can also do the following.
|u • v|2 |u|2 |v|2 sin 2 2
|u|2 |v|2 Ÿ1 " cos 2 2 |u|2 |v|2 " |u|2 |v|2 cos 2 2
|u|2 |v|2 " Ÿu v 2
which implies the expression equals 0.
17. 105]
Solution:
For u, v, w functions of t, show the product rules
Ÿu • v U u U • v u • v U
Ÿu v U u U v u v U
You can do this more elegantly by repeating the proof of the product rule given in calculus and
using the properties of the two products. However, I will show it using the summation convention
and permutation symbol.
Ÿu • v U
U
i
q ŸŸu • v i Ÿ/ ijk u j v k U
/ ijk u Uj v k / ijk u k v Uk Ÿu U •v u • v U i
and so Ÿu • v U u U •v u • v U . Similar but easier reasoning shows the next one.
18. 106]
Solution:
If u is a function of t, and the magnitude |uŸt | is a constant, show from the above problem that
the velocity u U is perpendicular to u.
u u c where c is a constant. Differentiate both sides.
uU u u uU 0
Hence u U u 0. For example, if you move about on the surface of a sphere, then your velocity
will be perpendicular to the position vector from the center of the sphere.
19. 107]
Solution:
When you have a rotating rigid body with angular velocity vector (, then the velocity vector
v q u U is given by
v (•u
where u is a position vector. The acceleration is the derivative of the velocity. Show that if ( is a
constant vector, then the acceleration vector a v U is given by the formula
a ( •Ÿ( • u .
Now simplify the expression. It turns out this is centripetal acceleration.
a v U ( • u U ( •Ÿ( • u Of course you can simplify the right side. It equals
a Ÿ( u ( "Ÿ( ( u
Also, its magnitude is |(||( • u| |(|2 |u|sin 2 and is perpendicular to (.
20. 108]
Solution:
Verify directly that the coordinate description of the cross product, a • b has the property that it is
perpendicular to both a and b. Then show by direct computation that this coordinate description
satisfies
|a • b|2 |a|2 |b|2 " Ÿa b 2
|a|2 |b|2 Ÿ1 " cos 2 Ÿ2 where 2 is the angle included between the two vectors. Explain why |a • b| has the correct
magnitude. All that is missing is the material about the right hand rule. Verify directly from the
coordinate description of the cross product that the right thing happens with regards to the vectors
i, j, k. Next verify that the distributive law holds for the coordinate description of the cross
product. This gives another way to approach the cross product. First define it in terms of
coordinates and then get the geometric properties from this. However, this approach does not yield
the right hand rule property very easily.
From the coordinate description,
a • b a / ijk a j b k a i "/ jik a j b k a i "/ jik b k a i a j "a • b a
and so a • b is perpendicular to a. Similarly, a • b is perpendicular to b. The Problem above
shows that
|a • b|2 |a|2 |b|2 Ÿ1 " cos 2 2 |a|2 |b|2 sin 2 2
and so |a • b| |a||b|sin 2, the area of the parallelogram determined by a, b. Only the right hand
rule is a little problematic. However, you can see right away from the component definition that
the right hand rule holds for each of the standard unit vectors. Thus i • j k etc.
i j k
1 0 0
0 1 0
k
Systems Of Equations
1. 109]
Solution:
Do the three lines, 2y " x 2, 3x 4y 2, and "24x " 22y "8 have a common point of
intersection? If so, find the point and if not, tell why they don’t have such a common point of
intersection.
2y " x 2
3x 4y 2
, Solution is: x " 25 , y 4
5
2. 110]
Solution:
Do the three lines, 3x 3y 3, 2x " y 3, and 16y " 5x "12 have a common point of
intersection? If so, find the point and if not, tell why they don’t have such a common point of
intersection.
3x 3y 3
2x " y 3
, Solution is: x 4
3
, y " 13
3. 111]
Solution:
Do the three lines, 3x 3y 4, 2x y 2, and 2y " 5x "1 have a common point of
intersection? If so, find the point and if not, tell why they don’t have such a common point of
intersection.
There is no point of intersection.
4. 112]
Solution:
Do the three planes,
x 2z 0,
y " 2x " 3z "1,
y"x 0
have a common point of intersection? If so, find one and if not, tell why there is no such point.
The point of intersection is Ÿx, y, z Ÿ"2, "2, 1 .
5. 113]
Solution:
Do the three planes,
x " 5z 10,
y z 2,
x y " 3z 10
have a common point of intersection? If so, find one and if not, tell why there is no such point.
The point of intersection is Ÿx, y, z Ÿ0, 4, "2 .
6. 114]
Solution:
Do the three planes,
x " z 3,
4x y " 3z 11,
5x y " 3z 15
have a common point of intersection? If so, find one and if not, tell why there is no such point.
The point of intersection is Ÿx, y, z Ÿ4, "2, 1 .
7. 115]
Solution:
You have a system of k equations in two variables, k u 2. Explain the geometric significance of
a. No solution.
b. A unique solution.
c. An infinite number of solutions.
The lines don’t intersect, they intersect in a single point, they intersect in a whole line.
8. 116]
Solution:
Here is an augmented matrix in which ' denotes an arbitrary number and Q denotes a nonzero
number. Determine whether the given augmented matrix is consistent. If consistent, is the solution
unique?
Q ' ' ' ' | '
0 Q ' ' 0 | '
0
0 Q ' ' | '
0
0
0 0 Q | '
It appears that the solution exists but is not unique.
9. 117]
Solution:
Here is an augmented matrix in which ' denotes an arbitrary number and Q denotes a nonzero
number. Determine whether the given augmented matrix is consistent. If consistent, is the solution
unique?
Q ' ' | '
0 Q ' | '
0
0 Q | '
It appears that there is a unique solution.
10. 118]
Solution:
Here is an augmented matrix in which ' denotes an arbitrary number and Q denotes a nonzero
number. Determine whether the given augmented matrix is consistent. If consistent, is the solution
unique?
Q ' ' ' ' | '
0 Q 0 '
0 | '
0
0 0 Q ' | '
0
0 0 0 Q | '
It appears that there is a solution but it is not unique.
11. 119]
Solution:
Here is an augmented matrix in which ' denotes an arbitrary number and Q denotes a nonzero
number. Determine whether the given augmented matrix is consistent. If consistent, is the solution
unique?
Q ' ' ' ' | '
0 Q ' ' 0 | '
0
0 0 0 Q | 0
0
0 0 0 ' | Q
There might be a solution. If so, there are infinitely many.
12. 120]
Solution:
Suppose a system of equations has fewer equations than variables. Must such a system be
consistent? If so, explain why and if not, give an example which is not consistent.
No. Consider x y z 2 and x y z 1.
13. 121]
Solution:
If a system of equations has more equations than variables, can it have a solution? If so, give an
example and if not, tell why not.
These can have a solution. For example, x y 1, 2x 2y 2, 3x 3y 3 even has an infinite
set of solutions.
14. 122]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 "2
x2 4
1 "2 2x 2 3
Inconsistent: x £1, "1¤ Consistent: x £1, "1¤.
15. 123]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 2
x2
1 2 2x 2 " 4
Inconsistent: x £2, "2¤ Consistent: x £2, "2¤.
16. 124]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 x 1 "1
2 x2
3
Consistent: x p 0, Inconsistent: x 0.
17. 125]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 3 1 x4
1 3 1 x4
1 3 1 2x 2
Consistent: x 2, Inconsistent: x p 2
18. 126]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 1 1 x2 6
1 1 1 x2 6
1 1 1 2x 2 5
Consistent: x £1, "1¤, Inconsistent: x £1, "1¤
19. 127]
Solution:
Find x such that the following augmented matrix is the augmented matrix of an inconsistent
system. Then find values of x such that the augmented matrix pertains to a consistent system.
1 3 4 x2 2
1 3 4 x2 2
1 3 4 2x 2 " 2
Consistent: x £"2, 2¤, Inconsistent: x £"2, 2¤
20. 128]
Solution:
Choose h and k such that the augmented matrix shown has one solution. Then choose h and k
such that the system has no solutions. Finally, choose h and k such that the system has infinitely
many solutions.
1 h 2
.
1 2 k
If 2 " h p 0 there is only one solution. If 2 " h 0 and k " 2 p 0 there is no solution. If
2 " h 0 and k " 2 p 0, then there are infinitely many.
21. 129]
Solution:
Choose h and k such that the augmented matrix shown has one solution. Then choose h and k
such that the system has no solutions. Finally, choose h and k such that the system has infinitely
many solutions.
4
h
2
.
28 56 k
If 8 " h p 0 there is only one solution. If 8 " h 0 and k " 14 p 0 there is no solution. If
8 " h 0 and k " 14 p 0, then there are infinitely many.
22. 130]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
x y z
2s 3t " 5 s t
"1
2
1
"2 "3 "5
"2
4
3
6
5
10
, t, s R
23. 131]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
3 8 3 46
1 3 1 17
2 7 3 37
x1 x2 x3
4 5 "2
24. 132]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
3 "6 3 "18
1 "2 1
"6
2 "4 2 "12
x y z
2s " t " 6 s t
, t, s R
25. 133]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
"1 "4 "10 "27
1
3
8
22
"2 "5 "14 "38
Inconsistent.
26. 134]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
0
"1 0
1
3
1 "8
"1 "2 0
x1 x2 x3
4
8
0 "4 4
27. 135]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
0
"1 "1
"4
1
3
19
6
"1 "2 "5 "15
x y z
7 " 3t 4 " t t
,t R
28. 136]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
1 2 "2 15
1 3 "4 20
0 1 "2
x y z
5 " 2t 2t 5 t
5
,t R
29. 137]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
0
"1 0
1
3
1 "6
"1 "2 0
x1 x2 x3
1
0
2 "1 "5
30. 138]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
1 2 1 5
1 3 1 8
0 1 1 7
x1 x2 x3
"5 3 4
31. 139]
Solution:
Give the complete solution to the system of equations.
3x 1 8x 2 3x 3 " 5x 4 " 11x 5 " 16 0
x 1 3x 2 x 3 " 2x 4 " 4x 5 " 6 0
2x 1 7x 2 3x 3 " 7x 4 " 8x 5 " 15 0
3x 1 7x 2 3x 3 " 4x 4 " 10x 5 " 15 0
32. 140]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
2
0 5
"8
15
1
0 3
"5
9
1
0 4
"7
12
"2 0 4 "10 12
x y z w
"t s 2t 3 t
, t, s R
33. 141]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
"1 "4 "1
13
"25
1
1
"11
19
"2 "5 "1
15
"27
"5 "1
15
"51
6
3
, inconsistent.
34. 142]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
2
"4
5 15 14
1
"2
3
1
"2
4 12 10
9
8
6 "12 4 12 20
x y z w
, t, s R
2t 2 t 2 " 3s s
35. 143]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
1
1
0 "1
4
1
3
1 "2
5
3
10 4 "5 17
"1 10 4 "3
x y z w
3
4 1 0 1
36. 144]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
1
3
2 "5
5
1
3
3 "8
8
0
0
1 "3
3
4 12 1
x y z w
"3s " t " 1 s 3t 3 t
1
"1
, t, s R
37. 145]
Solution:
Give the complete solution to the system of equations.
4x 1 11x 2 4x 3 " 33x 4 " 29x 5 " 29 0
x 1 3x 2 x 3 " 9x 4 " 8x 5 " 8 0
3x 1 10x 2 4x 3 " 32x 4 " 23x 5 " 29 0
3x 1 10x 2 4x 3 " 32x 4 " 23x 5 " 29 0
x1 x2 x3 x4 x5
38. 146]
3s " 2t " 3 3s 3t 3 2t " 4s 2 t s
, t, s R
Solution:
Give the complete solution to the system of equations.
6x 5 " 4x 2 " x 3 " x 1 6 0
x 1 3x 2 x 3 x 4 " 5x 5 " 5 0
2x 4 " 5x 2 " x 3 " 2x 1 5x 5 9 0
4x 4 " 5x 2 " x 3 " 4x 1 x 5 13 0
x1 x2 x3 x4 x5
t " 2s 2 s t 1 4s " 5t t s
, t, s R
39. 147]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
3 8 3 "19 "19 10
1 3 1 "7
"7
4
2 7 3 "16 "15 10
1 7 3 "15 "13 12
x1 x2 x3 x4 x5
2s t " 2 2s 2t 2 "s t s
, t, s R
40. 148]
Solution:
Give the complete solution to the system of equations whose augmented matrix is.
"1 "4 "1
8
"6
1
"7
4
3
1
"2 "5 "1 10 "8
x y z w
4
"5 "1
t t 2 3t " 2 t
,t R
4
"8
41. 149]
Solution:
Four times the weight of Gaston is 150 pounds more than the weight of Ichabod. Four times the
weight of Ichabod is 660 pounds less than seventeen times the weight of Gaston. Four times the
weight of Gaston plus the weight of Siegfried equals 290 pounds. Brunhilde would balance all
three of the others. Find the weights of the four sisters.
4g " I 150
4I " 17g "660
4g s 290
, Solution is : £g 60, I 90, b 200, s 50¤
gIs"b 0
42. 150]
Solution:
The steady state temperature, u in a plate solves Laplace’s equation, u 0. One way to
approximate the solution which is often used is to divide the plate into a square mesh and require
the temperature at each node to equal the average of the temperature at the four adjacent nodes.
This procedure is justified by the mean value property of harmonic functions. In the following
picture, the numbers represent the observed temperature at the indicated nodes. Your task is to
find the temperature at the interior nodes, indicated by x, y, z, and w. One of the equations is z 1
0 w x .
4 Ÿ10
You need
1
30 w x " y 0
4 Ÿ20
1
30 0 z " w 0
4 Ÿy
1
y z 10 " x 0
4 Ÿ20
1
w 0 10 " z 0
4 Ÿx
, Solution is: ¡w 15, x 15, y 20, z 10¢.
43. 151]
Solution:
Consider the following diagram of four circuits.
Those jagged places denote resistors and the numbers next to them give their resistance in ohms,
written as (. The breaks in the lines having one short line and one long line denote a voltage
source which causes the current to flow in the direction which goes from the longer of the two
lines toward the shorter along the unbroken part of the circuit. The current in amps in the four
circuits is denoted by I 1 , I 2 , I 3 , I 4 and it is understood that the motion is in the counter clockwise
direction if I k ends up being negative, then it just means it moves in the clockwise direction. Then
Kirchhoff’s law states that
The sum of the resistance times the amps in the counter clockwise direction around a loop equals
the sum of the voltage sources in the same direction around the loop.
In the above diagram, the top left circuit should give the equation
2I 2 " 2I 1 5I 2 " 5I 3 3I 2 5
For the circuit on the lower left, you should have
4I 1 I 1 " I 4 2I 1 " 2I 2 "10
Write equations for each of the other two circuits and then give a solution to the resulting system
of equations. You might use a computer algebra system to find the solution. It might be more
convenient than doing it by hand.
The other two equations are
6I 3 " 6I 4 I 3 I 3 5I 3 " 5I 2 "20
2I 4 3I 4 6I 4 " 6I 3 I 4 " I 1 0
Then the system is
2I 2 " 2I 1 5I 2 " 5I 3 3I 2 5
4I 1 I 1 " I 4 2I 1 " 2I 2 "10
6I 3 " 6I 4 I 3 I 3 5I 3 " 5I 2 "20
, Solution is:
2I 4 3I 4 6I 4 " 6I 3 I 4 " I 1 0
£¡I 1 "2. 010 7, I 2 "1. 269 9, I 3 "2. 735 5, I 4 "1. 535 3¢¤
44. 152]
Solution:
Consider the following diagram of four circuits.
Those jagged places denote resistors and the numbers next to them give their resistance in ohms,
written as (. The breaks in the lines having one short line and one long line denote a voltage
source which causes the current to flow in the direction which goes from the longer of the two
lines toward the shorter along the unbroken part of the circuit. The current in amps in the four
circuits is denoted by I 1 , I 2 , I 3 and it is understood that the motion is in the counter clockwise
direction. If I k ends up being negative, then it just means the current flows in the clockwise
direction. Then Kirchhoff’s law states that
The sum of the resistance times the amps in the counter clockwise direction around a loop equals
the sum of the voltage sources in the same direction around the loop. Find I 1 , I 2 , I 3 .
You have
2I 1 5I 1 3I 1 " 5I 2 "10
I 2 3I 2 7I 2 5I 2 " 5I 1 "12
2I 3 4I 3 4I 3 I 3 " I 2 0
Simplifying this yields
10I 1 " 5I 2 "10
16I 2 " 5I 1 "12
11I 3 " I 2 0
Then you just find the solution.
I 1 " 44 , I 2 " 34 , I 3 " 34
27
297
27
Thus all currents flow in the clockwise direction.
Matrices
1. 153]
Solution:
Here are some matrices:
"2 2 "3
A
"3 1
2
2 "2
C
3
,
"1 0 1
2 0
,D 1
"3 2 2
,B "2
,E 2 1
.
0
Find if possible 3A, 3B " A, AC, CB, AE, EA. If it is not possible explain why.
"6 6 "9
"9 3
6
,
"7
4
0
"1 1
9
"4 4 2
, Not possible,
, Not possible, Not possible.
"10 6 7
2. 154]
Solution:
Here are some matrices:
2
A
C
2 "3
"1 1
2
2 2
,D 1 1
"3
,B 2
2
,
"1 "1 1
2 "1
2
1
,E 2
.
"1
Find if possible A, B " A, AC, CB, AE, EA. If it is not possible explain why.
2
2 "3
"1 1
2
,
"5
0
0
"2 "1
5
"8 2 6
, Not possible,
, Not possible, Not possible.
"4 1 3
3. 155]
Solution:
Here are some matrices:
A
C
0 2 "2
1 1
2 0
2 1
1
,B ,D "2 2 2
"1 2 1
1 2
2 1
,E ,
0
2
.
Find if possible AB T , BA T " C, CA, E T A, E T E, EE T . If it is not possible explain why.
0 2
2 2
2 2 2
4. 156]
,
"2 2
0
, Ÿ4 ,
1
,
0 0
0 4
0 4 "4
1 5 "3
Solution:
Here are some matrices:
"2 3
A
3
"2 1 "1
3 "2
C
3
,D 1
3
,B 3
2
"1 "3 1
"1 "3
2
,
"2
,E 1
.
"3
Find if possible AB T , BA T " C, CA, E T A, E T E, EE T . If it is not possible explain why.
9
"4
,
"5 "2
10 "9 "3
"3
6
"2
,
"7 "3
7
11
"8 10
8
4 6
, Ÿ13 ,
6 9
5. 157]
Solution:
Here are some matrices:
4
A
1 0
"3 1 2
1 4
C
2 1
0
,B ,D 1
2
"1 "2 1
2 "2
2
1
,E ,
4
"2
.
Find if possible AB T , BA T " C, CA, E T A, E T E, EE T . If it is not possible explain why.
1 "6
5
3
22 2 "4
,
0
1
"8 2
, Ÿ20 ,
,
"8 5 8
5
3 2
16 "8
"8
4
6. 158]
Solution:
Suppose A and B are square matrices of the same size. Which of the following are correct?
a. ŸA " B 2 A 2 " 2AB B 2
b. ŸAB 2 A 2 B 2
c. ŸA B 2 A 2 2AB B 2
d. ŸA B 2 A 2 AB BA B 2
e. A 2 B 2 AŸAB B
f. ŸA B 3 A 3 3A 2 B 3AB 2 B 3
g. ŸA B ŸA " B A 2 " B 2
Part d,e.
7. 159]
Solution:
Let
A
2 2
1 1
Find all 2 • 2 matrices B such that AB 0.
B
x
y
"x "y
8. 160]
Solution:
Let A 2
3
"1 "3
what should k equal?
. Is it possible to choose k such that AB BA? If so,
1 k
3k 6
7
Would need
2 3
,B "5 "3k " 3
1
"3
2 " k 3 " 3k
and this cannot happen. No such k.
9. 161]
Solution:
Let A 2 3
0 1
should k equal?
You need
4 3k 6
0
2 3
,B k
0 k
. Is it possible to choose k such that AB BA? If so, what
4 9
0 k
and so you must have k 1.
10. 162]
Solution:
Let A be an n • n matrix. Show A equals the sum of a symmetric and a skew symmetric matrix.
(M is skew symmetric if M "M T . M is symmetric if M T M.) Hint: Show that 12 ŸA T A is
symmetric and then consider using this as one of the matrices.
T
T
A AA
A"A
.
2
2
11. 163]
Solution:
Show every skew symmetric matrix has all zeros down the main diagonal. The main diagonal
consists of every entry of the matrix which is of the form a ii . It runs from the upper left down to
the lower right.
If A "A T , then a ii "a ii and so each a ii 0.
12. 164]
Solution:
Suppose M is a 3 • 3 skew symmetric matrix. Show there exists a vector ( such that for all
u R3
Mu ( • u
Hint: Explain why, since M is skew symmetric it is of the form
M
0
"F 3
F2
F3
0
"F 1
"F 2
F1
0
where the F i are numbers. Then consider F 1 i F 2 j F 3 k.
Since M is skew symmetric, it is of the form mentioned above. Now it just remains to verify that
i
j
k
( qF 1 i F 2 j F 3 k is of the right form. But F 1 F 2 F 3 u1 u2 u3
iF 2 u 3 " iF 3 u 2 " jF 1 u 3 jF 3 u 1 kF 1 u 2 " kF 2 u 1 . In terms of matrices, this is
F2u3 " F3u2
. If you multiply by the matrix, you get
"F 1 u 3 F 3 u 1
F1u2 " F2u1
0
"F 3
F2
u1
F3
0
"F 1
u2
"F 2
F1
0
u3
F2u3 " F3u2
F3u1 " F1u3
F1u2 " F2u1
which is the same thing.
13. 165]
Solution:
Using only the vector space axioms, show that 0 is unique.
Suppose 0 U is another one. Then 0 U 0 U 0 0.
14. 166]
Solution:
Using only the vector space axioms, show that "A is unique.
Suppose B also works. Then
" A "A ŸA B Ÿ"A A B 0 B B
15. 167]
Solution:
Using only the vector space axioms, show that 0A 0. Here the 0 on the left is the scalar 0 and
the 0 on the right is the zero for m • n matrices.
0A Ÿ0 0 A 0A 0A. Now add "Ÿ0A to both sides. Then 0 0A.
16. 168]
Solution:
Using only the vector space axioms, show that Ÿ"1 A "A.
A Ÿ"1 A Ÿ1 Ÿ"1 A 0A 0. Therefore, from the uniqueness of the additive inverse
proved above, it follows that "A Ÿ"1 A.
17. 169]
Solution:
Prove that for A, B m • n matrices, and a, b scalars, it follows that
ŸaA bB T aA T bB T
Ÿ)A *B T
Ÿ)A *B ji )A ji *B ji
ij
)ŸA T ij Ÿ*B Tij Ÿ)A T *B T ij
18. 170]
Solution:
Prove that I m A A where A is an m • n matrix.
ŸI m A ij q ! j - ik A kj A ij .
19. 171]
Solution:
Show that if A "1 exists for an n • n matrix, then it is unique. That is, if BA I and AB I, then
B A "1 .
A "1 A "1 I A "1 ŸAB ŸA "1 A B IB B.
20. 172]
Solution:
Give an example of matrices, A, B, C such that B p C, A p 0, and yet AB AC.
An example, not the only one is
1
"1
1 1
"1
1
1 1
1
"1
2 2
"1
1
2 2
0 0
0 0
0 0
0 0
21. 173]
Solution:
Suppose AB AC and A is an invertible n • n matrix. Does it follow that B C? Explain why or
why not. What if A were a non invertible n • n matrix?
Yes. Multiply on the left by A "1 . This does not follow if A "1 does not exist. You could have
1 1
2
"2
1 1
"2
2
1 1
1
"1
1 1
"1
1
22. 174]
Solution:
Find your own examples:
a. 2 • 2 matrices, A and B such that A p 0, B p 0 with AB p BA.
i.
0 1
1 2
1 0
3 4
1 2
0 1
3 4
1 0
3 4
2 1
1 2
4 3
b. 2 • 2 matrices, A and B such that A p 0, B p 0, but AB 0.
0 0
0 0
1
"1
1 1
"1
1
1 1
i.
0 0
0 0
c. 2 • 2 matrices, A, D, and C such that A p 0, C p D, but AC AD.
1
"1
1 1
"1
1
1 1
i.
1
"1
2 2
"1
1
2 2
0 0
0 0
0 0
0 0
23. 175]
Solution:
Give an example of a matrix, A such that A 2 I and yet A p I and A p "I.
1
0
0
0 "1 0
0
0
1
24. 176]
Solution:
Let
A
4
2
"4 "2
Find A "1 if it exists and if it does not, explain why.
The row reduced echelon form of
4
2
"4 "2
1
is:
0 0
25. 177]
Solution:
Let
A
0
"4
"1 "1
Find A "1 if it exists and if it does not, explain why.
0
"4
"1 "1
"1
1
4
" 14
"1
0
26. 178]
Solution:
Let
A
1
2
3
1
"3 "2
Find A "1 if it exists and if it does not, explain why.
and so there is no inverse.
"1
3
1
"3 "2
2
3
1
3
"1 "1
27. 179]
Solution:
Let
A
"3 3
1
1
Find A "1 if it exists and if it does not, explain why.
"3 3
1
"1
1
" 16
1
6
1
2
1
2
28. 180]
Solution:
Let
A
"2 "1
4
2
Find A "1 if it exists and if it does not, explain why.
The row reduced echelon form of
"2 "1
4
2
is:
1
1
2
and so there is no inverse.
0 0
29. 181]
Solution:
Let A be a 2 • 2 matrix which has an inverse. Say A a b
c d
. Find a formula for A "1 in
terms of a, b, c, d.
"1
a b
c d
d
ad"bc
c
" ad"bc
b
" ad"bc
a
ad"bc
30. 182]
Solution:
Prove that if A "1 exists and Ax 0 then x 0.
Multiply on both sides on the left by A "1 . Thus
0 A "1 0 A "1 ŸAx ŸA "1 A x Ix x
31. 183]
Solution:
Let
3 5 12
A
1 2 5
1 2 6
Find A "1 if possible. If not possible, explain why.
"1
3 5 12
1 2
5
1 2
6
2
"6
1
"1
6
"3
0
"1
1
32. 184]
Solution:
Let
1 0 "3
A
1 1
0
1 1
0
Find A "1 if possible. If not possible, explain why.
There is no inverse. The row reduced echelon form is
1 0 "3
0 1
3
0 0
0
33. 185]
Solution:
Let
1 1 2
A
1 2 7
2 4 15
Find A "1 if possible. If not possible, explain why.
"1
1 1
2
1 2
7
2 4 15
2
"7
3
"1 11 "5
0
"2
1
34. 186]
Solution:
Let
1 0 4
A
0 1 4
1 1 8
Find A "1 if possible. If not possible, explain why.
There is no inverse. The row reduced echelon form is
1 0 4
0 1 4
0 0 0
35. 187]
Solution:
Let
1 0 "1
A
3 1
0
1 1
2
Find A "1 if possible. If not possible, explain why.
There is no inverse. The row reduced echelon form is
1 0 "1
0 1
3
0 0
0
36. 188]
Solution:
Let
"1 "3
8
1
"3
A
2
"3 "6 10
Find A "1 if possible. If not possible, explain why.
"1
"1 "3
8
1
"3
2
"3 "6 10
2
"18 "7
"1
14
5
0
3
1
37. 189]
Solution:
"2x 1 " 3x 2 " 3x 3
2x 1 3x 3
Write
0
x 1 " 2x 2 2x 4
"2 "3 "3 0
A
2
0
3
0
0
0
0
0
1
"2
0
2
x1
in the form A
x2
x3
x4
where A is an appropriate matrix.
38. 190]
Solution:
4x 1 4x 2 " x 3
2x 1 x 3
Write
3x 3
x1
in the form A
x1 x2 x4
x2
where A is an appropriate matrix.
x3
x4
4 4 "1 0
A
2 0
1
0
0 0
3
0
1 1
0
1
39. 191]
Solution:
Show that if A is an n • n invertible matrix and x is a n • 1 matrix such that Ax b for b an n • 1
matrix, then x A "1 b.
Multiply on the left of both sides by A "1 .
40. 192]
Solution:
Using the inverse of the matrix, find the solution to the systems
2
"1 "4
x
"1
1
3
y
1
0
0
z
3
2
"1 "4
x
1
"1
1
3
y
1
0
0
z
1
2
0
,
,
2
"1 "4
x
"1
1
3
y
1
0
0
z
2
"1 "4
x
"1
1
3
y
1
0
0
z
1
Now give the solution in terms of l, m, and n to
2
"1 "4
x
"1
1
3
y
1
0
0
z
l
m
n
.
2
1
0
3
"1
"2
.
3
0
,
5
0
"3
1
"2
,
1
,
10
9
"4
0
n
3l 4m " 2n
For the general solution, it is
n"m"l
41. 193]
Solution:
Using the inverse of the matrix, find the solution to the systems
1 "2
x
"1 1 "5
y
1
0
4
z
3
0
1 "2
x
1
"1 1 "5
y
1
z
0
0
4
1 "2
x
"1 1 "5
y
1
z
1
0
,
2
,
1
4
1 "2
x
"1 1 "5
y
1
z
0
0
0
0
4
Now give the solution in terms of l, m, and n to
1 "2
x
"1 1 "5
y
1
z
0
0
4
"13
,
"1
1
22
,
0
4l " 4m " 3n
2m " l 2n
m"ln
m
n
0
4
For the general solution, it is
4
9
1
l
"9
"6
,
.
2
1
0
3
"1
"2
.
42. 194]
Solution:
Using the inverse of the matrix, find the solution to the system
"8
3 0 1
x
"21 3 2 5
y
"5
1 1 1
z
"8
1 1 2
w
l " 3m " 2n 8r
l
m
n
.
r
x
2l " 5m " 3n 13r
y
n " m 2r
z
3l " 9m " 7n 25r
w
43. 195]
Solution:
Using the inverse of the matrix, find the solution to the system
l " 2m 7n " 5r
m " 3n 2r
3m " l " 8n 6r
2m " l " 8n 6r
2
2
0
1
x
0
0
1 "1
y
2 "1 1
1
z
3 "1 1
2
w
l
m
n
.
r
x
y
z
w
44. 196]
Solution:
"1
T
Show that if A is an invertible n • n matrix, then so is A T and ŸA T ŸA "1 .
You just need to show that ŸA "1 T acts like the inverse of A T because from uniqueness in the
above problem, this will imply it is the inverse. But from properties of the transpose,
A T ŸA "1 T ŸA "1 A T I T I
T
T
ŸA "1 A T ŸAA "1 I T I
Hence ŸA "1 T ŸA T "1 and this last matrix exists.
45. 197]
Solution:
Show that ŸABC "1 C "1 B "1 A "1 by verifying that ŸABC ŸC "1 B "1 A "1 I and ŸC "1 B "1 A "1 ŸABC I.
From the above problem, ŸBC "1 exists and equals ŸBC "1 C "1 B "1 Therefore, ŸABC "1 exists
and equals ŸBC "1 A "1 C "1 B "1 A "1 .
46. 198]
Solution:
"1
2
If A is invertible, show ŸA 2 ŸA "1 .
2
A 2 ŸA "1 AAA "1 A "1 AIA "1 AA "1 I
2
ŸA "1 A 2 A "1 A "1 AA A "1 IA A "1 A I
47. 199]
Solution:
If A is invertible, show ŸA "1 "1 A.
"1
A "1 A AA "1 I and so by uniqueness, ŸA "1 A.
48. 200]
Solution:
Let A and be a real m • n matrix and let x R n and y R m . Show ŸAx, y R m Ÿx, A T y R n
where Ÿ, R k denotes the dot product in R k . In the notation above, Ax y x A T y. Use the
definition of matrix multiplication to do this.
Ax y ! k ŸAx k y k ! k ! i A ki x i y k ! i ! k A Tik x i y k Ÿx, A T y 49. 201]
Solution:
Use the fact that ŸAx y Ÿx A T y to verify directly that ŸAB T B T A T without making any
reference to subscripts.
ŸABx, y ŸBx, A T y Ÿx, B T A T y ŸABx, y x, ŸAB T y
Since this is true for all x, it follows that, in particular, it holds for
x B T A T y "ŸAB T y
and so from the axioms of the dot product,
B T A T y "ŸAB T y, B T A T y "ŸAB T y 0
and so B T A T y "ŸAB T y 0. However, this is true for all y and so B T A T "ŸAB T 0.
50. 202]
Solution:
A matrix A is called a projection if A 2 A. Here is a matrix.
3
"6
"2
"1
4
1
6
"18 "5
Show that this is a projection. Show that a vector in the column space of a projection is left
unchanged by multiplication by A.
3
"6
"2
"1
4
1
6
"18 "5
2
3
"6
"2
"1
4
1
6
"18 "5
So it is a projection. In general, a typical vector in the column space of A is Ax. Therefore,
AŸAx A 2 x Ax
Thus A leaves vectors in the column space unchanged.
51. 203]
Solution:
Suppose A is an n • n matrix and for each j,
n
!|A ij | 1
i1
.
k0
Show that the infinite series ! A converges in the sense that the ij th entry of the partial sums
n
converge for each ij. Hint: Let R q max j ! i1 |A ij |. Thus R 1. Show that ŸA 2 ij t R 2 . Then
generalize to show that ŸA m ij t R m . Use this to show that the ij th entry of the partial sums is a
Cauchy sequence. From calculus, these converge by completeness of the real or complex
.
numbers. Next show that ŸI " A "1 ! k0 A k . The Leontief model in economics involves solving
an equation for x of the form
x Ax b, or ŸI " A x b
The vector Ax is called the intermediate demand and the vectors A k x have economic meaning as
does the series which is also called the Neuman series.
! k A ik A kj t ! k |A ik ||A kj | t R ! k |A kj | t R 2 . Thus ŸA 2 ij t R 2 . Suppose ŸA m ij t R m for
all ij. Then
k
!ŸA m ik A kj
ŸA m1 ij t
k
!|ŸA m ik ||A kj | t R m !|A kj | t R m1
k
k
It follows that for N M,
N
M
! Ak
"
k0
N
!
A
ij
! ŸA
kM1
ij
.
N
k
ij t
kM1
ij
k0
N
k
! Ak
!
ŸA ij t
k
kM1
! Rk kM
RM
1"R
and so the partial sums for each ij are a Cauchy sequence. Therefore, the partial sums converge.
Also
N
ŸI " A !A
N
k
k0
and it was just shown that A
N1
! Ak
ŸI " A I " A N1
k0
v 0 as N v .. Therefore, passing to a limit, it follows that
.
ŸI " A !A
k0
.
k
! Ak
k0
ŸI " A I
Solutions
1. 1]
Solution:
Find the determinants of the following matrices.
"5 "3 "4
0
"4
0
4
6
3
"4
2. 2]
Solution:
Find the determinants of the following matrices.
"12 "10 "7
0
"3
0
9
10
4
"45
3. 3]
Solution:
Find the determinants of the following matrices.
10
18
9
"3
"6
"3
"11 "16 "10
12
4. 4]
Solution:
Find the determinants of the following matrices.
2
"1
1
"4 "1 "4
5
5
6
9
5. 5]
Solution:
Find the determinant of the following matrix.
"2
0
2
"3
6
11 12
6
"2 "4 "4 "2
1
"4 "7
2
6
6. 6]
Solution:
Find the determinant of the following matrix.
"5 "14 "16 "6
7
10
7
7
"6
"9
"9
"6
2
11
16
3
18
7. 7]
Solution:
Find the determinant of the following matrix.
20
31
34
17
"3
"6
"5
"3
5
7
7
5
"21 "31 "35 "18
"6
8. 8]
Solution:
Find the determinant of the following matrix.
"6 "9 "12 "7
11 18
21
11
"5 "7
"7
"5
3
1
1
4
3
3
6
1
6
13
14
6
0
"3
"3
0
18
9. 9]
Solution:
Find the determinant of the following matrix.
"8 "12 "16 "6
6
10. 10]
Solution:
Find the determinant of the following matrix.
"6 "7 "10 "5
"3 "9 "11 "3
"2
0
0
"2
10 15
20
9
4
11. 11]
Solution:
Find the determinant of the following matrix.
"5 "8 "5 "6
11 23 25 11
"3 "8 "8 "3
"4 "9
0
1
30
12. 12]
Solution:
Find the following determinant by expanding along the first row and then by expanding along the
second column.
"7 "14
2
8
13
2
"6
"4
"9
15
13. 13]
Solution:
Find the following determinant by expanding along the first row and then by expanding along the
second column.
40
52
20
"20 "26 "10
"20 "26 "10
0
14. 14]
Solution:
Find the following determinant by expanding along the first row and then by expanding along the
second column.
23
29
11
"8
"10
"3
"20 "24 "13
"30
15. 15]
Solution:
Find the following determinant by expanding along the first row and then by expanding along the
second column.
"34 "40 "20
17
20
10
20
23
13
0
16. 16]
Solution:
Find the following determinant by cofactor expansion. Pick whatever is easiest.
"2 "7 "4
1
1
10 13
1
2
"1 "1
2
0
"1 "7 "3
"27
17. 17]
Solution:
Find the following determinant by cofactor expansion. Pick whatever is easiest.
"18 "26 "38 "23
20
25
28
20
"11 "12 "12 "11
10
14
23
15
15
18. 18]
Solution:
Find the following determinant by cofactor expansion. Pick whatever is easiest.
"23 "32 "40 "25
19
28
31
19
"11 "14 "14 "11
18
21
26
20
54
19. 19]
Solution:
Find the following determinant by cofactor expansion. Pick whatever is easiest.
"14 "23 "24 "9
1
8
11
1
"1
"3
"3
"1
15
19
17
10
"30
20. 20]
Solution:
An operation is done to get from the first matrix to the second. Identify what was done and tell
how it will affect the value of the determinant.
a b
c d
a c
,
b d
It does not change the determinant. This was just taking the transpose.
21. 21]
Solution:
Suppose
u v w
q r s
2
x y z
Find
ux
det
vy
wz
q 3x r 3y s 3z
x
y
z
The value is unchanged. It is still 2.
22. 22]
Solution:
Suppose that A, B are n • n matrices and detŸA 4 while detŸB 2. Find detŸAB "1 .
2
23. 23]
Solution:
Suppose
u v w
"9
q r s
x y z
Find
u
det
v
w
q 3x r 3y s 3z
x
y
z
The value is unchanged. It is still "9.
24. 24]
Solution:
Suppose that A is an n • n matrix and A 2 A. Find the possible values for detŸA .
You have detŸA 2 detŸA and so you have detŸA ŸdetŸA " 1 0 so detŸA 0 or 1.
25. 25]
Solution:
An operation is done to get from the first matrix to the second. Identify what was done and tell
how it will affect the value of the determinant.
a b
c d
,
b a
d c
In this case the two columns were switched so the determinant of the second is "1 times the
determinant of the first.
26. 26]
Solution:
Suppose
u v w
q r s
"9
x y z
Find
u v w
det
x y z
q r s
It is 9
27. 27]
Solution:
Suppose
u v w
10
q r s
x y z
Find
u " 3x v " 3y w " 3z
det
x
y
z
q
r
s
It is "10
28. 28]
Solution:
An operation is done to get from the first matrix to the second. Identify what was done and tell
how it will affect the value of the determinant.
a b
c d
,
a
b
ac bd
It is unchanged. It just adds the top row to the bottom.
29. 29]
Solution:
Suppose that A, B are n • n matrices and detŸA 5 while detŸB 3. Find detŸAB T .
15
30. 30]
Solution:
Suppose
u v w
q r s
"7
x y z
Find
v u w " 3u
det
r q s " 3q
y x z " 3x
It is 7
31. 31]
Solution:
Suppose
u v w
3
q r s
x y z
Find
"3u v 3w w
det
"3q r 3w s
"3x 3w y z
It is 9
32. 32]
Solution:
An operation is done to get from the first matrix to the second. Identify what was done and tell
how it will affect the value of the determinant.
a b
c d
,
a
b
2c 2d
The second row was multiplied by 2 so the determinant of the result is 2 times the original
determinant.
33. 33]
Solution:
Suppose that A is a complex n • n matrix and A m I. Find the possible values for detŸA .
You have detŸA m 1 and so detŸA is an m th root of 1.
34. 34]
Solution:
Suppose
u v w
q r s
"9
x y z
Find
u v 3u w
det
q r 3q s
x y 3x z
It is unchanged. Still "9.
35. 35]
Solution:
An operation is done to get from the first matrix to the second. Identify what was done and tell
how it will affect the value of the determinant.
a b
c d
c d
,
a b
In this case two rows were switched and so the resulting determinant is "1 times the first.
36. 36]
Solution:
Suppose
u v w
10
q r s
x y z
Find
u"x v"y w"z
det
"q
"r
"s
x
y
z
It is 10
37. 37]
Solution:
Verify detŸAB detŸA detŸB , detŸA T detŸA , and that det is multilinear in the rows
(columns) for 2 • 2 matrices.
Here is the verification of the formula for products.
det
x y
a b
z w
c d
det
ax cy bx dy
cw az dw bz
adwx " bcwx " adyz bcyz
det
x y
det
z w
a b
c d
Ÿad " bc Ÿwx " yz adwx " bcwx " adyz bcyz
which is the same thing.
38. 38]
Solution:
ad ae af
Find det
bd be bf
and explain why.
cd ce cf
a
The column space is just the span of
b
c
39. 39]
Solution:
and so the determinant is 0.
'
1 1
Find det
a b
, det
1
1
1
a
b
c
2
2
2
a
b
c
. Then conjecture a formula for
1 C 1
1
a0 a1 C an
det
B
B
B
a n0 a n1 C a nn
Finally prove your formula. This is called a Vandermonde determinant. Hint: The formula you
should get is
Ÿa j " a i 0tijtn
which means to take all possible products of terms of the form a j " a i where i j. To verify this
formula, consider
pŸt q det
1
1
C
1
a0
a1
C
t
B
B
a n"1
0
a n0
a n"1
1
a n1
B
C t
n"1
C
tn
Explain why, using properties of determinants that pŸa j 0 for all j n. Therefore, explain why
n"1
pŸt c Ÿt " a i i0
Now note that pŸa n is the thing which is wanted. Of course c is the coefficient of t n and you
know what this is by induction. It equals the determinant of the upper left corner of the matrix
whose determinant equals pŸt in the above formula.
det
1 1
1
det
b"a
a b
1
1
a1 a2 a3
" a 21 a 2 a 21 a 3 a 1 a 22 " a 1 a 23 " a 22 a 3 a 2 a 23 a 21 a 22 a 23
Ÿa 3 " a 2 Ÿa 3 " a 1 Ÿa 2 " a 1 By induction,
pŸt Ÿa j " a i Ÿt " a k 0tijtn"1
Hence
n"1
k0
pŸa n n"1
Ÿa j " a i Ÿa n " a k k0
0tijtn"1
Ÿa j " a i 0tijtn
40. 40]
Solution:
'
Suppose a 0 a 1 a 2 C a n and numbers b 0 , b 1 , C, b n are numbers. Show that there
exists a polynomial, pŸt of degree n such that pŸa i b i . This is known as Lagrange
interpolation.
Let pŸt ) 0 ) 1 x C ) n x n . Then you need to find the ) k such that for each a i ,
) 0 ) 1 a i C ) n a ni b i
In terms of matrices, you need to find the ) k such that
1 a 0 C a n0
)0
1 a 1 C a n1
)1
B
B
B
1 a n C a nn
B
b0
)n
b1
B
bn
However, from the above problem, the matrix has an inverse and so there exists a unique solution
to this problem.
41. 41]
Solution:
'
Suppose a 0 a 1 a 2 C a n and numbers b 0 , b 1 , C, b n are numbers. Show that there
exists a unique polynomial, pŸt of degree n such that pŸa i b i . In fact, show that the following
polynomial works.
n
Ÿt " a j ! b i j:jpiŸa i " a j pŸt i0
j:jpi
This obviously works. pŸa k b k
j:jpk Ÿa k "a j b k . This is because for
j:jpk Ÿa k "a j i p k, j:jpi Ÿa k " a j 0. Getting the solution to this is equivalent to finding ) k a solution to the
system
1 a 0 C a n0
)0
1 a1 C
a n1
)1
B
B
B
B
1 a n C a nn
)n
b0
b1
B
,
bn
the ) k being the coefficients of the desired polynomial. The above formula shows there exists a
solution to this for every choice of the b k and therefore, the matrix must have an inverse.
However, if this is so, then there is also at most one solution to the system. Hence the polynomial
is unique. Of course by an earlier problem, the matrix is known to have an inverse because it is the
transpose of the matrix of the one used in the Vandermonde determinant.
42. 42]
Solution:
An integration scheme is a formula
1
n
n
! cifi X ;x
xh
fŸt dt
i0
where X is in case fŸt 1, t, C, t n . Here f i is the value fŸx i for
x x 0 x 1 C x n x h, these points being equally spaced. Find the integration scheme
for n 1. You need to find the c i . Then give a technique based on this which will integrate every
function which is linear on the subintervals determined by the partition
a x 0 x 1 C x n b where x k " x k"1 h 0.
You need
c01 c11 ;x
1dt h
c0x c1b ;x
tdt 1 hŸ2x h 2
xh
xh
Then it follows that
c0 h , c1 h
2
2
and so the integration scheme is of the form
h fŸx h fŸx h X ; xh fŸt dt
2
2
x
Now suppose you have an interval ¡a, b¢. Apply the integration scheme on each interval of the
form ¡a kh, a Ÿk 1 h¢ to get something which will integrate exactly all functions which are
linear on each of these subintervals. Then
;a
b
n"1
fŸt dt ! ; xih
xŸih n"1
! h2 ŸfŸa ih fŸa Ÿi 1 h fŸt dt i0
h
2
h
2
n"1
! fŸa ih h2
i0
n"1
! fŸa ih h2
i0
i0
n"1
! fŸa Ÿi 1 h i0
n
! fŸa ih i1
h ŸfŸa 2fŸa h 2fŸa 2h C 2fŸa Ÿn " 1 h fŸb 2
This is just the trapezoid rule for numerical integration. When you let n 2, you get Simpson’s
rule and in fact the procedure integrates exactly polynomials of degree three.
43. 43]
Solution:
Let A be an r • r matrix and suppose there are r " 1 rows (columns) such that all rows (columns)
are linear combinations of these r " 1 rows (columns). Show detŸA 0.
If the determinant is non zero, then this will be unchanged with row operations applied to the
matrix. However, by assumption, you can obtain a row of zeros by doing row operations. Thus the
determinant must have been zero after all.
44. 44]
Solution:
Show detŸaA a n detŸA where here A is an n • n matrix and a is a scalar.
detŸaA detŸaIA detŸaI detŸA a n detŸA . The matrix which has a down the main
diagonal has determinant equal to a n .
45. 45]
Solution:
Is it true that detŸA B detŸA detŸB ? If this is so, explain why it is so and if it is not so,
give a counter example.
This is not true at all. Consider A 1 0
0 1
,B "1
0
0
"1
.
46. 46]
Solution:
An n • n matrix is called nilpotent if for some positive integer, k it follows A k 0. If A is a
nilpotent matrix and k is the smallest possible integer such that A k 0, what are the possible
values of detŸA ?
It must be 0 because 0 detŸ0 detŸA k ŸdetŸA k .
47. 47]
Solution:
A matrix is said to be orthogonal if A T A I. Thus the inverse of an orthogonal matrix is just its
transpose. What are the possible values of detŸA if A is an orthogonal matrix?
You would need detŸAA T detŸA detŸA T detŸA 2 1 and so detŸA 1, or "1.
48. 48]
Solution:
Fill in the missing entries to make the matrix orthogonal. A is orthogonal if A T A I.
"1
2
1
6
12
6
1
2
1
6
12
6
0
6
3
"
"1
2
1
6
12
6
1
2
_
_
_
6
3
_
.
12
6
49. 49]
Solution:
Let A and B be two n • n matrices. A L B (A is similar to B) means there exists an invertible
matrix, S such that A S "1 BS. Show that if A L B, then B L A. Show also that A L A and that if
A L B and B L C, then A L C.
If A S "1 BS, then SAS "1 B and so if A L B, then B L A. It is obvious that A L A because you
can let S I. Say A L B and B L C. Then A P "1 BP and B Q "1 CQ. Therefore,
A P "1 Q "1 CQP ŸQP "1 CŸQP and so A L C.
50. 50]
Solution:
Show that if A L B, then detŸA detŸB . To say A L B is to say there exists S such that
A S "1 BS. If detŸA detŸB , does it follow that A L B?
detŸA detŸS "1 BS detŸS "1 detŸB detŸS detŸB detŸS "1 S detŸB .
The other direction does not work. You could consider
4 0
0 1
2 0
,
0 2
These are not similar. If you had
"1
4 0
0 1
x y
2 0
x y
z w
0 2
z w
then you would have
x y
4 0
z w
0 1
2 0
x y
0 2
z w
2x 2y
and so
4x y
4z w
2z 2w
but there is no solution to this unless x, y, z, w are all zero and this would violate the existence of
"1
x y
z w
.
51. 51]
Solution:
Two n • n matrices, A and B, are similar if B S "1 AS for some invertible n • n matrix, S. Show
that if two matrices are similar, they have the same characteristic polynomials. The characteristic
polynomial of an n • n matrix, M is the polynomial, detŸ5I " M .
Say M S "1 NS. Then
detŸ5I " M detŸ5I " S "1 NS detŸ5S "1 S " S "1 NS detŸS "1 Ÿ5I " N S detŸ5I " N 52. 52]
Solution:
Recall that the volume of a parallelepiped determined by the vectors a, b, c in R 3 is the absolute
value of the box product a b • c. Show that this box product is the same as the determinant
aT
det
bT
det
a b c
cT
Also show that for a, b two vectors in R 2 the area of the parallelogram determined by these two
vectors is the absolute value of the determinant
det
aT
det
bT
a b
It can be shown that the only reasonable way to define the volume of a parallelepiped determined
by n vectors £a 1 , C, a n ¤ in R n is to take the absolute value of the determinant
det
a1 C an
The formula in three dimensions follows right away because the box product is of the form
i
j
k
b1 b2 b3
a1 a2 a3
a1i a2j a3k c1 c2 c3
b1 b2 b3
c1 c2 c3
In two dimensions, you just use the formula for the area of the parallelogram determined by
Ÿa 1 , a 2 , 0 and Ÿb 1 , b 2 , 0 . It reduces to the desired determinant.
53. 53]
Solution:
Suppose you have a box in three dimensions ¡a, b¢ • ¡c, d¢ • ¡e, f¢ R. Also suppose A is a 3 • 3
matrix. Show that the volume of AR is the absolute value of detŸA volume of R . Say
2
A
y2
2
a 1 a 2 a 3 . Then if you have an ellipsoid determined by ax 2 b 2 cz 2 t 1, show that
the volume of the ellipsoid is 43 =abc. Here each of a, b, c is positive. You can use the well known
formula for the volume of a ball of radius 1 which is just 43 =.
First note that R Ÿa, b, c R 0 where R 0 has the same length sides but the lower left corner is at
Ÿ0, 0, 0 . The lengths of the sides of both boxes are Ÿb " a , Ÿd " c , Ÿf " e . Then
a
AR A
b
AR 0
c
so the volume of AR is the same as the volume of AR 0 . Now let R 0 denote the box which has
lower left corner at Ÿ0, 0, 0 and sides of length 1. Then from matrix multiplication, the
parallelpiped determined by the columns of A is just AR 0 . Also, from geometric reasoning, if you
have a parallelpiped and you simply scale the sides by multiplying by constants, the volume of the
result is just the product of the constants times the volume of the unscaled parallelepiped. Now R 0
is just the scaled box R 0 the sides being multiplied by the above factors. You can verify this by
drawing a picture and using a little trigonometry. Thus, letting VŸS denote the volume of S,
VŸAR VŸAR 0 Ÿb " a Ÿd " c Ÿf " e VŸAR 0 Ÿb " a Ÿd " c Ÿf " e |detŸA | |detŸA |VŸR If you imagine stuffing various other sets U with non overlapping boxes, the same relation holds
|detŸA |VŸU VŸAŸU a 0 0
Let U be the unit ball. Then let A 0 b 0
. It follows that the ellipsoid is just AU and so
0 0 c
|detŸA |VŸU abc 4 =
3
54. 54]
Solution:
Tell whether the statement is true or false.
a. If A is a 3 • 3 matrix with a zero determinant, then one column must be a multiple of some
other column.
1
i. False. Consider
1 2
"1 5 4
0
3 3
b. If any two columns of a square matrix are equal, then the determinant of the matrix equals
zero.
i. True
c. For A and B two n • n matrices, detŸA B detŸA detŸB .
i. False
d. For A an n • n matrix, detŸ3A 3 detŸA i. False
e. If A "1 exists then detŸA "1 detŸA "1 .
i. True
f. If B is obtained by multiplying a single row of A by 4 then detŸB 4 detŸA .
i. True
g. For A an n • n matrix, detŸ"A Ÿ"1 n detŸA .
i. True
h. If A is a real n • n matrix, then detŸA T A u 0.
i. True
i. Cramer’s rule is useful for finding solutions to systems of linear equations in which there is
an infinite set of solutions.
i. False
j. If A k 0 for some positive integer, k, then detŸA 0.
i. True
k. If Ax 0 for some x p 0, then detŸA 0.
i. True
55. 55]
Solution:
Use Cramer’s rule to find the solution to
x " 3y 3
x " 2y 1
Ÿ"3, "2 56. 56]
Solution:
Use Cramer’s rule to find the solution to
x " 3y "3
4x " 11y "11
Ÿ0, 1 57. 57]
Solution:
Use Cramer’s rule to find the solution to
x " 4y 0
5x " 19y 1
Ÿ4, 1 58. 58]
Solution:
Use Cramer’s rule to find the solution to
"3x " 2y " 2z " 3 0
"4x " 3y " 3z " 3 0
2x 2y 3z 2 0
Ÿ"3, 5, "2 59. 59]
Solution:
Use Cramer’s rule to find the solution to
6x 7y 7z 16 0
5x 6y 6z 14 0
3x 3y 4z 6 0
Ÿ2, "4, 0 60. 60]
Solution:
Use Cramer’s rule to find the solution to
4x 5y 5z " 31 0
3x 4y 4z " 24 0
3x 3y 4z " 20 0
Ÿ4, 4, "1 61. 61]
Solution:
Here is a matrix,
4 5 5
6 8 8
5 5 6
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"1
4 5 5
6 8 8
4
" 52
0
2
" 12
5
2
"1
"5
5 5 6
1
62. 62]
Solution:
Here is a matrix,
2 3 3
4 5 8
1 1 2
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"1
2 3 3
4 5 8
1 1 2
2
"3
9
0
1
"4
"1
1
"2
63. 63]
Solution:
Here is a matrix,
1 2 2
0 1 1
3 3 4
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"1
1 2 2
0 1 1
3 3 4
1
"2
3
"2 "1
"3
3
0
1
64. 64]
Solution:
Here is a matrix,
"2 "1 "1
"3 "2 "2
1
1
2
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"2 "1 "1
"1
1
1
4
"3 "1
"1
1
"3 "2 "2
1
"2
2
0
1
65. 65]
Solution:
Here is a matrix,
1
2
2
0 "2 "2
5
5
6
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"1
1
2
2
0 "2 "2
5
5
6
1
1
0
5
2
"1
"5 " 52
1
66. 66]
Solution:
Here is a matrix,
3 4 4
2 3 3
2 2 3
Determine whether the matrix has an inverse by finding whether the determinant is non zero. If
the determinant is nonzero, find the inverse using the formula for the inverse which involves the
cofactor matrix.
"1
3
"4
0
0
1
"1
"2
2
1
3 4 4
2 3 3
2 2 3
67. 67]
Solution:
Here is a matrix,
e 2t
2e
2t
cos t
sin t
" sin t
cos t
4e 2t " cos t " sin t
Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
e 2t
"1
cos t
sin t
2e 2t " sin t
cos t
1
5
4
5
4
5
4e 2t " cos t " sin t
e "2t
cos t sin t "
1
5
0
2
sin t
5
2
cos t
5
" sin t
cos t
e "2t
2
sin t " 15 cos t
5
" 25 cos t " 15 sin t
68. 68]
Solution:
Here is a matrix,
et
" sin 2t
cos 2t
e t "2 sin 2t "2 cos 2t
e t "4 cos 2t
4 sin 2t
Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
et
cos 2t
" sin 2t
e t "2 sin 2t "2 cos 2t
e t "4 cos 2t
4 sin 2t
"1
4
5
1
5
2
5
e "t
cos 2t cos 2t "
2
5
1
5
1
5
0
sin 2t " 12 sin 2t
sin 2t " 12 cos 2t
1
10
1
10
e "t
sin 2t "
cos 2t 1
cos 2t
5
1
sin 2t
5
69. 69]
Solution:
Here is a matrix,
e t cosh t sinh t
e t sinh t cosh t
e t cosh t sinh t
Does there exist a value of t for which this matrix fails to have an inverse? Explain.
e t cosh t sinh t
det
e t sinh t cosh t
e t cosh t sinh t
0 and so this matrix fails to have a nonzero determinant at any
value of t.
70. 70] Here is a matrix,
"1
t
t2
0
"1 2t
t
0
4
Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
"1
Solution:
"1
t2
t
0
"1 2t
t
0
2
" 3t 344 "4 3t 3t4 3 3tt3 4
2 3tt3 4
" 3tt 34
4
3
2 3t 3t4
t
3t 3 4
t2
3t 3 4
1
3t 3 4
2
4
provided 3t 3 4 p 0.
71. 71]
Solution:
Here is a matrix,
Ÿcos 2t e "2t
e 5t
5e 5t
"e "2t sin 2t
"2e "2t Ÿcos 2t sin 2t "2e "2t Ÿcos 2t " sin 2t 8e "2t sin 2t
25e 5t
8Ÿcos 2t e "2t
Determine whether the matrix has an inverse by finding whether the determinant is non zero.
determinant is " 106
1 Ÿ"4t e4
72. 72]
Solution:
Here is a matrix,
1 t
e "2t
0 1 "2e "2t
0 0
4e "2t
Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
1 t
e "2t
0 1 "2e "2t
0 0
"1
1 "t " 12 t "
4e "2t
1
4
1
2
0 1
1
4
0 0
e 2t
73. 73]
Solution:
Here is a matrix,
1
0
0 "3e
0
Ÿcos 3t e
"3t
0
"3t
Ÿcos 3t sin 3t "3e
"e
"3t
"3t
sin 3t
Ÿcos 3t " sin 3t Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
"1
1
0
0
0
Ÿcos 3t e "3t
"e "3t sin 3t
0 "3e "3t Ÿcos 3t sin 3t "3e "3t Ÿcos 3t " sin 3t 1
0
0
0
e Ÿcos 3t " sin 3t 3t
"
0 "e 3t Ÿcos 3t sin 3t "
sin3t
3e "3t cos2 3t3e "3t sin 2 3t
cos3t
3e "3t cos2 3t3e "3t sin 2 3t
74. 74]
Solution:
Here is a matrix,
et
Ÿcos 4t e 2t
"e 2t sin 4t
et
2e 2t Ÿcos 4t " 2 sin 4t "2e 2t Ÿ2 cos 4t sin 4t e t "12Ÿcos 4t e 2t " 16e 2t sin 4t 12e 2t sin 4t " 16Ÿcos 4t e 2t
Determine whether the matrix has an inverse by finding whether the determinant is non zero.
determinant is "68e 5t
75. 75]
Solution:
Here is a matrix,
1
0
Ÿcos t e
0
0
"t
Ÿsin t e "t
0 "e "t Ÿcos t sin t e "t Ÿcos t " sin t Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
"1
1
0
0
1
0
Ÿcos t e "t
Ÿsin t e "t
0 "e "t Ÿcos t sin t e "t Ÿcos t " sin t 0
0
0 e t Ÿcos t " sin t "e t sin t
0 e t Ÿcos t sin t Ÿcos t e t
76. 76]
Solution:
Here is a matrix,
e "3t
te "3t
1
"3e "3t "e "3t Ÿ3t " 1 0
9e "3t
3e "3t Ÿ3t " 2 0
Determine whether the matrix has an inverse by finding whether the determinant is non zero. Then
find its inverse if it has one.
e "3t
te "3t
"1
"3e "3t "e "3t Ÿ3t " 1 0
9e "3t
0
1
3e "3t Ÿ3t " 2 0
1
3
e 3t Ÿ3t " 2 1
9
e 3t Ÿ3t " 1 0
"e 3t
" 13 e 3t
1
2
3
1
9
77. 77]
Solution:
Show that if detŸA p 0 for A an n • n matrix, it follows that if Ax 0, then x 0.
If detŸA p 0, then A "1 exists and so you could multiply on both sides on the left by A "1 and
obtain that x 0.
78. 78]
Solution:
Suppose A, B are n • n matrices and that AB I. Show that then BA I. Hint: You might do
something like this: First explain why detŸA , detŸB are both nonzero. Then ŸAB A A and then
show BAŸBA " I 0. From this use what is given to conclude AŸBA " I 0.
You have 1 detŸA detŸB . Hence both A and B have inverses. Letting x be given,
AŸBA " I x ŸAB Ax " Ax Ax " Ax 0
and so it follows from the above problem that ŸBA " I x 0. Since x is arbitrary, it follows that
BA I.
79. 79]
Solution:
Suppose A is an upper triangular matrix. Show that A "1 exists if and only if all elements of the
main diagonal are non zero. Is it true that A "1 will also be upper triangular? Explain. Is everything
the same for lower triangular matrices?
The given condition is what it takes for the determinant to be non zero. Recall that the determinant
of an upper triangular matrix is just the product of the entries on the main diagonal. The inverse
will also be upper triangular. This follows from the method of row operations applied to finding
the inverse.
80. 80]
Solution:
If A, B, and C are each n • n matrices and ABC is invertible, why are each of A, B, and C
invertible.
This is obvious because detŸABC detŸA detŸB detŸC and if this product is nonzero, then
each determinant in the product is nonzero and so each of these matrices is invertible.
81. 81]
Solution:
Let FŸt det
aŸt bŸt cŸt dŸt . Verify
F U Ÿt det
Now suppose
a U Ÿt b U Ÿt cŸt dŸt det
aŸt bŸt c U Ÿt d U Ÿt .
aŸt bŸt cŸt FŸt det
.
dŸt eŸt fŸt gŸt hŸt iŸt Use Laplace expansion and the first part to verify F U Ÿt a U Ÿt b U Ÿt c U Ÿt det
det
dŸt eŸt fŸt gŸt hŸt iŸt aŸt det
bŸt cŸt d U Ÿt e U Ÿt f U Ÿt gŸt aŸt bŸt cŸt dŸt eŸt hŸt iŸt .
fŸt g U Ÿt h U Ÿt i U Ÿt Conjecture a general result valid for n • n matrices and explain why it will be true. Can a similar
thing be done with the columns?
Yes, a similar thing can be done with the columns because the determinant of the transpose is the
same as the determinant of the matrix.
82. 82]
Solution:
Let Ly y Ÿn a n"1 Ÿx y Ÿn"1 C a 1 Ÿx y U a 0 Ÿx y where the a i are given continuous
functions defined on a closed interval, Ÿa, b and y is some function which has n derivatives so it
makes sense to write Ly. Suppose Ly k 0 for k 1, 2, C, n. The Wronskian of these functions,
y i is defined as
WŸy 1 , C, y n Ÿx q det
y 1 Ÿx C
y n Ÿx y U1 Ÿx C
y Un Ÿx B
B
y Ÿn"1 Ÿx 1
C
y Ÿn"1 Ÿx n
Show that for WŸx WŸy 1 , C, y n Ÿx to save space,
U
W Ÿx det
y 1 Ÿx C
y n Ÿx y U1 Ÿx C
y Un Ÿx B
B
.
Ÿn y Ÿn 1 Ÿx C y n Ÿx Now use the differential equation, Ly 0 which is satisfied by each of these functions, y i and
properties of determinants presented above to verify that W U a n"1 Ÿx W 0. Give an explicit
solution of this linear differential equation, Abel’s formula, and use your answer to verify that
the Wronskian of these solutions to the equation, Ly 0 either vanishes identically on Ÿa, b or
never. Hint: To solve the differential equation, let A U Ÿx a n"1 Ÿx and multiply both sides of the
differential equation by e AŸx and then argue the left side is the derivative of something.
The last formula above follows because W U Ÿx equals the sum of determinants of matrices which
have two equal rows except for the last one in the sum which is the displayed expression. Now let
m i Ÿx " a n"1 Ÿx y Ÿn"1 C a 1 Ÿx y Ui a 0 Ÿx y i
i
Since each y i is a solution to Ly 0, it follows that y Ÿn i Ÿt m i Ÿt . Now from the properties of
determinants, being linear in each row,
W U Ÿx "a n"1 Ÿx det
y 1 Ÿx C
y n Ÿx y U1 Ÿx C
y Un Ÿx B
B
y Ÿn"1 Ÿx 1
C
y Ÿn"1 Ÿx n
"a n"1 Ÿx W U Ÿx Now let A U Ÿx a n"1 Ÿx . Then
d Ÿe AŸx WŸx 0
dx
and so WŸx Ce "AŸx . Thus the Wronskian either vanishes for all x or for no x.
83. 83]
Solution:
Find the determinant of the following matrix.
"2 i 1 4i
3i
det
"2 i 1 4i
3i
0
0
12 " 3i
84. 84]
Solution:
Find the determinant of the following matrix.
22 " 24i
32 " 34i
16 " 10i
"13 14i "19 20i "9 6i
"8 10i "12 14i "6 4i
8
85. 85]
Solution:
Find the determinant of the following matrix.
19 " 36i
21 " 51i
5 " 15i
"5 21i
"3 30i "1 9i
"14 15i "18 21i "4 6i
0
Rank, Dimension, Subspaces
1. 86]
Solution:
Let £u 1 , C, u n ¤ be vectors in R n . The parallelepiped determined by these vectors PŸu 1 , C, u n is
defined as
n
PŸu 1 , C, u n q
! t k u k : t k ¡0, 1¢ for all k
.
k1
Now let A be an n • n matrix. Show that
£Ax : x PŸu 1 , C, u n ¤
is also a parallelepiped.
n
By definition, AŸPŸu 1 , C, u n ! k1 t k Au k : t k ¡0, 1¢ for all k
which equals
PŸAu 1 , C, Au n which is a parallelepiped.
2. 87]
Solution:
Draw PŸe 1 , e 2 where e 1 , e 2 are the standard basis vectors for R 2 . Thus e 1 Ÿ1, 0 , e 2 Ÿ0, 1 .
Now suppose
E
1 1
0 1
where E is the elementary matrix which takes the third row and adds to the first. Draw
£Ex : x PŸe 1 , e 2 ¤.
In other words, draw the result of doing E to the vectors in PŸe 1 , e 2 . Next draw the results of
doing the other elementary matrices to PŸe 1 , e 2 .
For E the given elementary matrix, the result is as follows.
It is called a shear. Note that it does not change the area.
In caseE 1 0
, the given square PŸe 1 , e 2 becomes a rectangle. If ) 0, the square is
0 )
stretched by a factor of ) in the y direction. If ) 0 the rectangle is reflected across the x axis and
) 0
in addition is stretched by a factor of |)| in the y direction. When E the result is
0 1
similar only it features changes in the x direction. These elementary matrices do change the area
unless |)| 1 when the area is unchanged. The permutation
0 1
simply switches e 1 and
1 0
e 2 so the result appears to be a square just like the one you began with.
3. 88]
Solution:
Either draw or describe the result of doing elementary matrices to PŸe 1 , e 2 , e 3 , the parallelepiped
determined by the three listed vectors. Describe geometrically the fact proved in the chapter that
every invertible matrix is a product of elementary matrices.
It is the same sort of thing. The elementary matrix either switches the e i about or it produces a
shear or a magnification in one direction.
4. 89]
Solution:
Explain why the elementary matrix which comes from the row operation which replaces a row
with a constant times another row added to it does not change the volume of a box which has the
sides parallel to the coordinate axes.
It is because the change happens in only two dimensions where a rectangle becomes a
parallelogram which has the same base and height as the rectangle. Thus the area of this face does
not change and so the volume of the resulting sheared box is the same as the volume of the box.
5. 90]
Solution:
Determine which matrices are in row reduced echelon form.
a.
1 2 0
0 1 7
i. This one is not.
1 0 0 0
b.
0 0 1 2
0 0 0 0
i. This one is.
1 1 0 0 0 5
c.
0 0 1 2 0 4
0 0 0 0 1 3
This one is.
6. 91]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
"1 "1 "2 "6
1
1
1
3
1
1
0
0
1 1 0 0
row echelon form:
0 0 1 3
0 0 0 0
7. 92]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
"1
1
2
1
1
"2
"7
"3
1
"3 "12 "5
1 0 3 1
row echelon form:
0 1 5 2
0 0 0 0
8. 93]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
3 11 "24 11
1 0
row echelon form:
3
1
4
"9
4
1
3
"6
3
0
0 1 "3 1
0 0
0
0
9. 94]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
"2 "4 "1 "1
1 2 0 "1
row echelon form:
10. 95]
Solution:
0 0 1
3
0 0 0
0
"1
1
2
0
1
2
"1 "4
Row reduce the following matrix to obtain the row reduced echelon form.
1 0
row echelon form:
0
"2 "9
27
"29
1
4
"12
13
1
3
"9
10
1
0 1 "3 3
0 0
0
0
11. 96]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
3 "4 21 "5
1 0
row echelon form:
3
1 "1
6
"1
1 "2
9
"3
1
0 1 "3 2
0 0
0
0
12. 97]
Solution:
Row reduce the following matrix to obtain the row reduced echelon form.
"1
1
"1 "5
1
"1
0
1
"1 "1 "1
2
1 "1 0 2
row echelon form:
0
0
1 3
0
0
0 0
13. 98]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
1 "4 1 "1
1 "4 2
1
1 "4 1 "1
1 "4 0 "3
row echelon form:
0
0
1
2
0
0
0
0
1
Rank is 2 Basis for column space is
1
,
2
1
0 0 1 2
,
Basis for row space is
1
1
1 "4 0 "3
14. 99]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
"2
4
1
"2 "2
2
7
1
"2 "2
1
3
"3
6
"3
"9
4
6
"5 "18
1 "2 "2 0 "1
row echelon form:
"2
1
1
"3
0
0
0
1
4
0
0
0
0
0
0
0
0
0
0
Then a basis for the column space is
"5
,
2
1
Rank 2 Basis for row space is
"3
1 "2 "2 0 "1
,
0 0 0 1 4
15. 100]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
1 "2 0 "2
1 "2 1
0
1 "2 0 "2
1 "2 0 "2
row echelon form:
0
0
1
2
0
0
0
0
0
Rank is 2 Basis for column space is
1
,
1
0
0 0 1 2
,
Basis for row space is
1
1
1 "2 0 "2
16. 101]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
3 9 "4 "14
1 3 "1
"4
1 3 "2
"6
1 3 0 "2
row echelon form:
0 0 1
2
0 0 0
0
"4
Rank is 2 Basis for column space is
3
"1
,
"2
0 0 1 2
,
1
Basis for row space is
1
1 3 0 "2
17. 102]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
4 8 "16 "9 6
1 2
"4
"2 2
1 2
"4
"3 0
3 6 "12 "9 0
1 2 "4 0 6
row echelon form:
0 0
0
1 2
0 0
0
0 0
0 0
0
0 0
Then a basis for the column space is
"9
4
1
1
"2
,
Rank 2 Basis for row space is
"3
1 2 "4 0 6
,
0 0 0 1 2
"9
3
18. 103]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
"1 "1
1
1
1
"2 "5 "32
1
1
"3 "4 "24
"2 "2
1 1 0 0
"1
1
1
"2
0 0 0 1
7
0 0 0 0
0
1
,
48
Rank 3 Basis for column space is
5
"2
,
"3
"5
Basis for row space is
"4
6
1 1 0 0 1
8
33
1
0 0 1 0 "1
row echelon form:
6
5
8
,
0 0 1 0 "1
,
0 0 0 1 7
19. 104]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
2 "5 23
1 "2 10
1 "3 13
1 "3 13
1 0
row echelon form:
4
0 1 "3
0 0
0
0 0
0
Then a basis for the column space is
"5
2
1
Basis for column space:
"2
,
1
"3
1
1 0 4
Rank 2 Basis for row space:
"3
0 1 "3
,
20. 105]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
1 2 "1 0 "4
1 2
0
0 "5
1 2 "1 1
2
0 0
0
0
0
1 2 0 0 "5
0 0 1 0 "1
row echelon form:
"1
1
1
1
0
,
0
"1
6
0 0 0 0
0
Rank 3 Basis for column space is
0
0
,
1
0
1 2 0 0 "5
0 0 0 1
Basis for row space is
0
,
0 0 1 0 "1
,
0 0 0 1 6
21. 106]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
"1 "5 "15
1
4
12
1
3
9
"2 "6 "18
1 0 0
row echelon form:
0 1 3
0 0 0
0 0 0
Then a basis for the column space is
"1
"5
1
Basis for column space:
4
,
1
"2
1 0 0
,
Rank 2 Basis for row space:
3
"6
0 1 3
22. 107]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
2 4 5 8 35
1 2 3 4 16
1 2 2 5 26
1 2 2 5 26
1 2 0 0 "3
0 0 1 0 "3
row echelon form:
2
1
1
1
5
,
3
2
7
0 0 0 0
0
Rank 3, Basis for column space is
8
4
,
Basis for row space is
5
2
1 2 0 0 "3
0 0 0 1
5
0 0 1 0 "3
,
,
0 0 0 1 7
23. 108]
Solution:
Find the rank of the following matrix. Also find a basis for the row and column spaces.
1 "1 "1 0
row echelon form:
"2
2
2
"1
0
1
"1 "1
0
1
1
"1 "1 "1
3
"3
3
"9
3
3
1
0
0
0
1 "2
0
0
0
0
0
0
0
0
0
0
Then a basis for the column space is
"2
"1
1
,
1
"3
0
Rank 2 Basis for row space is
"1
3
1 "1 "1 0 1
0 0 0 1 "2
,
24. 109]
Solution:
Suppose A is an m • n matrix. Explain why the rank of A is always no larger than minŸm, n .
It is because you cannot have more than minŸm, n nonzero rows in the row reduced echelon
form.
25. 110]
Solution:
5
1
Let H denote span
1
14
,
4
of H and determine a basis.
It is dimension 3 and a basis is
5
1
1
14
,
4
3
2
3
2
33
,
7
5
15
,
3
3
8
20
12
19
62
5
5
,
1
2
. Find the dimension
8
5
,
8
1
2
8
26. 111]
Solution:
5
1
Let H denote span
1
,
4
determine a basis.
It is dimension 3 and a basis is
5
1
1
19
,
4
27. 112]
Solution:
4
3
12
5
,
1
2
8
4
3
12
,
13
10
40
,
1
2
8
. Find the dimension of H and
"1
0
1
Let H denote span
,
0
"1
5
0
"1
"4
dimension of H and determine a basis.
It is dimension 3 and a basis is
"1
0
1
,
0
"1
3
0
3
,
"2
0
1
,
0
,
0
"2
"2
7
0
. Find the
"5
0
1
,
0
"2
"2
28. 113]
Solution:
"7
3
Let H denote span
1
1
"2
,
"3
8
,
11
"6
2
"8
27
"2
,
. Find the dimension of H and
"4
"8
22
determine a basis.
It is dimension 2 and a basis is
"7
3
1
1
"2
,
"3
"6
2
29. 114]
Solution:
"2
Let H denote span
1
1
"3
and determine a basis.
It is dimension 2 and a basis is
"2
1
1
"9
,
"3
30. 115]
Solution:
4
3
"9
"9
,
4
3
"9
"33
,
15
12
"36
"2
,
1
1
"3
. Find the dimension of H
2
1
Let H denote span
2
27
,
1
of H and determine a basis.
It is dimension 3 and a basis is
2
1
2
7
,
1
4
6
7
15
,
24
4
6
6
,
3
8
12
3
4
3
12
12
16
,
9
14
. Find the dimension
7
6
,
3
3
8
4
31. 116]
Solution:
4
1
Let H denote span
1
,
3
1
0
,
3
3
,
3
4
0
9
12
2
6
2
. Find the dimension of H and
determine a basis.
It is dimension 3 and a basis is
4
1
1
3
,
3
1
0
0
12
,
3
4
12
32. 117]
Solution:
1
Let H denote span
2
1
,
0
determine a basis.
It is dimension 2 and a basis is
1
2
1
2
,
0
33. 118]
Solution:
6
2
0
6
2
0
,
16
6
0
,
4
2
0
. Find the dimension of H and
34.
35.
36.
37.
Let M £u Ÿu 1 , u 2 , u 3 , u 4 R 4 : sinŸu 1 1¤. Is M a subspace? Explain.
Not a subspace. Ÿ =2 , 0, 0, 0 is in it. 4Ÿ =2 , 0, 0, 0 is not.
119]
Solution:
Let M u Ÿu 1 , u 2 , u 3 , u 4 R 4 : u i u 0 for each i 1, 2, 3, 4 . Is M a subspace? Explain.
Not a subspace. Ÿ1, 1, 1, 1 is in it. However, Ÿ"1 Ÿ1, 1, 1, 1 is not.
120]
Solution:
Let w R 4 and let M £u Ÿu 1 , u 2 , u 3 , u 4 R 4 : w u 0¤. Is M a subspace? Explain.
Subspace
121]
Solution:
Let M £u Ÿu 1 , u 2 , u 3 , u 4 R 4 : |u 1 | t 4¤. Is M a subspace? Explain.
This is not a subspace. Ÿ1, 1, 1, 1 is in it, but 20Ÿ1, 1, 1, 1 is clearly not.
122]
Solution:
Let w, w 1 be given vectors in R 4 and define
M u Ÿu 1 , u 2 , u 3 , u 4 R 4 : w u 0 and w 1 u 0 .
Is M a subspace? Explain.
Sure. This is a subspace. It is obviously closed with respect to vector addition and scalar
multiplication.
38. 123]
Solution:
Prove the following theorem: If A, B are n • n matrices and if AB I, then BA I and B A "1 .
Hint: First note that if AB I, then it must be the case that A is onto. Explain why this requires
span columns of A F n . Now explain why, using the corollary that this requires A to be one to
one. Next explain why AŸBA " I 0 and why the fact that A is one to one implies BA I.
If AB I, then B must be one to one. Otherwise there exists x p 0 such that Bx 0. But then
you would have
x Ix ABx A0 0
In particular, the columns of B are linearly independent. Therefore, B is also onto. Also,
ŸBA " I Bx BŸAB x "Bx 0
Since B is onto, it follows that BA " I maps every vector to 0 and so this matrix is 0. Thus BA I.
39. 124]
Solution:
Let M £u Ÿu 1 , u 2 , u 3 , u 4 R 4 : u 3 u u 1 ¤. Is M a subspace? Explain.
This is not a subspace. Ÿ0, 0, 1, 0 is in it. But Ÿ0, 0, "1, 0 is not.
40. 125]
Solution:
Suppose £x 1 , C, x k ¤ is a set of vectors from F n . Show that 0 is in spanŸx 1 , C, x k .
! ki1 0x k 0.
41. 126]
Solution:
Explain why if A is m • n where m n, then A cannot be one to one.
The columns cannot be a basis for F m because there are too many of them. Therefore, they are
dependent. Hence there exists x p 0 such that Ax 0.
42. 127]
Solution:
Let M £u Ÿu 1 , u 2 , u 3 , u 4 R 4 : u 3 u 1 0¤. Is M a subspace? Explain
Yes. This is a subspace. It is closed with respect to vector addition and scalar multiplication.
43. 128]
Solution:
Prove the following theorem: If A, B are n • n matrices and if AB I, then BA I and B A "1 .
Hint: First note that if AB I, then it must be the case that A is onto. Explain why this requires
span columns of A F n . Now explain why, using the corollary that this requires A to be one to
one. Next explain why AŸBA " I 0 and why the fact that A is one to one implies BA I.
If AB I, then B must be one to one. Otherwise there exists x p 0 such that Bx 0. But then
you would have
x Ix ABx A0 0
In particular, the columns of B are linearly independent. Therefore, B is also onto. Also,
ŸBA " I Bx BŸAB x "Bx 0
Since B is onto, it follows that BA " I maps every vector to 0 and so this matrix is 0. Thus BA I.
44. 129]
Solution:
Study the definition of span. Explain what is meant by the span of a set of vectors. Include
pictures.
The span of vectors is a long flat thing. In three dimensions, it would be a plane which goes
through the origin.
45. 130]
Solution:
Let £v 1 , C, v m ¤ be vectors. Prove that spanŸv 1 , C, v m is a subspace.
m
m
Say a, b are scalars. Let ! k1 ) k v k and ! k1 * k v k are two things in spanŸv 1 , C, v m . Then
m
m
k1
k1
a ! )kvk b ! *kvk m
!Ÿa) k * k b v k spanŸv 1 , C, v m k1
46. 131]
Solution:
Let A be an m • n matrix. Then
NŸA q kerŸA q £x F n : Ax 0¤
Show that NŸA is a subspace of F n .
Say Ax Ay 0. Then if a, b are scalars,
AŸax by aAx bAy a0 b0 0
47. 132]
Solution:
Let A be an m • n matrix. Then the image of A is defined as
ImŸA q AŸF n q £Ax : x F n ¤
Show that NŸA is a subspace of F m . Note that this is really the column space. It is done in the
book but you should do it here without dealing with columns.
Let a, b be scalars. Let Ax, Ay be two things in ImŸA . Then is it the case that aAx bAy is in
ImŸA ?
aAx bAy AŸax by ImŸA 48. 133]
Solution:
Suppose you have a subspace V of F n . Is it the case that V equals NŸA kerŸA for some m • n
matrix?
You know that V has a basis £v 1 , C, v m ¤, m t n. If m n, then V F n and you take A 0 n•n .
Otherwise, choose vectors u m1 , C, u n such that u k v j 0 and u j u i - ij . How do you do
this?
v '1
B
v 'm
is longer than it is tall. Hence there exists u 1 p 0 such that this matrix times u 1 to equal 0. Next,
to get u 2 , do the same thing with
v '1
B
v 'm
u '1
and continue till the matrix has the same height as length. Each time divide by the magnitude of
the vector obtained. Consider the Ÿn " m • n matrix.
u 'm1
Aq
B
u 'n
By construction, it sends each v k to 0 and so kerŸA — spanŸv 1 , C, v m . Now suppose
x
m
n
i1
jm1
m
n
n
i1
jm1
jm1
! a i v i ! b ju j kerŸA Then
0 Ax ! a i Av i ! b jAu j ! b ju j
and so each b j 0. Hence x spanŸv 1 , C, v m .
49. 134]
Solution:
Suppose you have a subspace V of F n . Is it the case that V equals ImŸA for some matrix A?
Yes, this is obvious. You just form the matrix which has a basis for V as its columns.
50. 135]
Solution:
Consider the vectors of the form
v
u"v"w
: u, v, w R
.
w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 3 and a basis is
0
1
1
"1
,
0
0
0
,
"1
1
51. 136]
Solution:
Consider the vectors of the form
2u v " 5w
4v " 2u 10w
: u, v, w R
.
2v " 4u 14w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
2
"2
"4
1
,
4
2
52. 137]
Solution:
Consider the vectors of the form
3u v 5w
7u 3v 13w
: u, v, w R
.
15u 6v 27w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
3
1
,
7
15
3
6
53. 138]
Solution:
Consider the vectors of the form
2u v
w"v"u
: u, v, w R
.
w " 2v " 2u
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 3 and a basis is
2
1
"1
,
"2
"1
"2
0
,
1
1
54. 139]
Solution:
Consider the vectors of the form
4u 12v 15w
u 3v 4w
4u 12v 12w
: u, v, w R
.
3u 9v 9w
Is this set of vectors a subspace of R 4 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
4
1
4
3
15
,
4
12
9
55. 140]
Solution:
Consider the vectors of the form
3u v
5u v " w
: u, v, w R
.
w " 2v " 7u
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 3 and a basis is
3
1
,
5
1
"7
"2
0
,
"1
1
56. 141]
Solution:
Consider the vectors of the form
3u v " w
3u 2v w
: u, v, w R
.
uvw
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
3
3
1
,
1
2
1
57. 142]
Solution:
Consider the vectors of the form
2u v 7w
2u " 4v 2w
: u, v, w R
.
"12v " 12w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
2
2
0
1
,
"4
"12
58. 143]
Solution:
Consider the vectors of the form
3u v
6u 6v 2w
: u, v, w R
.
2u 4v 2w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 3 and a basis is
3
6
1
,
0
,
6
2
4
2
2
59. 144]
Solution:
Consider the vectors of the form
3u v 5w
0
: u, v, w R
.
"6u " 4v " 14w
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
3
1
,
0
0
"6
"4
60. 145]
Solution:
Consider the vectors of the form
4u 4v 7w
0
4u 4v 4w
: u, v, w R
.
3u 3v 3w
Is this set of vectors a subspace of R 4 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
4
0
4
3
7
,
0
4
3
61. 146]
Solution:
Consider the vectors of the form
"w
2u 2v 4w
0
: u, v, w R
.
"u " v " w
Is this set of vectors a subspace of R 4 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 2 and a basis is
"1
0
2
0
,
"1
4
0
"1
62. 147]
Solution:
Consider the vectors of the form
2u v
w " 2v " 3u
: u, v, w R
.
2w " 6v " 8u
Is this set of vectors a subspace of R 3 ? If so, explain why, give a basis for the subspace and find
its dimension.
It is of dimension 3 and a basis is
2
"3
"8
1
,
"2
"6
0
,
1
2
63. 148]
Solution:
If you have 5 vectors in F 5 and the vectors are linearly independent, can it always be concluded
they span F 5 ? Explain.
Yes. If not, there would exist a vector not in the span. But then you could add in this vector and
obtain a linearly independent set of vectors with more vectors than a basis.
64. 149]
Solution:
If you have 6 vectors in F 5 , is it possible they are linearly independent? Explain.
No. If so, you would have a longer independent set than a spanning set.
65. 150]
Solution:
Suppose A is an m • n matrix and £w 1 , C, w k ¤ is a linearly independent set of vectors in
AŸF n – F m . Now suppose AŸz i w i . Show £z 1 , C, z k ¤ is also independent.
k
Say ! i1 c i z i 0. Then you can do A to it.
k
! c i Az i i1
k
! ci wi 0
i1
and so, by linear independence of the w i , it follows that each c i 0.
66. 151]
Solution:
Suppose V, W are subspaces of F n . Show V 9 W defined to be all vectors which are in both V and
W is a subspace also.
This is obvious. If x, y V 9 W, then for scalars ), *, the linear combination )x *y must be in
both V and W since they are both subspaces.
67. 152]
Solution:
Suppose V and W both have dimension equal to 7 and they are subspaces of F 10 . What are the
possibilities for the dimension of V 9 W? Hint: Remember that a linear independent set can be
extended to form a basis.
See the next problem.
68. 153]
Solution:
Suppose V has dimension p and W has dimension q and they are each contained in a subspace, U
which has dimension equal to n where n maxŸp, q . What are the possibilities for the dimension
of V 9 W? Hint: Remember that a linear independent set can be extended to form a basis.
Let £x 1 , C, x k ¤ be a basis for V 9 W. Then there is a basis for V and W which are respectively
£x 1 , C, x k , y k1 , C, y p ¤, £x 1 , C, x k , z k1 , C, z q ¤
It follows that you must have k p " k q " k t n and so you must have
pq"n t k
69. 154]
Solution:
If b p 0, can the solution set of Ax b be a plane through the origin? Explain.
No. It can’t. It does not contain 0.
70. 155]
Solution:
Suppose a system of equations has fewer equations than variables and you have found a solution
to this system of equations. Is it possible that your solution is the only one? Explain.
No. There must then be infinitely many solutions. If the system is Ax b, then there are infinitely
many solutions to Ax 0 and so the solutions to Ax b are a particular solution to Ax b added
to the solutions to Ax 0 of which there are infinitely many.
71. 156]
Solution:
Suppose a system of linear equations has a 2 • 4 augmented matrix and the last column is a pivot
column. Could the system of linear equations be consistent? Explain.
No. This would lead to 0 1.
72. 157]
73.
74.
75.
76.
Solution:
Suppose the coefficient matrix of a system of n equations with n variables has the property that
every column is a pivot column. Does it follow that the system of equations must have a solution?
If so, must the solution be unique? Explain.
Yes. It has a unique solution.
158]
Solution:
Suppose there is a unique solution to a system of linear equations. What must be true of the pivot
columns in the augmented matrix.
The last one must not be a pivot column and the ones to the left must each be pivot columns.
159]
Solution:
State whether each of the following sets of data are possible for the matrix equation Ax b. If
possible, describe the solution set. That is, tell whether there exists a unique solution no solution
or infinitely many solutions.
a. A is a 5 • 6 matrix, rankŸA 4 and rankŸA|b 4. Hint: This says b is in the span of four
of the columns. Thus the columns are not independent.
i. Infinite solution set.
b. A is a 3 • 4 matrix, rankŸA 3 and rankŸA|b 2.
i. This surely can’t happen. If you add in another column, the rank does not get smaller.
c. A is a 4 • 2 matrix, rankŸA 4 and rankŸA|b 4. Hint: This says b is in the span of the
columns and the columns must be independent.
i. You can’t have the rank equal 4 if you only have two columns.
d. A is a 5 • 5 matrix, rankŸA 4 and rankŸA|b 5. Hint: This says b is not in the span of
the columns.
i. In this case, there is no solution to the system of equations represented by the augmented
matrix.
e. A is a 4 • 2 matrix, rankŸA 2 and rankŸA|b 2.
In this case, there is a unique solution since the columns of A are independent.
160]
Solution:
Suppose A is an m • n matrix in which m t n. Suppose also that the rank of A equals m. Show
that A maps F n onto F m . Hint: The vectors e 1 , C, e m occur as columns in the row reduced
echelon form for A.
This says that the columns of A have a subset of m vectors which are linearly independent.
Therefore, this set of vectors is a basis for F m . It follows that the span of the columns is all of F m .
Thus A is onto.
161]
Solution:
Suppose A is an m • n matrix in which m u n. Suppose also that the rank of A equals n. Show
that A is one to one. Hint: If not, there exists a vector, x such that Ax 0, and this implies at least
one column of A is a linear combination of the others. Show this would require the column rank to
be less than n.
The columns are independent. Therefore, A is one to one.
77. 162]
Solution:
Explain why an n • n matrix, A is both one to one and onto if and only if its rank is n.
The rank is n is the same as saying the columns are independent which is the same as saying A is
one to one which is the same as saying the columns are a basis. Thus the span of the columns of A
is all of F n and so A is onto. If A is onto, then the columns must be linearly independent since
otherwise the span of these columns would have dimension less than n and so the dimension of F n
would be less than n.
78. 163]
Solution:
Suppose A is an m • n matrix and B is an n • p matrix. Show that
dimŸkerŸAB t dimŸkerŸA dimŸkerŸB .
Hint: Consider the subspace, BŸF p 9 kerŸA and suppose a basis for this subspace is
£w 1 , C, w k ¤. Now suppose £u 1 , C, u r ¤ is a basis for kerŸB . Let £z 1 , C, z k ¤ be such that
Bz i w i and argue that
kerŸAB – spanŸu 1 , C, u r , z 1 , C, z k .
Here is how you do this. Suppose ABx 0. Then Bx kerŸA 9 BŸF p and so Bx ! i1 Bz i
showing that
k
k
x " ! z i kerŸB .
i1
Consider BŸF 9 kerŸA and let a basis be £w 1 , C, w k ¤. Then each w i is of the form Bz i w i .
Therefore, £z 1 , C, z k ¤ is linearly independent and ABz i 0. Now let £u 1 , C, u r ¤ be a basis for
k
kerŸB . If ABx 0, then Bx kerŸA 9 BŸF p and so Bx ! i1 c i Bz i which implies
p
k
x " ! c i z i kerŸB i1
and so it is of the form
k
x " ! cizi i1
r
! d juj
j1
It follows that if ABx 0 so that x kerŸAB , then
x spanŸz 1 , C, z k , u 1 , C, u r .
Therefore,
dimŸkerŸAB t k r dimŸBŸF p 9 kerŸA dimŸkerŸB t dimŸkerŸA dimŸkerŸB 79. 164]
Solution:
Explain why Ax 0 always has a solution even when A "1 does not exist.
a. Just let x 0. Then this solves the equation.
b. What can you conclude about A if the solution is unique?
i. You can conclude that the columns of A are linearly independent and so A "1 exists.
c. What can you conclude about A if the solution is not unique?
i. You can conclude that the columns of A are dependent and so A "1 does not exist. Thus
A is not one to one and A is not onto.
80. 165]
Solution:
Suppose detŸA " 5I 0. Show there exists x p 0 such that ŸA " 5I x 0.
If detŸA " 5I 0 then ŸA " 5I "1 does not exist and so the columns are not independent which
means that for some x p 0, ŸA " 5I x 0.
81. 166]
Solution:
Let A be an n • n matrix and let x be a nonzero vector such that Ax 5x for some scalar 5.
When this occurs, the vector x is called an eigenvector and the scalar, 5 is called an eigenvalue.
It turns out that not every number is an eigenvalue. Only certain ones are. Why? Hint: Show that
if Ax 5x, then ŸA " 5I x 0. Explain why this shows that ŸA " 5I is not one to one and not
onto. Now argue detŸA " 5I 0. What sort of equation is this? How many solutions does it
have?
If A is n • n, then the equation is a polynomial equation of degree n.
82. 167]
Solution:
Let m n and let A be an m • n matrix. Show that A is not one to one. Hint: Consider the n • n
matrix, A 1 which is of the form
A
A1 q
0
where the 0 denotes an Ÿn " m • n matrix of zeros. Thus det A 1 0 and so A 1 is not one to one.
Now observe that A 1 x is the vector,
Ax
A1x 0
which equals zero if and only if Ax 0. Do this using the Fredholm alternative.
Since A 1 is not one to one, it follows there exists x p 0 such that A 1 x 0. Hence Ax 0
although x p 0 so it follows that A is not one to one. From another point of view, if A were one to
one, then kerŸA H R n and so by the Fredholm alternative, A T would be onto R n . However, A T
has only m columns so this cannot take place.
83. 168]
Solution:
Let A be an m • n real matrix and let b R m . Show there exists a solution, x to the system
A T Ax A T b
Next show that if x, x 1 are two solutions, then Ax Ax 1 . Hint: First show that ŸA T A T A T A.
Next show if x kerŸA T A , then Ax 0. Finally apply the Fredholm alternative. This will give
existence of a solution.
That ŸA T A T A T A follows from the properties of the transpose. Therefore,
ker ŸA T A T
H
ŸkerŸA T A H
Suppose A T Ax 0. Then ŸA T Ax, x ŸAx, Ax and so Ax 0. Therefore,
ŸA T b, x Ÿb, Ax Ÿb, 0 0
It follows that A T b ker ŸA T A T
H
and so there exists a solution x to the equation
A T Ax A T b
by the Fredholm alternative.
84. 169]
Solution:
Show that in the context of the above problem that if x is the solution there, then
|b " Ax| t |b " Ay| for every y. Thus Ax is the point of AŸR n which is closest to b of every point
in AŸR n .
|b "Ay|2 |b "Ax Ax "Ay|2
|b "Ax|2 |Ax " Ay|2 2Ÿb "Ax, AŸx " y |b "Ax|2 |Ax " Ay|2 2ŸA T b "A T Ax, Ÿx " y |b "Ax|2 |Ax " Ay|2
and so, Ax is closest to b out of all vectors Ay.
85. 170]
Solution:
2
Let A be an n • n matrix and consider the matrices I, A, A 2 , C, A n . Explain why there exist
scalars, c i not all zero such that
n2
! c i A i 0.
i1
Then argue there exists a polynomial, pŸ5 of the form
5 m d m"1 5 m"1 C d 1 5 d 0
such that pŸA 0 and if qŸ5 is another polynomial such that qŸA 0, then qŸ5 is of the form
pŸ5 lŸ5 for some polynomial, lŸ5 . This extra special polynomial, pŸ5 is called the minimal
2
polynomial. Hint: You might consider an n • n matrix as a vector in F n .
2
The dimension of F n is n 2 . Therefore, there exist scalars c k such that
n2
! ckAk 0
k0
Let pŸ5 be the monic polynomial having smallest degree such that pŸA 0. If qŸA 0 then
from the Euclidean algorithm,
qŸ5 pŸ5 lŸ5 rŸ5 where the degree of rŸ5 is less than the degree of pŸ5 or else rŸ5 equals 0. However, if it is not
zero, you could plug in A and obtain
0 qŸA 0 rŸA and this would contradict the definition of pŸ5 as being the polynomial having smallest degree
which sends A to 0. Hence qŸ5 pŸ5 lŸ5 .
Linear Transformations
1. 171]
Solution:
Show the map T : R n v R m defined by TŸx Ax where A is an m • n matrix and x is an m • 1
column vector is a linear transformation.
This is obvious from the properties of matrix multiplication.
2. 172]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
" 14 =.
1
2
" 12
2
2
1
2
1
2
2
2
3. 173]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
5=/12. Hint: Note that 5=/12 2=/3 " =/4.
1
4
1
4
6 "
2 1
4
1
4
2 " 14 2 "
6
1
4
6 "
1
4
1
4
6
2
4. 174]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
=.
"1
0
0
"1
5. 175]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
3
=.
2
0
1
"1 0
6. 176]
Solution:
Show the map T : R n v R m defined by TŸx Ax where A is an m • n matrix and x is an m • 1
column vector is a linear transformation.
This is obvious from the properties of matrix multiplication.
7. 177]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
" 16 =.
1
2
1
2
3
" 12
1
2
3
8. 178]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
4
=.
3
2
1
4
3 1
4
2
1
4
3 "
1
4
2 "
1
4
1
4
2
1
4
2 3
3 1
4
9. 179]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
1
= and then reflects through the y axis.
6
"1 0
0
1
2
3
1
2
1
" 12
1
2
3
" 12 3
1
2
1
2
1
2
3
10. 180]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
2
= and then reflects through the y axis.
3
1
0
0 "1
" 12
1
2
" 12 3
3
" 12
" 12
" 12 3
" 12 3
1
2
11. 181]
Solution:
Find the matrix for the linear transformation which reflects through the y axis and then rotates
every vector in R 2 through an angle of 12 =.
0 "1
"1 0
1
0
0
1
0
"1
"1
0
12. 182]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
2
= and then reflects through the y axis.
3
"1 0
0
1
" 12
1
2
3
" 12 3
" 12
1
2
1
2
1
2
3
3
" 12
13. 183]
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle of
" 12 = and then reflects through the x axis.
1
0
0
0 "1
1
0 1
"1 0
1 0
14. 184]
Solution:
Find the matrix for the linear transformation which reflects through the x axis and then rotates
every vector in R 2 through an angle of 34 =.
" 12 2 " 12 2
1
2
1
" 12 2
0 "1
" 12 2
2
0
1
2
1
2
1
2
2
2
2
15. 185]
Solution:
Find the matrix for the linear transformation which reflects through the x axis and then rotates
every vector in R 2 through an angle of 13 =.
" 12 3
1
2
1
2
1
0 "1
1
2
3
0
1
2
1
2
1
2
3
" 12
3
16. 186]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
" 16 = about the positive z axis when viewed from high on the positive z axis and then reflects
through the xy plane.
1 0
0
1
2
0 1
0
" 12
0 0 "1
0
1
2
3
1
2
3 0
0
1
2
0
1
1
2
3
" 12
0
1
2
0
3
0
0
"1
17. 187]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
1
= about the positive x axis when viewed from the positive x axis and then reflects through the yz
3
plane.
"1 0 0
1
0
0
0
1 0
0
1
2
" 12 3
0
0 1
0
1
2
3
1
2
"1
0
0
0
1
2
" 12 3
0
1
2
3
1
2
18. 188]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
1
= about the positive z axis when viewed from high on the positive z axis and then reflects
2
through the xy plane.
1 0
0
0 "1 0
0 1
0
1
0
0
0 0 "1
0
0
1
0 "1
0
1
0
0
0
0
"1
19. 189]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
3
= about the positive z axis when viewed from high on the positive z axis and then reflects
2
through the xy plane.
1 0
0
0
0 1
0
"1 0 0
0 0 "1
0
1 0
0
0 1
1
0
"1 0
0
0
0 "1
20. 190]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
1
= about the positive z axis when viewed from high on the positive z axis and then reflects
2
through the yz plane.
"1 0 0
0 "1 0
0
1 0
1
0
0
0
0 1
0
0
1
0 1 0
1 0 0
0 0 1
21. 191]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which reflects every vector in R 3 through the xy plane
and then rotates every vector in R 3 through an angle of 14 = about the positive y axis when viewed
from the positive y axis and then reflects through the xz plane.
1
0
0
1
2
0 "1 0
0
0
1
2
0
0
1
2
1
" 12 2 0
2
0
1
2
1 0
0
0 1
0
0 0 "1
2
0
" 12 2
0
"1
0
" 12 2
0
" 12 2
1
2
2
22. 192]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which rotates every vector in R 3 through an angle of
1
= about the positive y axis when viewed from the positive y axis and then reflects through the xz
2
plane.
1
0
0
0
0 1
0 "1 0
0
1 0
0
"1 0 0
0
1
0
0
1
0
"1 0
"1
0
0
23. 193]
Solution:
Recall the special vectors i, j, k such that i points in the direction of the positive x axis, j in the
direction of the positive y axis and k in the direction of the positive z axis. The vectors i, j, k form
a right handed system as shown in the following picture.
Find the matrix of the linear transformation which reflects every vector in R 3 through the xz plane
and then rotates every vector through an angle of 14 = about the positive z axis when viewed from
high on the positive z axis.
1
2
2 " 12 2 0
1
2
2
1
2
0
2
0
1
0
0
0
0 "1 0
1
0
0
1
2
2
1
2
2 " 12 2 0
1
1
2
0
2
0
0
1
24. 194]
Solution:
T
Find the matrix for proj u Ÿv where u Columns are
1
11
1
11
3
11
1
11
1
11
3
11
.
2 2 6
3
11
3
11
9
11
1
11
1
11
3
11
, concatenate:
1
11
1
11
3
11
3
11
3
11
9
11
25. 195]
Solution:
T
Find the matrix for proj u Ÿv where u Columns are
0
0
0
0
4
5
2
5
2
5
1
5
0
0 2 1
.
0 0
0
4
5
2
5
2
5
1
5
0
, concatenate:
0
26. 196]
Solution:
Find the matrix for proj u Ÿv where u Columns are
2
7
1
7
" 37
1
7
1
14
3
" 14
"2 "1 3
T
.
" 37
3
" 14
, concatenate:
9
14
2
7
1
7
" 37
" 37
1
7
1
14
3
" 14
3
" 14
9
14
27. 197]
Solution:
Find the matrix for proj u Ÿv where u Columns are
4
9
" 29
" 49
2 "1 "2
" 29
" 49
1
9
2
9
2
9
4
9
T
.
, concatenate:
4
9
" 29
" 49
" 29 " 49
1
9
2
9
2
9
4
9
28. 198]
Solution:
Let M be the matrix for proj u Ÿv , u R n , n 1. Explain why detŸM 0
The columns of M are all multiples of a single vector so they are linearly dependent.
29. 199]
Solution:
Show that the function T u defined by T u Ÿv q v " proj u Ÿv is also a linear transformation.
Ÿav bw u u
|u|2
Ÿv u Ÿw u av " a
u bw "b
u
2
|u|
|u|2
aT u Ÿv bT u Ÿw T u Ÿav bw av bw "
30. 200]
Solution:
Show that Ÿv " proj u Ÿv , u 0 and conclude every vector in R n can be written as the sum of
two vectors, one which is perpendicular and one which is parallel to the given vector.
Ÿv " proj u Ÿv , u Ÿv, u "
Ÿvu |u|2
Ÿv, u " Ÿv, u 0.
u, u
Therefore, v v " proj u Ÿv proj u Ÿv . The first is perpendicular to u and the second is a
multiple of u so it is parallel to u.
31. 201]
Solution:
Here are some descriptions of functions mapping R n to R n .
a. T multiplies the j th component of x by a nonzero number b.
b. T replaces the i th component of x with b times the j th component added to the i th component.
c. T switches two components.
Show these functions are linear and describe their matrices.
Each of these is an elementary matrix. The first is the elementary matrix which multiplies the j th
diagonal entry of the identity matrix by b. The second is the elementary matrix which takes b
times the j th row and adds to the i th row and the third is just the elementary matrix which switches
the i th and the j th rows where the two components are in the i th and j th positions.
32. 202]
Solution:
Let u Ÿa, b be a unit vector in R 2 . Find the matrix which reflects all vectors across this vector.
Hint: You might want to notice that Ÿa, b Ÿcos 2, sin 2 for some 2. First rotate through "2.
Next reflect through the x axis which is easy. Finally rotate through 2.
a.
cosŸ2 " sinŸ2 1
sinŸ2 0 "1
cosŸ2 0
cos 2 2 " sin 2 2
2 cos 2 sin 2
2 cos 2 sin 2
sin 2 2 " cos 2 2
cosŸ"2 " sinŸ"2 sinŸ"2 cosŸ"2 Now to write in terms of Ÿa, b , note that a/ a 2 b 2 cos 2, b/ a 2 b 2 sin 2. Now plug this
in to the above. The result is
a 2 "b 2
a 2 b 2
2
ab
a 2 b 2
2 a 2ab
b 2
b 2 "a 2
a 2 b 2
1
a2 b2
Since this is a unit vector, a 2 b 2 1 and so you get
a2 " b2
2ab
2ab
b " a2
2
a2 " b2
2ab
2ab
b " a2
2
33. 203]
Solution:
Let u be a unit vector. Show the linear transformation of the matrix I " 2uu T preserves all
distances and satisfies
T
ŸI " 2uu T ŸI " 2uu T I.
This matrix is called a Householder reflection. More generally, any matrix Q which satisfies
Q T Q QQ T is called an orthogonal matrix. Show the linear transformation determined by an
orthogonal matrix always preserves the length of a vector in R n . Hint: First show that for any
matrix A,
˜Ax, y™ ˜x, A T y™
T
ŸI " 2uu T ŸI " 2uu T ŸI " 2uu T ŸI " 2uu T 1
¥
I " 2uu T " 2uu T 4u u T u u T I
Now, why does this matrix preserve distance? For short, call it Q and note that Q T Q I. Then
|x|2 ŸQ T Qx, x ŸQx, Qx |Qx|2
and so Q preserves distances.
34. 204]
Solution:
Suppose |x| |y| for x, y R n . The problem is to find an orthogonal transformation Q
ŸQ T Q I which has the property that Qx y and Qy x. Show
x"y
Q q I"2
" y T
2 Ÿx
|x " y|
does what is desired.
From the above problem this preserves distances and Q T Q. Now do it to x.
x"y
QŸx " y x " y " 2
Ÿx " y, x " y y " x
|x " y|2
x"y
QŸx y x y " 2
" y T Ÿx y 2 Ÿx
|x " y|
x"y
2
" |y|2 x y
xy"2
2 Ÿ|x|
|x " y|
and so
Qx " Qy y " x
Qx Qy x y
Hence, adding these yields Qx y and then subtracting them gives Qy x.
35. 205]
Solution:
Let a be a fixed vector. The function T a defined by T a v a v has the effect of translating all
vectors by adding a. Show this is not a linear transformation. Explain why it is not possible to
realize T a in R 3 by multiplying by a 3 • 3 matrix.
Linear transformations take 0 to 0. Also T a Ÿu v p T a u T a v.
36. 206]
Solution:
In spite of the above problem, we can represent both translations and rotations by matrix
multiplication at the expense of using higher dimensions. This is done by the homogeneous
coordinates. I will illustrate in R 3 where most interest in this is found. For each vector
v Ÿv 1 , v 2 , v 3 T , consider the vector in R 4 Ÿv 1 , v 2 , v 3 , 1 T . What happens when you do
1 0 0 a1
v1
0 1 0 a2
v2
0 0 1 a3
v3
0 0 0 1
1
?
Describe how to consider both rotations and translations all at once by forming appropriate 4 • 4
matrices.
That product above is of the form
a1 v1
a2 v2
a3 v3
1
If you just discard the one at the bottom, you have found a v. Then to do both rotations and
translations, you would look at matrices of the form
R 0
0 1
where R is a rotation matrix and for translation by a, you use
I a
0 1
To obtain a rotation followed by a translation by a, you would just multiply these two matrices. to
get
If you did this to the vector
x
1
I a
R 0
0 1
0 1
, you would get
R a
0 1
Rx a
1
. Now discard the 1 and you
have what you want.
37. 207]
Solution:
You want to add
3 "1 "2
to every point in R 3 and then rotate about the z axis counter
clockwise through an angle of
1
4
2 " 12 2 0 0
1
2
1
2
1
2
2
2
=. Find what happens to the point
1 0 0
3
1
0 0
0 1 0 "1
3
0
0
1 0
0 0 1 "2
4
0
0
0 1
0 0 0
1
1
1 3 4
.
2
3 2
and so the answer is
2
1
2
.
3 2
2
38. 208]
Solution:
You want to add
clockwise through an angle of
" 12
to every point in R 3 and then rotate about the z axis counter
"1 "1 "3
2
3
" 12 3 0 0
=. Find what happens to the point
1 0 0 "1
3
3 1 2
.
"1
" 12
0 0
0 1 0 "1
1
0
0
1 0
0 0 1 "3
2
0
0
0 1
0 0 0
1
1 1 "2
to every point in R 3 and then rotate about the z axis counter
1
2
3
1
3
"1
and so the answer is
1
"1
.
3
"1
39. 209]
Solution:
You want to add
clockwise through an angle of
1
2
=. Find what happens to the point
0 "1 0 0
1 0 0
1
3
"3
1
0
0 0
0 1 0
1
2
4
0
0
1 0
0 0 1 "2
3
0
0
0 1
0 0 0
1
1
1
1
3 2 3
.
"3
and so the answer is
4
1
40. 210]
Solution:
You want to rotate about the z axis counter clockwise through an angle of 13 = and then add
3 3 "2 to every point in R 3 . Find what happens to the point 1 2 1 .
.
1 0 0
3
0 1 0
3
" 12 3 0 0
1
2
1
2
3
1
2
0 0
2
0 0 1 "2
0
0
1 0
1
0 0 0
0
0
0 1
1
1
" 3
7
2
1
2
1
3 4
"1
.
1
41. 211]
Solution:
You want to rotate about the z axis counter clockwise through an angle of 54 = and then add
.
"3 0 "1 to every point in R 3 . Find what happens to the point
2 1 4
1 0 0 "3
" 12 2
0 1 0
" 12 2
0
1
2
" 12
2
0 0
2
2 0 0
1
0 0 1 "1
0
0
1 0
4
0 0 0
0
0
0 1
1
1
" 12 2 " 3
" 32 2
3
1
42. 212]
Solution:
Give the general solution to the following system of equations
wx"z 0
x"wy 0
x"wy 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
wx"z3 0
x"wy2 0
x"wy2 0
The solution to the first system of equations is
x
y
z
"1
1
"1
s
1
w
2
t
, s, t R
0
0
1
The general solution to the second system is
x
y
z
w
"1
1
s
"1
1
0
t
2
0
1
"3
1
0
0
.
43. 213]
Solution:
Give the general solution to the following system of equations
x " 3w 3z 0
x " 4w y 7z 0
x " 4w y 7z 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
x " 3w 3z 2 0
x " 4w y 7z 1 0
x " 4w y 7z 1 0
The solution to the first system of equations is
"3
x
y
z
"4
s
1
w
3
1
t
, s, t R
0
0
1
The general solution to the second system is
"3
x
y
z
w
s
"4
1
0
"2
3
t
1
0
1
1
0
0
44. 214]
Solution:
Give the general solution to the following system of equations
x " w " 2z 0
x " 2w y 2z 0
x " 2w y 2z 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
x " w " 2z 1 0
x " 2w y 2z " 2 0
x " 2w y 2z " 2 0
The solution to the first system of equations is
x
y
z
2
"4
s
1
w
1
1
t
, s, t R
0
0
1
The general solution to the second system is
x
2
y
"4
s
z
1
w
"1
1
t
0
1
3
0
0
1
0
45. 215]
Solution:
Give the general solution to the following system of equations
3w x " z 0
8w x y 3z 0
8w x y 3z 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
3w x " z " 2 0
8w x y 3z " 4 0
8w x y 3z " 4 0
The solution to the first system of equations is
x
y
z
"3
1
s
"4
1
w
t
"5
0
0
, s, t R
1
The general solution to the second system is
x
y
z
w
46. 216]
Solution:
"3
1
s
"4
1
0
t
"5
0
1
2
2
0
0
Give the general solution to the following system of equations
3w x " 2z 0
8w x y " 4z 0
8w x y " 4z 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
3w x " 2z " 1 0
8w x y " 4z " 2 0
8w x y " 4z " 2 0
The solution to the first system of equations is
x
y
z
"3
2
2
s
1
w
t
"5
0
0
, s, t R
1
The general solution to the second system is
x
y
z
w
"3
2
s
2
1
0
t
"5
0
1
1
1
0
0
47. 217]
Solution:
Give the general solution to the following system of equations
3w x " z 0
5w x y 0
5w x y 0
by considering it as finding the solution to a system of the form
Ax 0
Now, using what you just did, find the general solution to the system
3w x " z " 1 0
5w x y " 3 0
5w x y " 3 0
The solution to the first system of equations is
x
y
z
"3
1
"1
s
1
w
"2
t
0
0
, s, t R
1
The general solution to the second system is
x
y
z
"3
1
s
w
"1
1
1
"2
t
0
0
1
2
0
0
48. 218]
Solution:
Find kerŸA NŸA where
2 1 3 "3
6
3 2 7 "4 10
A
2 1 3 "3
6
3 1 2 "5
8
x1
1
2
"2
x2
"5
"1
"2
x3
s
1
t
r
0
0
x4
0
1
0
x5
0
0
1
49. 219]
Solution:
Find kerŸA NŸA where
2 1 11 "2
A
6
3 2 19 "2 10
2 1 11 "2
6
3 1 14 "4
8
x1
"3
2
"2
x2
"5
"2
"2
x3
s
t
1
r
0
0
x4
0
1
0
x5
0
0
1
50. 220]
Solution:
Find kerŸA NŸA where
2 1 5
A
9
0
3 2 7 15 1
2 1 5
9
1
3 1 8 12 1
x1
"3
"3
x2
1
"3
x3
s
t
1
0
x4
1
1
x5
0
0
51. 221]
Solution:
Find kerŸA NŸA where
A
2 1
0
10 0
3 2
1
17 1
2 1
0
10 1
3 1 "1 13 1
x1
1
"3
x2
"2
"4
x3
52. 222]
Solution:
Find kerŸA NŸA where
s
1
t
0
x4
1
1
x5
0
0
2 1 "1 "3 0
3 2
A
1
"3 1
2 1 "1 "3 1
3 1 "4 "6 1
x1
3
3
x2
"5
"3
x3
s
1
t
0
x4
1
1
x5
0
0
53. 223]
Solution:
Find kerŸA NŸA where
"5
7
3 2 15 "7
9
7
2 1
A
8
2 1
8
"5
3 1
9
"8 12
x1
"1
3
"5
x2
"6
"1
3
x3
s
1
t
r
0
0
x4
0
1
0
x5
0
0
1
54. 224]
Solution:
Find kerŸA NŸA where
2 1 10 "2
A
0
3 2 17 "1 "1
2 1 10 "2
0
3 1 13 "5
1
x1
"3
3
"1
x2
"4
"4
2
x3
s
1
t
0
r
0
x4
0
1
0
x5
0
0
1
55. 225]
Solution:
Suppose Ax b has a solution. Explain why the solution is unique precisely when Ax 0 has
only the trivial (zero) solution.
If not, then there would be a infintely many solutions to Ax 0 and each of these added to a
solution to Ax b would be a solution to Ax b.
56. 226]
Solution:
Show that if A is an m • n matrix, then kerŸA is a subspace.
If x, y kerŸA then
AŸax by aAx bAy a0 b0 0
and so kerŸA is closed under linear combinations. Hence it is a subspace.
57. 227]
Solution:
Give an example of a 2 • 3 matrix which maps R 3 onto R 2 . Is it possible that there could exist an
example of such a matrix which is one to one?
There is no one to one such matrix because there must be a free variable. Also, to be one to one,
the columns would have to be an independent set and this is not possible.
Factorizations
1. 228]
Solution:
Find an LU factorization for the matrix
5
1
2
2
5
3
7
3
"5
3
9
1
10 10 29 12
and use to compute the determinant of the matrix.
"10
2. 229]
Solution:
Find an LU factorization for the matrix
1
1
2
1
1
"3 "1 "1 "2 "1
"3 "1 "4 "1 "4
2
1
0 0 0
1 1
2
1
1
"3 1 0 0
0 2
5
1
2
"3 1 1 0
0 0 "3
1
"3
2
0 0
"3
1
4 5 1
0
10
9
8
"4
3. 230]
Solution:
Find an LU factorization for the matrix
1
2
"3
"5 "2
5
4
5
20
7
"5 "5
1
0
0 0
5
1
2
"3
"1
1
0 0
0 "1
7
1
4
"3 1 0
0
0
"1
1
"1
4
0
0
0
4
7 1
4. 231]
Solution:
Find an LU factorization for the matrix
"14 "14
19
18
5
1
2
1
1
5
4
7
2
3
"15 0
15
1
0 0 0
5 1 2
1
1
1
1 0 0
0 3 5
1
2
"3 1 1 0
0 0 1
1
"3
3
0 0 0 "3
1
2 5 1
0 "1 "4
9 21
"7
7
5. 232]
Solution:
Find an LU factorization for the matrix
4
1
2
3
"8 "4
2
"5
"4 "7
14
1
"8 "12 14
4
and use to compute the determinant of the matrix.
"16
6. 233]
Solution:
Find an LU factorization for the matrix
3
1
2
"3
6
2
9
"5
18 6 "1 "20
0
1
0
0 0
3 1 2 "3
2
1
0 0
0 0 5
1
6 "3 1 0
0 0 2
1
0
0 0 0
6
4
5 1
7. 234]
Solution:
Find an LU factorization for the matrix
0 30
15
4
1
2
"3
"4
"4
2
4
4
10
"11 "5
"12 "21
1
0
0 0
4
1
2
"3
"1
1
0 0
0 "3
4
1
1
"3 1 0
0
0
"1
1
"3
6
0
0
0
1
4 1
14
20
8. 235]
Solution:
Find an LU factorization for the matrix
4
1
2
3
"4
"1
5
"2
"12 "3 14 "5
0
1
0 0 0
4 1
2
3
"1 1 0 0
0 0
7
1
"3 3 1 0
0 0 "1
1
0
0 0
"3
7 7 1
0
0
42 11
9. 236]
Solution:
Is there only one LU factorization for a given matrix? Hint: Consider the equation
0 1
0 1
1 0
0 1
1 1
0 0
.
Sometimes there is more than one LU factorization as is the case in this example. The above
equation clearly gives an LU factorization. However, it appears that
0 1
0 1
1 0
0 1
0 1
0 1
also. Therefore, this is another.
10. 237]
Solution:
Computer algebra systems are very good at finding PLU factorizations. Describe how to use a
PLU factorization to compute the determinant of a matrix.
You let m be the number of switches used to form P and then the determinant is just Ÿ"1 m times
the product of the diagonal entries of U. This is because if A PLU, then
detŸA detŸP detŸL detŸU Ÿ"1 m • 1 •above product of diagonal entries.
11. 238]
Solution:
1 2 1
Find a PLU factorization of
.
1 2 2
2 1 1
1 2 1
1 2 2
2 1 1
1 0 0
1 0 0
1
2
1
0 0 1
2 1 0
0 "3 "1
0 1 0
1 0 1
0
0
1
12. 239]
Solution:
1 2 1
1 2 2
Find a PLU factorization of
.
2 4 1
3 2 1
1 2 1
1 2 2
2 4 1
3 2 1
1 0 0 0
1 0
0
0
1
2
1
0 0 0 1
3 1
0
0
0 "4 "2
0 0 1 0
2 0
1
0
0
0
"1
0 1 0 0
1 0 "1 1
0
0
0
Here are steps for doing this. First use the top row to zero out the entries in the first column
which are below the top row. This yields
1
2
1
0
0
1
0
0
"1
0 "4 "2
Obviously there will be no way to obtain an LU factorization because a switch of rows must be
done. Switch the second and last row in the original matrix. This will yield one which does have
an LU factorization.
1 0 0 0
1 2 1
0 0 0 1
1 2 2
0 0 1 0
2 4 1
0 1 0 0
3 2 1
1 2 1
3 2 1
2 4 1
1 2 2
1 0
0
0
1
2
1
3 1
0
0
0 "4 "2
2 0
1
0
0
0
"1
1 0 "1 1
0
0
0
Then the original matrix is
1 2 1
1 2 2
2 4 1
3 2 1
1 0 0 0
1 0
0
0
1
2
0 0 0 1
3 1
0
0
0 "4 "2
0 0 1 0
2 0
1
0
0
0
"1
0 1 0 0
1 0 "1 1
0
0
0
13. 240]
Solution:
1 2 1
2 4 1
Find a PLU factorization of
and use it to solve the systems
1 0 2
2 2 1
1 2 1
a.
2 4 1
1 0 2
2 2 1
1 2 1
2 4 1
1 0 2
2 2 1
z
1 0 0 0
1
2
1
0 0 1 0
1 1 0 0
0 "2
1
0 1 0 0
2 0 1 0
0
0
"1
0 0 0 1
2 1 2 1
0
0
0
1
x
y
1 0 0 0
2
1
1
a. First solve
1 0 0 0
1 0 0 0
t
0 0 1 0
1 1 0 0
u
0 1 0 0
2 0 1 0
v
0 0 0 1
2 1 2 1
w
1
t
2 0 1 0
u
1 1 0 0
v
2 1 2 1
w
1
2
1
1
Thus t 1, v 0, u 0, w "1. Next solve
1
2
1
0 "2
1
0
0
"1
0
0
0
1
x
y
z
1
1
This is not too hard because it is the same as solving
1 0 0 0
2
0
0
"1
1
which clearly has no solution. Thus the original problem has no solution either. Note that the
augmented matrix has the following row reduced echelon form.
1 2 1 1
2 4 1 2
1 0 2 1
1 0 0 0
0 1 0 0
, row echelon form:
0 0 1 0
2 2 1 1
0 0 0 1
i. Thus there is no solution.
1 2 1
b.
2 4 1
1 0 2
2 2 1
a
x
b
y
c
z
d
There might not be any solution from part a. Thus suppose there is. First solve
t
c.
u
v
w
1 0 0 0
t
2 0 1 0
u
1 1 0 0
v
2 1 2 1
w
a
b
c
d
a
c"a
. Next solve
b " 2a
3a " 2b " c d
1
2
1
0 "2
1
0
0
"1
0
0
0
a
x
c"a
y
b " 2a
z
3a " 2b " c d
Note that there will be no solution unless 3a " 2b " c d 0, but if this condition holds,
then the solution to the problem is
2b " 4a c
x
y
z
To check this, note that
3
2
a"
1
2
b"
2a " b
1
2
c
1 2 1
a
2b " 4a c
2 4 1
3
2
1 0 2
a"
1
2
b"
1
2
b
c
c
2a " b
2 2 1
2b " 3a c
where the bottom entry on the right equals d if there is any solution.
14. 241]
Solution:
Find a QR factorization for the matrix
1 1 1
0
0 1 0
1
1 0 1 "1
1 1 1
0
0 1 0
1
1
2
2
0
1 0 1 "1
1
2
2
1
6
1
3
" 16
2 3
2 3
2 3
1
3
" 13
" 13
3
2
3
0
3
0
2
1
1
2
1
2
2
2
2 3
0
0
0
" 12 2
1
2
2 3
0
15. 242]
Solution:
Find a QR factorization for the matrix
1
3 "2 1
1
1
2
3 "2 1
1
0
1
11
3
11
1
11
1
2
11
11
11
13
66
5
" 66
1
33
2 11
0
2
" 16 2
2 11 " 16 2
2 11
2
3
2
11
4
" 11
11
0
6
11
0
2 11
0
16. 243]
Solution:
Find a QR factorization for the matrix
1 2 1
0 1 1
1 0 2
1
2
1 2 1
0 1 1
1 0 2
17. 244]
2
0
1
2
2
1
3
1
3
" 13
3
" 16 6
3
1
3
3
1
6
2
2
2 3
0
3
6
0
0
3
2
2
0
1
2
2 3
6
11
2
11
11
2 11
2
Solution:
Find a QR factorization for the matrix
1 1 1
0 1 0
1 0 2
1
2
1 1 1
0 1 0
1 0 2
1
3
0
1
2
1
6
2
2
6
2 3
" 16 6
" 13 3
1
3
1
3
1
2
2
3
0
3
0
1
2
2
2 3
0
3
2
" 16
1
3
2
6
3
18. 245]
Solution:
If you had a QR factorization, A QR, describe how you could use it to solve the equation
Ax b. This is not usually the way people solve this equation. However, the QR factorization is
of great importance in certain other problems, especially in finding eigenvalues and eigenvectors.
You would have QRx b and so then you would have Rx Q T b. Now R is upper triangular
and so the solution of this problem is fairly simple.
19. 246]
Solution:
Starting with an independent set of n vectors in R n £v 1 , C, v n ¤, form the matrix and QR
factorization
v1 C vn
q1 C qn
R
where Q q
q 1 C q n is the orthogonal matrix. Thus Q T Q I and the columns of Q form
an orthonormal set of vectors. Show that for each k t n,
span q 1 C q k
span v 1 C v k .
From the equation, v 1 r 11 q 1 , v 2 r 12 q 1 r 22 q 2 and so forth. Also R "1 must exist because the
determinant of the left side is nonzero, and from the usual process for finding it, this matrix is also
upper triangular. Thus
v1 C vn
R "1 q1 C qn
and now a similar thing can be said. Thus in any linear combination of the v i one can replace v k
with a suitable linear combination of the q j and in any linear combination of the q i one can replace
q k with a suitable linear combination of the v j . Hence the desired conclusion is obtained.
Linear Programming
1. 247]
Solution:
Maximize and minimize z x 1 " 2x 2 x 3 subject to the constraints
x 1 x 2 x 3 t 10, x 1 x 2 x 3 u 2, and x 1 2x 2 x 3 t 7 if possible. All variables are
nonnegative.
The constraints lead to the augmented matrix
1 1 1 1
0
0 10
1 1 1 0 "1 0
2
1 2 1 0
7
0
1
The obvious solution is not feasible. Do a row operation.
1 1 1
1
0 0 10
0 0 0
1
1 0
8
0 1 0 "1 0 1 "3
An obvious solution is still not feasible. Do another operation couple of row operations.
1 1 1 0 "1 0 2
0 0 0 1
1
0 8
0 1 0 0
1
1 5
At this point, you can spot an obvious feasible solution. Now assemble the simplex tableau.
1
1
1
0 "1 0 0 2
0
0
0
1
1
0 0 8
0
1
0
0
1
1 0 5
"1 2 "1 0
0
0 1 0
Now preserve the simple columns.
1 1 1 0 "1 0 0 2
0 0 0 1
1
0 0 8
0 1 0 0
1
1 0 5
0 2 0 0 "1 0 1 0
First lets work on minimizing this. There is a 2. The ratios are then 5, 2 so the pivot is the 1 on
the top of the second column. The next tableau is
1
1
1
0 "1 0 0
2
0
0
0
1
1
0 0
8
"1 0 "1 0
2
1 0
3
"2 0 "2 0
1
0 1 "4
There is a 1 on the bottom. The ratios of interest for that column are 3/2, 8, and so the pivot is the
2 in that column. Then the next tableau is
0
1 0
1
2
" 12
0
7
2
13
2
"1 0 "1 0 2
1
0
3
1
2
1
2
"
1
2
1
2
1
0
0 "
3
2
0 0
3
2
0 0 "
1 " 11
2
1
2
Now you stop because there are no more positive numbers to the left of 1 on the bottom row. The
minimum is "11/2 and it occurs when x 1 x 3 x 6 0 and x 2 7/2, x 4 13/2, x 6 "11/2.
Next consider maximization. The simplex tableau was
1
1
1
0 "1 0 0 2
0
0
0
1
1
0 0 8
0
1
0
0
1
1 0 5
"1 2 "1 0
0
0 1 0
This time you work on getting rid of the negative entries. Consider the "1 in the first column.
There is only one ratio to consider so 1 is the pivot.
1 1 1 0 "1 0 0 2
0 0 0 1
1
0 0 8
0 1 0 0
1
1 0 5
0 3 0 0 "1 0 1 2
There remains a "1. The ratios are 5 and 8 so the next pivot is the 1 in the third row and column 5.
1
2
1 0 0
1
0 7
0 "1 0 1 0 "1 0 3
0
1
0 0 1
1
0 5
0
4
0 0 0
1
1 7
Then no more negatives remain so the maximum is 7 and it occurs when
x 1 7, x 2 0, x 3 0, x 4 3, x 5 5, x 6 0.
2. 248]
Solution:
Maximize and minimize the following if possible. All variables are nonnegative.
a. z x 1 " 2x 2 subject to the constraints x 1 x 2 x 3 t 10, x 1 x 2 x 3 u 1, and
x 1 2x 2 x 3 t 7
i. an augmented matrix for the constraints is
1 1 1 1
0
0 10
1 1 1 0 "1 0 1
1 2 1 0
0
1
7
The obvious solution is not feasible. Do some row operations.
0
0
0
1
1
1
1
1
0 "1 0 1
"1 0 "1 0
0 9
2
1 5
Now the obvious solution is feasible. Then include the objective function.
0
0
0
1
1
0 0 9
1
1
1
0 "1 0 0 1
"1 0 "1 0
2
1 0 5
"1 2
0
0
0
0 1 0
1
First preserve the simple columns.
0
0
0
1
0 0
9
1
1
1
0 "1 0 0
1
"1 0 "1 0
2
1 0
"3 0 "2 0
2
0 1 "2
5
Lets try to maximize first. Begin with the first column. The only pivot is the 1. Use it.
0 0 0 1
1
0 0 9
1 1 1 0 "1 0 0 1
0 1 0 0
1
1 0 6
0 3 1 0 "1 0 1 1
There is still a -1 on the bottom row to the left of the 1. The ratios are 9 and 6 so the new
pivot is the 1 on the third row.
0 "1 0 1 0 "1 0 3
1
2
1 0 0
1
0 7
0
1
0 0 1
1
0 6
0
4
1 0 0
1
1 7
Then the maximum is 7 when x 1 7 and x 1 , x 3 0.
Next consider the minimum.
0
0
0
1
1
0 0
9
1
1
1
0 "1 0 0
1
"1 0 "1 0
2
1 0
"3 0 "2 0
2
0 1 "2
5
There is a positive 2 in the bottom row left of 1. The pivot in that column is the 2.
1 0 " 12 0
0 0
1
2
0
13
2
7
2
"1 0 "1 0 2
1
0
5
1
2
1
2
1
2
1
2
0
1
"2 0 "1 0 0 "1 1 "7
The minimum is "7 and it happens when x 1 0, x 2 7/2, x 3 0.
b. z x 1 " 2x 2 " 3x 3 subject to the constraints x 1 x 2 x 3 t 8, x 1 x 2 3x 3 u 1, and
x1 x2 x3 t 7
i. This time, lets use artificial variables to find an initial simplex tableau. Thus you add in
an artificial variable and then do a minimization procedure.
1 1 1 1
0
0
0
0 8
1 1 3 0 "1 0
1
0 1
1 1 1 0
0
1
0
0 7
0 0 0 0
0
0 "1 1 0
First preserve the seventh column as a simple column by a row operation.
1 1 1 1
0
0 0 0 8
1 1 3 0 "1 0 1 0 1
1 1 1 0
0
1 0 0 7
1 1 3 0 "1 0 0 1 1
Now use the third column.
0 " 13 0
2
3
2
3
1
1 3 0 "1 0
2
3
2
3
0
1
3
0 1
1
23
3
0
1
0
20
3
0 "1 1
0
0 0
1
3
1 "
0 0 0
0
1
3
It follows that a basic solution is feasible if
2
3
2
3
1
3
0
23
3
1
1 3 0 "1 0
1
2
3
2
3
20
3
0 1
1
3
0 0
1
Now assemble the simplex tableau
2
3
2
3
0 0
23
3
1
1 3 0 "1 0 0
1
2
3
2
3
0 1
1
3
0 0
1
3
1 0
20
3
"1 2 3 0
0
0 1
0
Preserve the simple columns by doing row operations.
1
3
2
3
2
3
1
1 3 0 "1 0 0
2
3
2
3
0 1
0 0
23
3
1
0 0
1
3
20
3
1 0
"2 1 0 0
1
0 1 "1
Lets do minimization first. Work with the second column.
0
0 "2 1
1
1
0
0 "2 0
1
1 0
"3 0 "3 0
2
0 1 "2
3
1
0 0
7
0 "1 0 0
1
6
Recall how you have to pick the pivot correctly. There is still a positive number in the
bottom row left of the 1. Work with that column.
0
0
0
1 0 "1 0
1
1
1
1
0 0
1
0
7
0
0 "2 0 1
1
0
6
"3 0
1
0 0 "2 1 "14
There is still a positive number to the left of 1 on the bottom row.
0
0
0 1 0 "1 0
1
1
1
1 0 0
1
0
7
2
2
0 0 1
3
0
20
"4 "1 0 0 0 "3 1 "21
It follows that the minimum is "21 and it occurs when x 1 x 2 0, x 3 7.
Next consider the maximum. The simplex tableau was
2
3
2
3
0 0
23
3
1
1 3 0 "1 0 0
1
2
3
2
3
20
3
0 1
1
3
0 0
1
3
1 0
"2 1 0 0
1
0 1 "1
0 0 "2 1
1
0 0 7
Use the first column.
1 1
3
0 "1 0 0 1
0 0 "2 0
0 3
6
1
1 0 6
0 "1 0 1 1
There is still a negative on the bottom row to the left of 1.
0 0
0
1 0 "1 0 1
1 1
1
0 0
1
0 7
0 0 "2 0 1
1
0 6
0 3
1
1 7
4
0 0
There are no more negatives on the bottom row left of 1 so stop. The maximum is 7 and
it occurs when x 1 7, x 2 0, x 3 0.
c. z 2x 1 x 2 subject to the constraints x 1 " x 2 x 3 t 10, x 1 x 2 x 3 u 1, and
x 1 2x 2 x 3 t 7.
i. The augmented matrix for the constraints is
1 "1 1 1
0
0 10
1
1
1 0 "1 0
1
1
2
1 0
7
0
1
The basic solution is not feasible because of that "1. Lets do a row operation to change
this. I used the 1 in the second column as a pivot and zeroed out what was above and
below it. Now it seems that the basic solution is feasible.
2
0
2
1 "1 0 11
1
1
1
0 "1 0
"1 0 "1 0
2
1
1
5
Assemble the simplex tableau.
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1
0
"1 0
2
1 0
5
0
0
0 1
0
"2 "1
0
1
Then do a row operation to preserve the simple columns.
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1 0 "1 0
"1 0
1
2
1
1 0
5
0 "1 0 1
1
Lets do minimization first. Work with the third column because there is a positive entry
on the bottom.
0
"2 0 1
1
1
1 0 "1 0 0 1
0
1
0 0
1
1 0 6
"2 "1 0 0
0
0 1 0
1
0 0 9
It follows that the minimum is 0 and it occurs when x 1 x 2 0, x 3 1.
Now lets maximize.
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1 0 "1 0
"1 0
1
2
1
1 0
5
0 "1 0 1
1
Lets begin with the first column.
0 "2 0 1
1
0 0 9
1
1
1 0 "1 0 0 1
0
1
0 0
0
1
2 0 "2 0 1 2
1
1 0 6
There is still a "2 to the left of 1 in the bottom row.
0 "3 0 1 0 "1 0 3
1
2
1 0 0
1
0
7
0
1
0 0 1
1
0
6
0
3
2 0 0
2
1 14
There are no negatives left so the maximum is 14 and it happens when
x 1 7, x 2 x 3 0.
d. z x 1 2x 2 subject to the constraints x 1 " x 2 x 3 t 10, x 1 x 2 x 3 u 1, and
x 1 2x 2 x 3 t 7.
i. The augmented matrix for the constraints is
1 "1 1 1
0
0 10
1
1
1 0 "1 0
1
1
2
1 0
7
0
1
Of course the obvious or basic solution is not feasible. Do a row operation involving a
pivot in the second row to try and fix this.
2
0
2
1 "1 0 11
1
1
1
0 "1 0
"1 0 "1 0
2
1
1
5
Now all is well. Begin to assemble the simplex tableau.
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1
0
"1 0
2
1 0
5
0
0
0 1
0
"1 "2
0
Do a row operation to preserve the simple columns.
1
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1 0 "1 0
1
0
2
2
1
1 0
5
0 "2 0 1
2
Next lets maximize. There is only one negative number in the bottom left of 1.
0
27
2
7
2
"1 0 "1 0 2 1 0
5
0
7
3
2
1
2
3
2
1
2
0
1
0
1
2
1
2
1 0
0 0
1
0
0 0 1 1
Thus the maximum is 7 and it happens when x 2 7/2, x 3 x 1 0.
Next lets find the minimum.
2
0
2
1 "1 0 0 11
1
1
1
0 "1 0 0
"1 0 "1 0
1
0
2
2
1
1 0
5
0 "2 0 1
2
Start with the column which has a 2.
0
"2 0 1
1
1
1 0 "1 0 0 1
0
1
0 0
1
1 0 6
"1 "2 0 0
0
0 1 0
1
0 0 9
There are no more positive numbers so the minimum is 0 when x 1 x 2 0, x 3 1.
3. 249]
Solution:
Consider contradictory constraints, x 1 x 2 u 12 and x 1 2x 2 t 5. You know these two
contradict but show they contradict using the simplex algorithm.
You can do this by using artificial variables, x 5 . Thus
1 1 "1 0
1
0 12
1 2
0
1
0
0 5
0 0
0
0 "1 1
0
Do a row operation to preserve the simple columns.
1 1 "1 0 1 0 12
1 2
0
1 0 0
5
1 1 "1 0 0 1 12
Next start with the 1 in the first column.
0 "1 "1 "1 1 0 7
1
2
0
1
0 0 5
0 "1 "1 "1 0 1 7
Thus the minimum value of z x 5 is 7 but, for there to be a feasible solution, you would need to
have this minimum value be 0.
4. 250]
Solution:
Find a solution to the following inequalities for x, y u 0 if it is possible to do so. If it is not
possible, prove it is not possible.
a.
6x 3y u 4
8x 4y t 5
i. Use an artificial variable. Let x 1 x, x 2 y and slack variables x 3 , x 4 with artificial
variable x 5 . Then minimize x 5 as described earlier.
6 3 "1 0
1
0 4
8 4
0
1
0
0 5
0 0
0
0 "1 1 0
Keep the simple columns.
6 3 "1 0 1 0 4
8 4
0
1 0 0 5
6 3 "1 0 0 1 4
Now proceed to minimize.
0 0 "1 " 34 1 0
8 4
0
1
1
4
0 0 5
0 0 "1 " 34 0 1
1
4
It appears that the minimum value of x 5 is 1/4 and so there is no solution to these
inequalities with x 1 , x 2 u 0.
6x 1 4x 3 t 11
b.
5x 1 4x 2 4x 3 u 8
6x 1 6x 2 5x 3 t 11
i. The augmented matrix is
6 0 4 1
0
0 11
5 4 4 0 "1 0 8
6 6 5 0
0
1 11
It is not clear whether there is a solution which has all variables nonnegative. However,
if you do a row operation using 5 in the first column as a pivot, you get
0 " 24
" 45 1
5
6
5
0
7
5
5
4
4
0 "1 0 8
0
6
5
1
5
0
6
5
1
7
5
and so a solution is x 1 8/5, x 2 x 3 0.
6x 1 4x 3 t 11
c.
5x 1 4x 2 4x 3 u 9
6x 1 6x 2 5x 3 t 9
i. The augmented matrix is
6 0 4 1
0
0 11
5 4 4 0 "1 0 9
6 6 5 0
0
1
9
Lets include an artificial variable and seek to minimize x 7 .
6 0 4 1
0
0
0
0 11
5 4 4 0 "1 0
1
0
9
6 6 5 0
0
1
0
0
9
0 0 0 0
0
0 "1 1
0
0
0 0 0 11
Preserving the simple columns,
6 0 4 1
5 4 4 0 "1 0 1 0
9
6 6 5 0
1 0 0
9
5 4 4 0 "1 0 0 1
9
0
Use the first column.
0 "6 "1 1
0 "1 "
6
6
1
6
5
0
"1 0 0 2
0 "1 " 56 1 0
0
0
1
3
2
0 0 9
0 "1 " 16 0 "1 " 56 0 1
3
2
It appears that the minimum value for x 7 is 3/2 and so there will be no solution to these
inequalities for which all the variables are nonnegative.
x1 " x2 x3 t 2
d.
x 1 2x 2 u 4
3x 1 2x 3 t 7
i. The augmented matrix is
1 "1 1 1
0
0 2
1
2
0 0 "1 0 4
3
0
2 0
0
1 7
Lets add in an artificial variable and set things up to minimize this artificial variable.
1 "1 1 1
0
0
0
0 2
1
2
0 0 "1 0
1
0 4
3
0
2 0
0
1
0
0 7
0
0
0 0
0
0 "1 1 0
1 "1 1 1
0
0 0 0 2
Then
1
2
0 0 "1 0 1 0 4
3
0
2 0
1
2
0 0 "1 0 0 1 4
0
1 0 0 7
Work with second column.
0 1 1 " 12 0
1
2
0 4
1 2 0 0 "1 0
1
0 4
3 0 2 0
0
1
0
0 7
0 0 0 0
0
0 "1 1 0
3
2
It appears the minimum value of x 7 is 0 and so this will mean there is a solution when
x 2 2, x 3 0, x 1 0.
5x 1 " 2x 2 4x 3 t 1
e.
6x 1 " 3x 2 5x 3 u 2
5x 1 " 2x 2 4x 3 t 5
i. The augmented matrix is
5 "2 4 1
0
0 1
6 "3 5 0 "1 0 2
5 "2 4 0
0
1 5
lets introduce an artificial variable x 7 and then minimize x 7 .
5 "2 4 1
0
0
0 1
6 "3 5 0 "1 0
1
0 2
5 "2 4 0
0
1
0
0 5
0
0
0 "1 1 0
0
0 0
Then, preserving the simple columns,
0
5 "2 4 1
0
0 0 0 1
6 "3 5 0 "1 0 1 0 2
5 "2 4 0
0
1 0 0 5
6 "3 5 0 "1 0 0 1 2
work with the first column.
5 "2
4
0 "
3
5
1
5
"
0
0
"1
0
1
5
"
"1 0 0 1
0
0 "
3
5
1
0
6
5
6
5
0 0 0 1
"1 0 1 0
4
5
1 0 0 4
4
5
There is still a positive entry.
"2 4
5
"
1
4
0
"
1
0 "
1
2
5
4
0 "1
0
0
0 0 0 1
"1 0 1 0
0
3
4
1 0 0 4
" 14 " 12 0 " 54 "1 0 0 1
3
4
It appears that there is no solution to this system of inequalities because the minimum
value of x 7 is not 0.
5. 251]
Solution:
Minimize z x 1 x 2 subject to x 1 x 2 u 2, x 1 3x 2 t 20, x 1 x 2 t 18. Change to a
maximization problem and solve as follows: Let y i M " x i . Formulate in terms of y 1 , y 2 .
You could find the maximum of 2M " x 1 " x 2 for the given constraints and this would happen
when x 1 x 2 is as small as possible. Thus you would maximize y 1 y 2 subject to the constraints
M " y1 M " y2 u 2
M " y 1 3ŸM " y 2 t 20
M " y 1 M " y 2 t 18
To simplify, this would be
2M " 2 u y 1 y 2
4M " 20 t y 1 3y 2
2M " 18 t y 1 y 2
You could simply regard M as large enough that y i u 0 and use the techniques just developed.
The augmented matrix for the constraints is then
0
0
2M " 2
1 3 0 "1
0
4M " 20
1 1 0
"1 2M " 18
1 1 1
0
Here M is large. Use the 3 as a pivot to zero out above and below it.
2
3
1
3
0 1
1 3 0 "1
2
3
M
14
3
4M " 20
0
"1
1
3
0 0
2
3
0
2
3
M"
34
3
Then it is still the case that the basic solution is not feasible. Lets use the bottom row and pick the
2/3 as a pivot.
0 0 1
0
1
16
0 3 0 " 32
3
2
3M " 3
2
3
1
3
0 0
"1
2
3
M"
34
3
Now it appears that the basic solution is feasible provided M is large. Then assemble the simplex
tableau.
0
0
1
0
1
0
16
0
3
0 " 32
3
2
0
3M " 3
2
3
0
0
1
3
"1 "1 0
0
"1 0
0
2
3
1
M"
34
3
0
Do row operations to preserve the simple columns.
0 0 1
0
1
0
16
0 3 0
3
2
0
3M " 3
0 0
" 32
1
3
"1 0
2
3
0 0 0
0
"1 1
2M " 18
2
3
M"
34
3
There is a negative number to the left of the 1 on the bottom row and we want to maximize so
work with this column. Assume M is very large. Then the pivot should be the top entry in this
column.
0 0
1
0 3 "
0
1 0
16
0 0
3M " 27
0 0
2
3
0 1
0
1
" 32
1
3
0 0
1
0
2
3
3
2
M
14
3
2M " 2
It follows that the maximum of y 1 y 2 is 2M " 2 and it happens when
y 1 M 7, y 2 M " 9, y 3 0. Thus the minimum of x 1 x 2 is
M " y 1 M " y 2 M " ŸM 7 M " ŸM " 9 2
Solutions
1
Solution:
State the eigenvalue problem from a geometric perspective.
It means that Ax is parallel to x.
2
Solution:
If A is an n × n matrix and c is a nonzero constant, compare the eigenvalues of A and cA.
Say Ax = λx. Then cAx = cλx and so the eigenvalues of cA are just cλ where λ is an eigenvalue
of A.
3
Solution:
If A is an invertible n × n matrix, compare the eigenvalues of A and A −1 . More generally, for m an
arbitrary integer, compare the eigenvalues of A and A m .
A m x = λ m x for any integer. In the case of −1, A −1 λx = AA −1 x = x so A −1 x = λ −1 x. Thus the
eigenvalues of A −1 are just λ −1 where λ is an eigenvalue of A.
4
Solution:
State the eigenvalue problem from an algebraic perspective.
It means Ax =λx where x ≠ 0.
5
Solution:
If A is the matrix of a linear transformation which rotates all vectors in R 2 through 60 ∘ , explain
why A cannot have any real eigenvalues. Is there an angle such that rotation through this angle
would have a real eigenvalue. What eigenvalues would be obtainable in this way?
If it did have λ ∈ R as an eigenvalue, then there would exist a vector x such that Ax = λx for λ a
real number. Therefore, Ax and x would need to be parallel. However, this doesn’t happen
because A rotates the vectors. You could let θ = 180 ∘ or θ = 0 ∘ . Then the eigenvalues would be 1
or −1.
6
Solution:
Let A, B be invertible n × n matrices which commute. That is, AB = BA. Suppose x is an
eigenvector of B. Show that then Ax must also be an eigenvector for B.
BAx = ABx = Aλx = λAx. Here is is assumed that Bx = λx.
7
Solution:
Suppose A is an n × n matrix and it satisfies A m = A for some m a positive integer larger than 1.
Show that if λ is an eigenvalue of A then |λ| equals either 0 or 1.
Let x be the eigenvector. Then A m x = λ m x, A m x = Ax = λx and so
λm = λ
Hence if λ ≠ 0, then
λ m−1 = 1
and so |λ| = 1.
8
Solution:
Show that if Ax = λx and Ay = λy, then whenever a, b are scalars,
Aax + by = λax + by.
Does this imply that ax + by is an eigenvector? Explain.
The formula is obvious from properties of matrix multiplications. However, this vector might not
be an eigenvector because it might equal 0 and
Eigenvectors are never 0
9
Solution:
Is it possible for a nonzero matrix to have only 0 as an eigenvalue?
Sure.
0 1
works.
0 0
10
Solution:
Find the eigenvalues and eigenvectors of the matrix
−17
17
28
−37 −32
−50
59
16
53
One eigenvalue is 5. Determine whether the matrix is defective.
−17
17
28
−37 −32
−50
59
16
, eigenvectors:
53
1
3
− 23
2
↔ −3,
↔5
−4
7
1
11
Solution:
Find the eigenvalues and eigenvectors of the matrix
5
1
0
0
5
0
−3 −91 2
One eigenvalue is 2. Determine whether the matrix is defective.
5
1
0
0
5
0
−3 −91 2
defective.
12
−1
, eigenvectors:
0
1
0
↔ 5,
0
1
↔ 2 The matrix is
Solution:
Find the eigenvalues and eigenvectors of the matrix
−23 −66 −88
−4
−13 −16
8
24
31
One eigenvalue is −3. Determine whether the matrix is defective.
−23 −66 −88
−4
−13 −16
8
24
, eigenvectors:
31
−4
− 114
−3
,
0
↔ −1,
1
1
↔ −3
− 12
0
1
The matrix is not defective.
13
Solution:
Find the eigenvalues and eigenvectors of the matrix
−12 45 15
−2
9
−8
24 11
2
One eigenvalue is 2. Determine whether the matrix is defective.
−12 45 15
−2
9
−8
24 11
1
, eigenvectors:
2
15
8
1
4
3
,
0
1
1
↔ 3,
0
↔2
1
The matrix is not defective.
14
Solution:
Find the eigenvalues and eigenvectors of the matrix
−11 −26
−8
6
14
4
−4
−17 −10
One eigenvalue is 1. Determine whether the matrix is defective.
−11 −26
−8
6
14
−4
−17 −10
defective
4
, eigenvectors:
4
3
− 23
1
3
2
↔ −4,
−1
1
↔1
15
Solution:
Find the eigenvalues and eigenvectors of the matrix
41
12
78
8
−3
12
−20 −4 −37
One eigenvalue is −3. Determine whether the matrix is defective.
41
12
78
8
−3
12
− 53
− 32
↔ −1,
− 23
, eigenvectors:
−20 −4 −37
−1
↔ −3,
1
1
−2
↔5
− 12
1
It is not defective.
16
Solution:
Find the eigenvalues and eigenvectors of the matrix
53
20 160
8
−1
22
−18 −6 −54
One eigenvalue is −3. Determine whether the matrix is defective.
53
20 160
8
−1
22
− 83
− 52
↔ −2,
− 23
, eigenvectors:
−18 −6 −54
1
1
−3
− 12
↔3
1
It is not defective.
17
Solution:
Find the eigenvalues and eigenvectors of the matrix
50
−1
114 −60
−10 −23
12
22
−26
51
One eigenvalue is −2. Determine whether the matrix is defective.
↔ −3,
11
5
− 25
114 −60
50
−10 −23
12
22
−26
51
, eigenvectors:
↔ 2,
1
9
4
− 12
↔ −2,
1
2
↔1
− 13
1
It is not defective.
18
Solution:
Find the eigenvalues and eigenvectors of the matrix
−2 −24 −12
0
6
4
−19 −14
2
One eigenvalue is −2. Determine whether the matrix is defective.
6
5
− 25
−2 −24 −12
0
2
6
4
, eigenvectors:
−19 −14
↔ −4,
1
5
4
− 12
↔ −2
1
19
Solution:
Find the eigenvalues and eigenvectors of the matrix
−15 −56
4
15
0
0
−12 −48 −1
One eigenvalue is 1. Determine whether the matrix is defective.
−15 −56
4
15
0
0
−4
0
, eigenvectors:
0
−12 −48 −1
,
1
1
0
The matrix is not defective.
20
Solution:
Find the eigenvalues and eigenvectors of the matrix
−7 −361 8
0
1
0
−4 −181 5
One eigenvalue is −3. Determine whether the matrix is defective.
↔ −1,
7
6
− 13
1
↔1
−7 −361 8
0
1
1
, eigenvectors:
0
2
↔ 1,
0
−4 −181 5
defective.
↔ −3 The matrix is
0
1
1
21
Solution:
Find the eigenvalues and eigenvectors of the matrix
−9 −6 −12
−4
1
−4
12
6
15
One eigenvalue is 3. Determine whether the matrix is defective.
− 45
−9 −6 −12
−4
1
−4
12
6
15
− 34
↔ 3,
− 25
, eigenvectors:
1
↔ 3,
− 12
1
−1
↔1
− 13
1
It is not defective.
22
Solution:
Find the eigenvalues and eigenvectors of the matrix
−13 −716 −48
0
−1
0
4
239
15
One eigenvalue is 3. Determine whether the matrix is defective.
−13 −716 −48
0
−1
0
−4
, eigenvectors:
−3
↔ −1,
0
4
239 15
matrix is defective.
1
0
1
23
Solution:
Find the eigenvalues and eigenvectors of the matrix
−8 −20
5
5
27
−5
5
85
−13
↔ 3 The
One eigenvalue is 2. Determine whether the matrix is defective. Hint: This one has some
complex eigenvalues.
−8 −20
5
5
27
−5
5
85
−13
2
29
5
29
−
+
5
29
2
29
, eigenvectors:
i
i
↔ 2 + 5i,
2
29
5
29
+
−
1
5
29
2
29
i
↔ 2 − 5i,
i
1
6
1
6
↔2
1
1
24
Solution:
Let A be a real 3 × 3 matrix which has a complex eigenvalue of the form a + ib where b ≠ 0.
Could A be defective? Explain. Either give a proof or an example.
The characteristic polynomial is of degree three and it has real coefficients. Therefore, there is a
real root and two distinct complex roots. It follows that A cannot be defective because it has three
distinct eigenvalues.
25
Solution:
Find the eigenvalues and eigenvectors of the matrix
−8
3
−14 21
−5
7
−66 92 −24
One eigenvalue is 2. Determine whether the matrix is defective. Hint: This one has some
complex eigenvalues.
−8
3
−14 21
−5
7
, eigenvectors:
−66 92 −24
1
− 10
−
1
5
−
1
5
1
10
− 101 +
i
i
↔ 1 + 4i,
1
5
+
1
1
5
1
10
i
1
↔ 1 − 4i,
i
1
↔2
1
1
26
Solution:
Find the eigenvalues and eigenvectors of the matrix
7
−15
6
−3
1
0
−18
24
−8
One eigenvalue is −2. Determine whether the matrix is defective. Hint: This one has some
complex eigenvalues.
7
−15
6
−3
1
0
−18
24
−8
1
− 10
−
3
10
−
, eigenvectors:
− 101 +
3
i
10
1
i
10
↔ 1 + 3i,
3
10
1
+
3
i
10
1
i
10
1
↔ 1 − 3i,
↔ −2
1
1
1
27
Solution:
Let T be the linear transformation which reflects all vectors in R 3 through the xy plane. Find a
matrix for T and then obtain its eigenvalues and eigenvectors.
A=
1 0
0
0 1
0
0 0 −1
1 0
0
0 1
0
0
, eigenvectors:
1
↔ −1,
0
0 0 −1
0
1
0
,
0
1
↔1
0
28
Solution:
Let T be the linear transformation which reflects vectors about the x axis. Find a matrix for T and
then find its eigenvalues and eigenvectors.
The matrix of T is
1
0
0 −1
1
0
0 −1
, eigenvectors:
.
0
1
1
↔ −1,
0
↔1
29
Solution:
Let A be the 2 × 2 matrix of the linear transformation which rotates all vectors in R 2 through an
angle of θ. For which values of θ does A have a real eigenvalue?
When you think of this geometrically, it is clear that the only two values of θ are 0 and π or these
added to integer multiples of 2π.
30
Solution:
Find the eigenvalues and eigenvectors of the matrix
0
−10
3
−12
32
−7
−61 175 −39
One eigenvalue is −1. Determine whether the matrix is defective. Hint: This one has some
complex eigenvalues.
0
−10
3
−12
32
−7
, eigenvectors:
−61 175 −39
1
− 26
−
5
26
−
− 261 +
5
i
26
1
i
26
↔ −3 + 5i,
5
26
1
+
5
i
26
1
i
26
↔ −3 − 5i,
1
3
1
3
↔ −1,
1
1
31
Solution:
Let T be the linear transformation which rotates all vectors in R 2 counterclockwise through an
angle of π/2. Find a matrix of T and then find eigenvalues and eigenvectors.
A=
0 −1
1
0
0 −1
1
−i
, eigenvectors:
0
↔ −i,
1
i
↔i
1
32
Solution:
Suppose A is a 3 × 3 matrix and the following information is available.
0
A
−1
0
=2
−1
−1
−4
,A
−1
−3
A
1
1
=1
1
1
1
1
−3
= −3
−3
−4
−3
1
Find A
−5
4
You have given that
0
1 −3
−1 1 −4
−1 1 −3
and so it follows that
−1
0
A
1 −3
−1 1 −4
−1 1 −3
=
2 0
0
0 1
0
0 0 −3
−12 12
1
A=
−1 −15 17
−1 −12 14
Therefore, the desired answer is given by
109
142
115
33
Solution:
Suppose A is a 3 × 3 matrix and the following information is available.
−2
A
−3
−2
=1
−3
−3
−6
,A
−3
−5
A
1
1
= −3
1
1
1
1
−5
= −1
−5
−6
−5
1
Find A
−3
4
You have given that
−2 1 −5
−3 1 −6
−1
−2 1 −5
A
−3 1 −5
1
=
−3 1 −6
−3 1 −5
and so it follows that
−11 10 −2
A=
−12
9
0
−12 10 −1
Therefore, the desired answer is given by
−49
−39
−46
34
Solution:
0
0
0 −3
0
0
−1
0
Suppose A is a 3 × 3 matrix and the following information is available.
−1
A
A
−2
−1
=2
−2
−2
−2
−2
−2
−5
1
=2
,A
1
=0
1
1
1
1
−5
−4
−4
3
Find A
−4
3
You have given that
−1 1 −2
−2 1 −5
−1
−1 1 −2
−2 1 −5
A
−2 1 −4
−2 1 −4
2 0 0
=
0 0 0
0 0 2
and so it follows that
−2 0 2
A=
−4 2 2
−4 0 4
Therefore, the desired answer is given by
0
−14
0
35
Solution:
Here is a symmetric matrix.
13
6
− 16
− 13
− 16
− 13
13
6
1
3
1
3
8
3
Find its largest eigenvalue and an eigenvector associated with this eigenvalue. This eigenvector
gives the direction of the principal stretch.
The largest eigenvalue is 3. The eigenvectors and eigenvalues are
−1
1
1
0
↔ 2,
1
2
1
↔ 3,
−1
1
↔2
36
Solution:
Here is a symmetric matrix.
24
11
− 112
− 116
− 112
− 116
24
11
6
11
6
11
40
11
Find its largest eigenvalue and an eigenvector associated with this eigenvalue. This eigenvector
gives the direction of the principal stretch.
The largest eigenvalue is 4. The eigenvectors and eigenvalues are
1
1
3
↔ 2,
1
↔ 2,
−3
0
↔4
−1
−3
2
37
Solution:
Here is a symmetric matrix.
13
6
5
6
2
3
5
6
13
6
− 23
2
3
− 23
11
3
Find its largest eigenvalue and an eigenvector associated with this eigenvalue. This eigenvector
gives the direction of the principal stretch.
The largest eigenvalue is 4. The eigenvectors and eigenvalues are
−4
1
1
↔ 3,
1
↔ 1,
4
0
↔4
−1
2
4
38
Solution:
Here is a symmetric matrix.
25
6
− 76
5
3
− 76
25
6
− 53
5
3
− 53
11
3
Find its largest eigenvalue and an eigenvector associated with this eigenvalue. This eigenvector
gives the direction of the principal stretch.
The largest eigenvalue is 7. The eigenvectors and eigenvalues are
−1
1
1
0
↔ 3,
1
2
1
↔ 2,
−1
1
↔7
39
Solution:
Here is a symmetric matrix.
25
6
− 76
5
3
− 76
25
6
− 53
5
3
− 53
11
3
Find its largest eigenvalue and an eigenvector associated with this eigenvalue. This eigenvector
gives the direction of the principal stretch.
The largest eigenvalue is 7. The eigenvectors and eigenvalues are
−1
1
1
0
↔ 3,
1
2
1
↔ 2,
↔7
−1
1
40
Solution:
Here is a Markov (migration matrix) for three locations
1
2
1
2
0
1
8
5
8
1
4
1
4
1
4
1
2
The total number of individuals in the migration process is
320
After a long time, how many are in each location?
80
160
80
41
Solution:
Here is a Markov (migration matrix) for three locations
1
10
2
5
1
2
1
2
1
6
1
3
2
9
2
3
1
9
The total number of individuals in the migration process is
949
After a long time, how many are in each location?
280
372
297
42
Solution:
Here is a Markov (migration matrix) for three locations
7
10
1
5
1
10
1
14
1
2
3
7
1
6
1
2
1
3
The total number of individuals in the migration process is
736
After a long time, how many are in each location?
200
308
228
43
Solution:
The following table describes the transition probabilities between the states rainy, partly cloudy
and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the
next day with probability 15 . If it starts off sunny, it ends up sunny the next day with probability
and so forth.
2
5
rains sunny p.c.
rains
sunny
p.c.
1
5
1
5
3
5
1
5
2
5
2
5
6
11
1
11
4
11
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
This is a vector in the direction
190
100
This vector is given by
242
5
14
25
133
121
266
44
Solution:
The following table describes the transition probabilities between the states rainy, partly cloudy
and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the
next day with probability 15 . If it starts off sunny, it ends up sunny the next day with probability
and so forth.
1
3
rains sunny p.c.
1
10
1
5
7
10
rains
sunny
p.c.
1
3
1
3
1
3
3
10
2
5
3
10
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
This is a vector in the direction
100
126
This vector is given by
160
50
193
63
193
80
193
45
Solution:
You own a trailer rental company in a large city and you have four locations, one in the South
East, one in the North East, one in the North West, and one in the South West. Denote these
locations by SE,NE,NW, and SW respectively. Suppose that the following table is observed to
take place.
SE NE NW SW
SE
NE
NW
SW
2
9
1
3
2
9
2
9
1
10
7
10
1
10
1
10
1
10
1
5
3
5
1
10
1
5
1
10
1
5
1
2
In this table, the probability that a trailor starting at NE ends in NW is 1/10, the probability that a
trailor starting at SW ends in NW is 1/5, and so forth. Approximately how many will you have in
each location after a long time if the total number of trailors is 463?
63
184
126
90
46
Solution:
The following table describes the transition probabilities between the states rainy, partly cloudy
and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the
next day with probability
and so forth.
1
5
. If it starts off sunny, it ends up sunny the next day with probability
1
5
rains sunny p.c.
1
10
1
5
7
10
rains
sunny
p.c.
2
5
1
5
2
5
3
8
3
8
1
4
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
This is a vector in the direction
180
165
This vector is given by
256
180
601
165
601
256
601
47
Solution:
You own a trailer rental company in a large city and you have four locations, one in the South
East, one in the North East, one in the North West, and one in the South West. Denote these
locations by SE,NE,NW, and SW respectively. Suppose that the following table is observed to
take place.
SE NE NW SW
SE
NE
NW
SW
1
3
1
3
1
6
1
6
1/4 1/10 1/5
1/4 1/5
1/10
1/4 3/5
1/5
1/4 1/10 1/2
In this table, the probability that a trailor starting at NE ends in NW is 1/10, the probability that a
trailor starting at SW ends in NW is 1/5, and so forth. Approximately how many will you have in
each location after a long time if the total number of trailors is 1030.
210
220
350
250
48
Solution:
In the city of Nabal, there are three political persuasions, republicans (R), democrats (D), and
neither one (N). The following table shows the transition probabilities between the political
parties, the top row being the initial political party and the side row being the political affiliation
the following year.
R
D
N
R
D
N
1
10
1
10
4
5
1
5
1
5
3
5
3
13
7
13
3
13
Find the probabilities that a person will be identified with the various political persuasions. Which
party will end up being most important?
This is a vector in the direction
190
330
which has the entries add to 1. This vector is given by
455
38
195
22
65
7
15
49
Solution:
The University of Poohbah offers three degree programs, scouting education (SE), dance
appreciation (DA), and engineering (E). It has been determined that the probabilities of
transferring from one program to another are as in the following table.
SE DA E
SE
.8
.1
.3
DA .1
.7
.5
E
.2
.2
.1
where the number indicates the probability of transferring from the top program to the program on
the left. Thus the probability of going from DA to E is .2. Find the probability that a student is
enrolled in the various programs.
8
10
−1
1
10
1
10
1 1 1
1
10
7
10
3
10
1
2
−1
2
10
2
10
14
5
13
5
0
, row echelon form:
0
−1 0
z
=
z
32
5
z
z
5
 14
 32
5
5
 13
5  32
5
32
=
7
16
13
32
5
32
0. 437 5
=
0. 406 25
0. 156 25
1 0 − 145
0
0 1 − 135
0
0 0
0
0
50
Solution:
Here is a Markov (migration matrix) for three locations
3
5
1
10
3
10
1
8
3
4
1
8
1
12
7
12
1
3
The total number of individuals in the migration process is
406
After a long time, how many are in each location?
90
232
84
51
Solution:
You own a trailer rental company in a large city and you have four locations, one in the South
East, one in the North East, one in the North West, and one in the South West. Denote these
locations by SE,NE,NW, and SW respectively. Suppose that the following table is observed to
take place.
SE NE NW SW
SE
NE
NW
SW
1
2
1
10
1
5
1
5
1
10
7
10
1
10
1
10
1
10
1
5
3
5
1
10
1
5
1
10
1
5
1
2
In this table, the probability that a trailor starting at NE ends in NW is 1/10, the probability that a
trailor starting at SW ends in NW is 1/5, and so forth. Approximately how many will you have in
each location after a long time if the total number of trailors is 350?
70
112
98
70
52
Solution:
Here is a Markov (migration matrix) for three locations
2
5
1
2
1
10
1
8
1
2
3
8
1
2
3
14
2
7
The total number of individuals in the migration process is
1000
After a long time, how many are in each location?
310
424
266
53
Solution:
Let A be the n × n, n > 1, matrix of the linear transformation which comes from the projection
v → proj w v. Show that A cannot be invertible. Also show that A has an eigenvalue equal to 1 and
that for λ an eigenvalue, |λ| ≤ 1.
Obviously A cannot be onto because the range of A has dimension 1 and the dimension of this
space should be 3 if the matrix is onto. Therefore, A cannot be invertible. Its row reduced echelon
form cannot be I since if it were, A would be onto. Aw = w so it has an eigenvalue equal to 1.
Now suppose Ax = λx. Thus, from the Cauchy Schwarz inequality,
|x, w|
|x||w|
|w| ≥
|w| = |λ||x|
|x| =
2
|w|2
|w|
and so |λ| ≤ 1.
54
Solution:
Consider the dynamical system in which A is an n × n matrix,
xn + 1 = Axn, x0 = x 0
Suppose the eigen pairs are
v i , λ i 
and suppose that v 1 , v 2 , ⋯, v n  is a basis for R n . If
n
x0 =
∑ civi,
i=1
show that the solution to the dynamical system is of the form
n
xn =
∑ c i λ ni v i
i=1
n
The solution is A x 0 and this reduces to
n
xn =
∑ ciA
i=1
n
n
vi =
∑ c i λ ni v i
i=1
55
Solution:
Consider the recurrance relation x n+2 = x n+1 + x n where x 1 and x 0 are given positive numbers.
1+ 5
Show that if lim n→∞ xxn+1
exists, then it must equal 2 .
n
You have
x n+2 = 1 + x n
x n+1
x n+1
Assuming this limit exists, and equals r, it must satisfy
r = 1 + 1r
and so r 2 − r − 1 = 0, Solution is: 12 5 + 12 , 12 − 12 5 . Since the solution is obviously positive,
it must be the first.
56
Solution:
Consider the dynamical system
xn + 1
yn + 1
.8
=
.8
xn
−. 8 . 8
yn
−i
eigenvalues and eigenvectors are 0. 8 + 0. 8i 
, 0. 8 − 0. 8i 
. Show the
i
. Find a formula
1
1
for the solutiont to the dynamical system for given initial condition x 0 , y 0  T . Show that the
magnitude of xn, yn T must diverge provided the initial condition is not zero. Next graph the
vector field for
.8
.8
x
−. 8 . 8
y
−
x
y
Note that this vector field seems to indicate a conclusion different than what you just obtained.
Therefore, in this context of discreet dynamical systems the consideration of such a picture is not
all that reliable.
It is straight forward to verify that these are the eigenvectors and eigenvalues. Thus the solution to
the dynamical system is of the form
a
−i
1
0. 8 + 0. 8i n + b
i
1
0. 8 − 0. 8i n
Now |0. 8 − 0. 8i| = 1. 131 4 = |0. 8 + 0. 8i| and so the magnitude of any solution having nonzero
initial conditions is unbounded as n → ∞. However, here is the graph of the vector field.
It does not appear to indicate that the solutions are getting larger in magnitude from one point to
the next and in fact, seems to indicate that they get smaller.
57
Solution:
Let v be a unit vector and let A = I − 2vv T . Show that A has an eigenvalue equal to −1.
Lets see what it does to v.
I − 2vv T v = v − 2vv T v = v − 2v =−1v
Yes. It has an eigenvector and the eigenvalue is −1.
58
Solution:
Let M be an n × n matrix and suppose x 1 , ⋯, x n are n eigenvectors which form a linearly
independent set. Form the matrix S by making the columns these vectors. Show that S −1 exists
and that S −1 MS is a diagonal matrix (one having zeros everywhere except on the main diagonal)
having the eigenvalues of M on the main diagonal. When this can be done the matrix is
diagonalizable.
Since the vectors are linearly independent, the matrix S has an inverse. Denoting this inverse by
w T1
S −1 =
⋮
w Tn
it follows by definition that
w Ti x j = δ ij .
Therefore,
w T1
S −1 MS = S −1 Mx 1 , ⋯, Mx n  =
⋮
λ 1 x 1 , ⋯, λ n x n 
w Tn
λ1
=
0
⋱
λn
0
59
Solution:
Show that a matrix, M is diagonalizable if and only if it has a basis of eigenvectors. Hint: The
first part is done earlier. It only remains to show that if the matrix can be diagonalized by some
matrix, S giving D = S −1 MS for D a diagonal matrix, then it has a basis of eigenvectors. Try using
the columns of the matrix S.
The formula says that
MS = SD
th
Letting x k denote the k column of S, it follows from the way we multiply matrices that
Mx k = λ k x k
th
where λ k is the k diagonal entry on D.
60
Solution:
Suppose A is an n × n matrix which is diagonally dominant. This means
|a ii | >
∑|a ij |.
j≠i
Show that A −1 must exist.
The diagonally dominant condition implies that none of the Gerschgorin disks contain 0.
Therefore, 0 is not an eigenvalue. Hence A is one to one, hence invertible.
61
Solution:
Let M be an n × n matrix. Then define the adjoint of M, denoted by M ∗ to be the transpose of the
conjugate of M. For example,
∗
2
i
1+i 3
=
2 1−i
−i
3
.
A matrix, M, is self adjoint if M ∗ = M. Show the eigenvalues of a self adjoint matrix are all real.
If the self adjoint matrix has all real entries, it is called symmetric.
First note that AB ∗ = B ∗ A ∗ . Say Mx = λx, x ≠ 0. Then
λ |x|2 = λ x ∗ x = λx ∗ x =Mx ∗ x = x ∗ M ∗ x
= x ∗ Mx = x ∗ λx = λ|x|2
Hence λ = λ .
62
Solution:
Suppose A is an n × n matrix consisting entirely of real entries but a + ib is a complex eigenvalue
having the eigenvector, x + iy. Here x and y are real vectors. Show that then a − ib is also an
eigenvalue with the eigenvector, x − iy. Hint: You should remember that the conjugate of a
product of complex numbers equals the product of the conjugates. Here a + ib is a complex
number whose conjugate equals a − ib.
Ax = a + ibx. Now take conjugates of both sides. Since A is real,
A x = a − ib x
63
Solution:
Recall an n × n matrix is said to be symmetric if it has all real entries and if A = A T . Show the
eigenvectors and eigenvalues of a real symmetric matrix are real.
These matrices are self adjoint by definition, so this follows from the above problems.
64
Solution:
Recall an n × n matrix is said to be skew symmetric if it has all real entries and if A = −A T . Show
that any nonzero eigenvalues must be of the form ib where i 2 = −1. In words, the eigenvalues are
either 0 or pure imaginary.
Suppose A is skew symmetric. Then what about iA?
iA ∗ = −iA ∗ = −iA T = iA
and so iA is self adjoint. Hence it has all real eigenvalues.
iAx = λx
where λ is real. Thus Ax = −iλx and therefore, the eigenvalues of A are all of the form iλ where λ
is real.
65
Solution:
A discreet dynamical system is of the form
xk + 1 = Axk, x0 = x 0
where A is an n × n matrix and xk is a vector in R n . Show first that
xk = A k x 0
for all k ≥ 1. If A is nondefective so that it has a basis of eigenvectors, v 1 , ⋯, v n  where
Av j = λ j v j
you can write the initial condition x 0 in a unique way as a linear combination of these
eigenvectors. Thus
n
x0 =
∑ a jv j
j=1
Now explain why
n
xk =
n
∑ a jA v j = ∑ a jλ kj v j
k
j=1
j=1
which gives a formula for xk, the solution of the dynamical system.
The first formula is obvious from induction. Thus
n
Akx0 =
∑ a jAk v j =
j=1
n
∑ a jλ kj v j
j=1
because if Av = λv, then if A k x = λ k x, do A to both sides. Thus A k+1 x = λ k+1 x.
66
Solution:
Suppose A is an n × n matrix and let v be an eigenvector such that Av = λv. Also suppose the
characteristic polynomial of A is
detλI − A = λ n + a n−1 λ n−1 + ⋯ + a 1 λ + a 0
Explain why
A n + a n−1 A n−1 + ⋯ + a 1 A + a 0 Iv = 0
If A is nondefective, give a very easy proof of the Cayley Hamilton theorem based on this. Recall
this theorem says A satisfies its characteristic equation,
A n + a n−1 A n−1 + ⋯ + a 1 A + a 0 I = 0.
A n + a n−1 A n−1 + ⋯ + a 1 A + a 0 Iv = λ n + a n−1 λ n−1 + ⋯ + a 1 λ + a 0 v = 0 If A is nondefective,
then it has a basis of eigenvectors v 1 , ⋯, v n  and so a typical vector w is of the form
n
w = ∑ i=1 c i v i and so, letting the characteristic polynomial be denoted as Pλ,
n
PAw = PA ∑ c i v i =
i=1
n
∑ c i PAv i = ∑ c i 0 = 0
i=1
i
Since w is arbitrary, this shows that PA = 0.
67
Solution:
Suppose an n × n nondefective matrix A has only 1 and −1 as eigenvalues. Find A 12 .
From the above formula,
A 12 x 0 =
Thus A
12
n
n
j=1
j=1
∑ a jλ 12j v j = ∑ a jv j = x 0
= I.
68
Solution:
Suppose the characteristic polynomial of an n × n matrix A is 1 − λ n . Find A mn where m is an
integer. Hint: Note first that A is nondefective. Why?
The eigenvalues are distinct because they are the n th roots of 1. Hence from the above formula, if
x is a given vector with
n
x=
∑ a jv j
j=1
then
n
A nm x = A nm ∑ a j v j =
j=1
so A
nm
n
∑ a jA nm v j =
j=1
n
∑ a jv j = x
j=1
= I.
69
Solution:
Sometimes sequences come in terms of a recursion formula. An example is the Fibonacci
sequence.
x 0 = 1 = x 1 , x n+1 = x n + x n−1
Show this can be considered as a discreet dynamical system as follows.
x n+1
=
xn
1 1
xn
1 0
x n−1
x1
,
1
=
x0
1
Now find a formula for x n .
What are the eigenvalues and eigenvectors of this matrix?
1 1
1 0
, eigenvectors:
1
2
−
1
2
5
1
2
↔ 1 − 1 5,
2
2
1
5 +
1
2
↔ 1 5 + 1
2
2
1
Now also
1 − 1 5
2
10
1
2
−
1
2
5
1
+
1 5 + 1
10
2
Therefore, the solution is of the form
x n+1
xn
=
1
2
5 +
1
1
2
=
1
1
1 − 1 5
2
10
+
1 5 + 1
10
2
1
2
n
1 − 1 5
2
2
−
1
2
5
1
n
1 5 + 1
2
2
1
2
5 +
1
2
1
In particular,
xn =
1 − 1 5
2
10
1 − 1 5
2
2
n
+
1 5 + 1
10
2
1 5 + 1
2
2
n
70
Solution:
Let A be an n × n matrix having characteristic polynomial
detλI − A = λ n + a n−1 λ n−1 + ⋯ + a 1 λ + a 0
Show that a 0 = −1 n detA.
The characteristic polynomial equals detλI − A. To get the constant term, you plug in λ = 0 and
obtain det−A = −1 n detA.
71
Solution:
Here is a matrix
A=
53
8
− 454
25
8
− 214
A=
19
9
− 209
2
3
− 13
Find lim n→∞ A n .
10
5
−18 −9
72
Solution:
Here is a matrix
Find lim n→∞ A n .
6
3
−10 −5
73
Solution:
Find the solution to the initial value problem
′
x
=
y
x0
xt
yt
x
6
y
5
3
=
y0
0 −1
3
12e 2t − 9e 2 2t
3
=
27e 2 2t − 24e 2t
3
74
Solution:
Find the solution to the initial value problem
′
x
=
y
x0
=
y0
xt
yt
−
1
e
1
=
e
3 2t
2
3 2t
2
−13 −5
x
30
12
y
7 −2
x
12 −3
y
2
2
6e 2 2t − 8
5
18e 2 2t − 16
5
75
Solution:
Find the solution to the initial value problem
′
x
y
x0
y0
xt
yt
=
=
=
3
2
7e 3t − 4e t
14e 3t − 12e t
Matrices And Inner Product
76
Solution:
Here are some matrices. Label according to whether they are symmetric, skew symmetric, or
orthogonal. If the matrix is orthogonal, determine whether it is proper or improper.
a.
1
0
0
1
2
0
1
2
0
−
1
2
1
2
This one is orthogonal and is a proper transformation.
b.
1
2 −3
2
1
4
−3 4
7
This is symmetric.
0 −2 −3
c.
2
0
−4
3
4
0
This one is skew symmetric.
77
Solution:
Here are some matrices. Label according to whether they are symmetric, skew symmetric, or
orthogonal. If the matrix is orthogonal, determine whether it is proper or improper.
a.
1
0
0
1
2
0
1
2
0
−
1
2
1
2
This one is orthogonal and is a proper transformation.
b.
1
2 −3
2
1
4
−3 4
7
This is symmetric.
0 −2 −3
c.
2
0
−4
3
4
0
This one is skew symmetric.
78
Solution:
Show that every real matrix may be written as the sum of a skew symmetric and a symmetric
matrix in a unique way. Hint: If A is an n × n matrix, show that B ≡ 12 A − A T  is skew
symmetric.
T
T
A = A+A
+ A−A
. There is only one way to do it. If A = S + W where S = S T , W = −W T , then
2
2
you also have A T = S − W and so 2S = A + A T while 2W = A − A T .
79
Solution:
Let x be a vector in R n and consider the matrix, I −
2xx T
|x|2
. Show this matrix is both symmetric and
orthogonal.
T
I−
2xx T
|x|2
= I−
2xx T
|x|2
so it is symmetric.
T
I − 2xx2
|x|
T
I − 2xx2
|x|
T
= I + 4 xx xx
|x|4
T
T
T
− 4 xx2
|x|
T
= I + 4 xx4 − 4 xx2 = I
|x|
|x|
because x T x
1
|x|2
= 1.
80
Solution:
For U an orthogonal matrix, explain why ||Ux|| = ||x|| for any vector, x. Next explain why if U is
an n × n matrix with the property that ||Ux|| = ||x|| for all vectors, x, then U must be orthogonal.
Thus the orthogonal matrices are exactly those which preserve distance.
First, suppose U is orthogonal.
‖Ux‖ 2 = Ux, Ux = U T Ux, x = Ix, x = ‖x‖ 2
Next suppose distance is preserved by U. Then
Ux + y, Ux + y = ‖Ux‖ 2 + ‖Uy‖ 2 + 2Ux, Uy
= ‖x‖ 2 + ‖y‖ 2 + 2U T Ux, y
But since U preserves distances, it is also the case that
Ux + y, Ux + y = ‖x‖ 2 + ‖y‖ 2 + 2x, y
Hence
x, y = U T Ux, y
and so
U T U − Ix, y = 0
Since y is arbitrary, it follows that U T U − I = 0. Thus U is orthogonal.
81
Solution:
A quadratic form in three variables is an expression of the form
a 1 x 2 + a 2 y 2 + a 3 z 2 + a 4 xy + a 5 xz + a 6 yz. Show that every such quadratic form may be written as
x
x y z
A
y
z
where A is a symmetric matrix.
a1
x y z
a 4 /2
a 4 /2 a 5 /2
a2
a 5 /2 a 6 /2
x
a 6 /2
y
a3
z
82
Solution:
Given a quadratic form in three variables, x, y, and z, show there exists an orthogonal matrix U
and variables x ′ , y ′ , z ′ such that
x′
x
y
z
=U
y′
z′
with the property that in terms of the new variables, the quadratic form is
λ 1 x ′  2 + λ 2 y ′  2 + λ 3 z ′  2
where the numbers, λ 1 , λ 2 , and λ 3 are the eigenvalues of the matrix, A in the above problem.
The quadratic form may be written as
x T Ax
where A = A T . By the theorem about diagonalizing a symmetric matrix, there exists an orthogonal
matrix U such that
U T AU = D, A = UDU T
Then the quadratic form is
x T UDU T x = U T x T DU T x
where D is a diagonal matrix having the real eigenvalues of A down the main diagonal. Now
simply let
x ′ ≡ U T x.
83
Solution:
If A is a symmetric invertible matrix, is it always the case that A −1 must be symmetric also? How
about A k for k a positive integer? Explain.
If A is symmetric, then A = U T DU for some D a diagonal matrix in which all the diagonal entries
are non zero. Hence A −1 = U −1 D −1 U −T . Now U −1 U −T = U T U −1 = I −1 = I and so
A −1 = QD −1 Q T , where Q is orthogonal. Is this thing on the right symmetric? Take its transpose.
This is QD −1 Q T which is the same thing, so it appears that a symmetric matrix must have
symmetric inverse. Now consider raising it to a power.
Ak = UTDkU
and the right side is clearly symmetric.
84
Solution:
If A, B are symmetric matrices, does it follow that AB is also symmetric?
2 1
Let A =
1 3
AB =
,B =
2 1
1 1
1 3
1 0
1 1
1 0
=
3 2
4 1
which is not symmetric.
85
Solution:
Suppose A, B are symmetric and AB = BA. Does it follow that AB is symmetric?
AB T = B T A T = BA = AB so the answer in this case is yes.
86
Solution:
Here are some matrices. What can you say about the eigenvalues of these matrices just by looking
at them?
0 0
a.
0
0 0 −1
0 1
0
The eigenvalues are all within 1 of 0, and pure imaginary or zero.
b.
1
2 −3
2
1
4
−3 4
7
The eigenvalues are in D1, 6 ∪ D7, 7. They are also real.
0 −2 −3
c.
2
0
−4
3
4
0
The eigenvalues are all imaginary or 0. They are no farther than 7 from 0.
1 2 3
d.
0 2 3
0 0 2
The eigenvalues are 1,2,2.
87
Solution:
c 0
Find the eigenvalues and eigenvectors of the matrix
0
0 0 −b
0 b
0
. Here b, c are real numbers.
c 0
0
0 0 −b
0 b
, eigenvectors:
0
1
0
↔ c,
0
0
↔ −ib,
−i
0
↔ ib
i
1
1
88
Solution:
c 0
Find the eigenvalues and eigenvectors of the matrix
0
0 a −b
0 b
. Here a, b, c are real
a
numbers.
c 0
0
0 a −b
0 b
, eigenvectors:
a
0
0
↔ a − ib,
−i
1
1
↔ a + ib,
i
↔c
0
1
0
89
Solution:
Determine which of the following sets of vectors are orthonormal sets. Justify your answer.
a. 1, 1, 1, −1
i. This one is not orthonormal.
1
2
b.
,
−1
2
, 1, 0
i. This one is not orthonormal.
c.  13 , 23 , 23 ,  −2
, −1
, 23 ,  23 , −2
,
3
3
3
1
3

T
1/3
i.
2/3
2/3
−2/3 −1/3 2/3
1/3
2/3
2/3
−2/3 −1/3 2/3
2/3 −2/3 1/3
2/3
orthonormal set of vectors.
−2/3 1/3
1 0 0
=
0 1 0
0 0 1
90
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
so this is an
A=
− 43
− 13
− 13
− 13
7
6
− 176
− 176
−
1
3
.
7
6
Hint: Two eigenvalues are −2, −1.
1
3
3
1
3
3
1
3
3
− 13 2 3
 −2,
1
6
2 3
1
6
2 3
0
 −1,
1
2
− 12
4
2
2
91
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
61
22
− 35
22
23
11
− 35
22
23
11
7
11
− 18
11
− 223
7
11
.
Hint: Two eigenvalues are −3, 0.
1
6
1
6
− 13
6
6
 −3,
6
1
66
7
66
2
33
− 113 11
6 11
6 11
 0,
1
11
− 111
6 11
4
11
11
92
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
1
−2
−2
−2
1
2
− 32
− 32
−2
.
1
2
Hint: Two eigenvalues are −3, 3.
1
3
1
3
1
3
− 13 2 3
3
3
3
 −3,
1
6
1
6
0
 3,
2 3
1
2
2
2
− 12 2
2 3
93
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
− 32
0
7
2
0
−5
0
7
2
0
− 32
.
Hint: Two eigenvalues are 2, −5.
− 32
0
7
2
0
−5
0
7
2
0
− 32
1
2
1
2
− 16 6
2
↔ 2,
0
, eigenvectors:
2
1
3
1
6
− 13 3
↔ −5,
6
↔ −5
− 13 3
1
3
6
3
94
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
20
11
− 116
9
11
A=
− 116
9
11
− 116
20
11
15
11
− 116
.
Hint: One eigenvalue is 1.
1
− 12 2
p 2 +2
p
p 2 +2
1
p 2 +2
 1,
 1,
0
1
2
2
3
22
− 111
2 11
3
22
2 11
3
2 11
95
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
1 0 0
A=
.
0 1 0
0 0 1
Hint: Two eigenvalues are 1, 1.
1
2
1 0 0
0 1 0
0 0 1
, eigenvectors:
↔ 1,
0
1
2
− 16 6
2
2
1
3
6
1
6
6
− 13 3
↔ 1,
96
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
− 13 3
1
3
3
↔1
3 0 0
A=
.
0 3 0
0 0 3
Hint: One eigenvalue is 3.
1
3
38
− 12 2
p 2 +2
p
 3,
p 2 +2
1
2
1
p 2 +2
 3,
0
2 38
3
38
2
3
− 381 2 38
2 38
97
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
3 1 1
A=
.
1 3 1
1 1 3
Hint: One eigenvalue is 2.
1
− 12 2
p 2 +2
p
 2,
p 2 +2
1
2
1
p 2 +2
 2,
0
2
1
3
1
3
3
1
3
3
5
3
98
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
3
2
A=
0
1
2
.
0 1 0
1
2
0
3
2
Hint: Two eigenvalues are 2, 1.
3
2
0
1
2
0 1 0
1
2
0
3
2
1
2
, eigenvectors:
↔ 2,
0
1
2
− 16 6
2
2
1
3
1
6
6
− 13 3
↔ 1,
6
99
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
− 13 3
1
3
3
↔1
A=
19
6
7
6
− 23
7
6
19
6
− 23
− 23
− 23
.
11
3
Hint: Two eigenvalues are 3, 2.
1
6
6
1
6
6
1
3
6
− 12 2
 3,
1
2
2
 2,
1
3
3
1
3
3
5
− 13 3
0
100
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
1
3
1
3
3
1
3
3
8
3
2
3
− 13
 3,
 3,
0
1
2
2
3
8
3
.
− 16 6
− 12 2
3
− 13
2
3
5
3
2
3
1
3
1
6
− 16 6
2
101
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
13
6
2
3
− 76
2
3
8
3
− 23
− 76
− 23
.
13
6
Hint: Two eigenvalues are 1, 2.
13
6
2
3
− 76
1
2
− 76
− 23
− 16 6
↔ 1,
2
, eigenvectors:
13
6
2
0
1
2
2
3
8
3
− 23
1
3
6
1
6
6
− 13 3
↔ 2,
− 13 3
1
3
↔4
3
102
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
137
54
5
27
− 25
54
− 25
54
5
27
79
27
5
27
5
27
137
54
.
Hint: One eigenvalue is 3.
1
p
p 2 +2
5
18
− 19
5
18
− 12 2
p 2 +2
 3,
1
2
1
p 2 +2
 3,
0
2
2 3
2
2 3
2 3
103
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A.
A=
1
3
1
3
1
3
− 12 2
3
3
3
 2,
 2,
0
1
2
2
5
2
−1
1
2
−1
4
−1
1
2
−1
5
2
1
6
− 13
1
6
.
6
6
5
6
104
Solution:
Show that if A is a real symmetric matrix and λ and μ are two different eigenvalues, then if x is
an eigenvector for λ and y is an eigenvector for μ, then x ⋅ y = 0. Also all eigenvalues are real.
First
CD T =D T C T
A is real
λ is eigenvalue
=
x T A x = x T Ax
=
xT λ x = λ xT x
λx T x = Ax T x
and so λ = λ . This shows that all eigenvalues are real. It follows all the eigenvectors can be
assumed real. Why?
Because A is real. Ax = λx, A x = λ x , so x + x is an eigenvector. Hence it can be assumed all
eigenvectors are real.
Now let x, y, μ and λ be given as above.
λx ⋅ y = λx ⋅ y = Ax ⋅ y = x ⋅ Ay = x ⋅μy = μx ⋅ y = μx ⋅ y
and so
λ − μx ⋅ y = 0.
Since λ ≠ μ, it follows x ⋅ y = 0.
105
Solution:
Suppose U is an orthogonal n × n matrix. Explain why rankU = n.
You could observe that detUU T  = detU 2 = 1 so detU ≠ 0.
106
Solution:
Show that the eigenvalues and eigenvectors of a real matrix occur in conjugate pairs.
This follows from the observation that ifAx = λx, then A x = λ x
107
Solution:
If a real matrix, A has all real eigenvalues, does it follow that A must be symmetric. If so, explain
why and if not, give an example to the contrary.
1 3
Certainly not.
0 2
108
Solution:
Suppose A is a 3 × 3 symmetric matrix and you have found two eigenvectors which form an
orthonormal set. Explain why their cross product is also an eigenvector.
There exists another eigenvector such that, with these two you have an orthonormal basis. Let
the third eigenvector be u and the two given ones v 1 , v 2 . Then u is perpendicular to the two given
vectors and so it has either the same or opposite direction to the cross product. Hence u =kv 1 × v 2
for some scalar k.
109
Solution:
Show that if u 1 , ⋯, u n  is an orthonormal set of vectors in F n , then it is a basis. Hint: It was
shown earlier that this is a linearly independent set. If you wish, replace F n with R n . Do this
version if you do not know the dot product for vectors in C n .
The vectors are linearly independent as shown earlier. If they do not span all of F n , then there is
a vector v ∉ spanu 1 , ⋯, u n . But then u 1 , ⋯, u n , v would be a linearly independent set which
has more vectors than a spanning set, e 1 , ⋯, e n  and this is a contradiction.
110
Solution:
Fill in the missing entries to make the matrix orthogonal.
−1
2
−1
6
1
3
−1
2
−1
6
1
3
1
2
−1
6
a
1
2
−1
6
a
0
6
3
b
0
6
3
b
−1
2
−1
6
1
3
1
2
_
_
_
6
3
_
T
.
=
3a −
1
3
1
3
3a −
1
3
1
a2 +
1
3
1
3
3b −
ab −
1
3
3b −
1
3
ab −
2
3
1
3
1
3
1
3
2
3
b2 +
Must have a = 1/ 3 , b = 1/ 3
−1
2
−1
6
1
2
0
T
1
3
−1
2
−1
6
1
3
−1
6
1/ 3
1
2
−1
6
1/ 3
6
3
1/ 3
0
6
3
1/ 3
1 0 0
=
0 1 0
0 0 1
111
Solution:
Fill in the missing entries to make the matrix orthogonal.
2
3
2
2
2
3
_
_
_
0
_
1
6
2
.
T
2
3
2
2
2
3
− 2
2
− 13
0
2
3
2
2
a
2
3
− 2
2
a
b
− 13
0
b
1
6
2
1
6
1
=
a=
1
6
1
6
2a −
2b −
a2 +
1
18
2
9
,b =
4
3 2
2
3
2
2
1
6
2
3
− 2
2
− 13
0
1
3 2
2a −
ab −
1
18
1
6
2
2b −
1
6
ab −
17
18
2
9
2
9
2
9
1
9
b2 +
T
2
3
2
2
3 2
2
3
− 2
2
1
3 2
4
3 2
− 13
0
4
3 2
2
1
1
6
2
1 0 0
=
112
Solution:
Fill in the missing entries to make the matrix orthogonal.
0 1 0
0 0 1
−
1
3
−
1
3
Try
2
5
2
3
0
2
3
1
5
=
2
3
0
_
_
c
1
3
d
2
3
0
2
3
1
5
4
15
5
_
_
c
d
4
15
5
cd +
2
9
4
15
5c−
8
45
cd +
2
9
d2 +
4
9
4
15
5d +
4
9
8
45
4
15
Would require that c =
−
2
5
2
3
0
2
3
1
5
5d +
2
3 5
2
4
9
1
−5
3 5
,d =
−
3 5
1
3
−5
3 5
2
3
0
2
3
1
5
4
15
5
5
T
2
5
41
45
5c−
.
4
15
c2 +
4
15
1
3
−
2
5
2
5
T
2
3 5
1 0 0
−5
3 5
4
15
=
0 1 0
so this seems to have
0 0 1
5
worked.
113
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
6
A=
0
0
0 −1
1
0
−1
1
.
T
1
0
0
0
− 12
1
2
0
2
2
1
2
1
2
6
0
1
0
2
0 −1
1
0
2
0
−1
0
1
0
− 12
1
2
0
1
2
1
2
2
2
6
2
2
=
0
0
0 −2 0
0
0
0
114
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
A=
13
1
4
1
13
4
4
4
10
.
Hint: Two eigenvalues are 12 and 18.
13
1
4
1
13
4
4
4
10
, eigenvectors:
− 16 6
− 16 6
1
3
− 12 2
↔ 6,
1
2
2 3
↔ 12,
2
0
1
3
1
3
3
1
3
3
3
↔ 18. The
matrix U has these as its columns.
115
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
A=
3 0
0
0
3
2
1
2
0
1
2
3
2
−
1
2
0
0
3
2
1
2
0
1
2
3
2
.
, eigenvectors:
0
1
2
3 0
0
2
↔ 1,
2
1
1
2
2
1
2
2
↔ 2,
0
↔ 3. These vectors are
0
the columns of the matrix U.
116
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
17 −7 −4
A=
−7 17 −4
−4 −4 14
Hint: Two eigenvalues are 18 and 24.
17 −7 −4
−7 17 −4
−4 −4 14
, eigenvectors:
.
1
3
3
1
3
1
3
3
− 16 6
↔ 6,
− 12 2
↔ 18,
− 16 6
1
3
3
1
2
2 3
↔ 24 . The
2
0
matrix U has these as its columns.
117
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
3
A=
0
0
0 −1
1
0
−1
1
.
T
1
0
0
0
− 12
1
2
0
2
2
1
2
1
2
3
0
1
0
2
0 −1
1
0
2
0
−1
0
1
0
− 12
1
2
0
1
2
1
2
2
2
3
=
2
0
0 −2 0
0
2
0
0
0
118
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
1
A=
0
0
0 −4
1
0
−4
1
.
T
1
0
0
0 − 12 2
0
1
2
2
1
2
1
2
0
0
1
2
0 −4
1
0 − 12 2
2
0
−4
0
1
1
0
1
2
0
1
2
1
2
2
1
=
2
2
0
0
0 −5
0
0
−3
0
119
Solution:
Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding
an orthogonal matrix, U and a diagonal matrix D such that U T AU = D.
3 0 0
A=
.
0 2 1
0 1 2
T
1
0
0
0 − 12 2
0
1
2
2
1
2
1
2
3 0 0
1
0
2
0 2 1
0 − 12 2
2
0 1 2
0
1
2
0
2
1
2
1
2
3 0 0
2
2
=
0 1 0
0 0 3
120
Solution:
Explain why a matrix, A is symmetric if and only if there exists an orthogonal matrix, U such
that A = U T DU for D a diagonal matrix.
If A is given by the formula, then
A T = U T D T U = U T DU = A
Next suppose A = A T . Then by the theorems on symmetric matrices, there exists an orthogonal
matrix U such that
UAU T = D
for D diagonal. Hence
A = U T DU
121
Solution:
The proof of a theorem concluded with the following observation. If −ta + t 2 b ≥ 0 for all t ∈ R
and b ≥ 0, then a = 0. Why is this so?
If a ≠ 0, then the derivative of the function ft = −ta + t 2 b is non zero at t = 0. However, this
requires that ft < 0 for values of t near 0.
122
Solution:
Using Schur’s theorem, show that whenever A is an n × n matrix, detA equals the product of
the eigenvalues of A.
There exists U unitary such that A = U ∗ TU such that T is uppser triangular. Thus A and T are
similar. Hence they have the same determinant. Therefore, detA = detT, but detT equals the
product of the entries on the main diagonal which are the eigenvalues of A.
123
Solution:
In the proof of a theorem, the following argument was used. If x ⋅ w = 0 for all w ∈ R n , then
x = 0. Why is this so?
Let w = x.
124
Solution:
Suppose A is a 3 × 2 matrix. Is it possible that A T is one to one? What does this say about A
being onto? Prove your answer.
A T is a 2 × 3 and so it is not one to one by an earlier corollary. Not every column can be a pivot
column.
125
Solution:
Find the least squares solution to the system
6y − 3x = 3
x + 3y = 2
x + 4y = 1
29
− 275
11
25
126
Solution:
Find the least squares solution to the system
3y − 3x = 3
2x + 3y = 2
x + 6y = 1
− 31
83
34
83
127
Solution:
Find the least squares solution to the system
2x + y = 3
x + 3y = 1
x + 4y = 1
118
75
4
− 25
128
Solution:
You are doing experiments and have obtained the ordered pairs,
0, 1, 1, 2, 2, 3. 5, 3, 4
Find m and b such that y = mx + b approximates these four points as well as possible. Now do the
same thing for y = ax 2 + bx + c, finding a, b, and c to give the best approximation.
Here is the second of the two. You really want the following system.
0a + 0b + c = 1
a+b+c = 2
4a + 2b + c = 3. 5
9a + 3b + c = 4
T
0 0 1
0 0 1
1 1 1
1 1 1
4 2 1
4 2 1
9 3 1
9 3 1
98 36 14
=
36 14
6
14
4
6
98 36 14
a
36 14
6
b
14
4
c
1 1 1 1
98 36 14
a
52. 0
36 14
6
b
14
4
c
6
6
1
0 1 4 9
=
0 1 2 3
=
52. 0
2
=
3. 5
10. 5
4
−0. 125
, Solution is:
21. 0
1. 425
10. 5
0. 925
y
-4
21. 0
5
-2
2
4
-5
x
129
Solution:
You are doing experiments and have obtained the ordered pairs,
0, 2. 0, 1, 1. 0, 2, 2. 0, 3, 2. 0
Find m and b such that y = mx + b approximates these four points as well as possible. Now do the
same thing for y = ax 2 + bx + c, finding a, b, and c to give the best approximation.
Here is the second of the two. You really want the following system.
0a + 0b + c = 2. 0
a + b + c = 1. 0
4a + 2b + c = 2. 0
9a + 3b + c = 2. 0
T
0 0 1
0 0 1
1 1 1
1 1 1
4 2 1
4 2 1
9 3 1
9 3 1
98 36 14
=
36 14
6
14
4
6
98 36 14
a
36 14
6
b
14
4
c
1 1 1 1
98 36 14
a
27. 0
36 14
6
b
14
4
c
6
6
0 1 4 9
=
=
0 1 2 3
11. 0
7. 0
2. 0
1. 0
2. 0
27. 0
=
7. 0
2. 0
0. 25
, Solution is:
11. 0
−0. 65
1. 85
0. 25x 2 − 0. 65x + 1. 85
y 10
5
-4
-2
0
2
4
x
130
Solution:
You are doing experiments and have obtained the ordered pairs,
0, 7. 0, 1, −3. 0, 2, 1. 0, 3, 1. 0
Find m and b such that y = mx + b approximates these four points as well as possible. Now do the
same thing for y = ax 2 + bx + c, finding a, b, and c to give the best approximation.
Here is the second of the two. You really want the following system.
0a + 0b + c = 7. 0
a + b + c = −3. 0
4a + 2b + c = 1. 0
9a + 3b + c = 1. 0
T
0 0 1
0 0 1
1 1 1
1 1 1
4 2 1
4 2 1
9 3 1
9 3 1
98 36 14
=
36 14
6
14
4
6
98 36 14
a
36 14
6
b
14
4
c
1 1 1 1
98 36 14
a
10. 0
36 14
6
b
14
4
c
6
6
7. 0
0 1 4 9
=
=
10. 0
−3. 0
0 1 2 3
=
1. 0
6. 0
1. 0
2. 5
−8. 9
, Solution is:
2. 0
6. 0
6. 1
2. 5x 2 − 8. 9x + 6. 1
y 100
50
-4
131
-2
0
2
2. 0
4
x
Solution:
Suppose you have several ordered triples, x i , y i , z i . Describe how to find a polynomial,
z = a + bx + cy + dxy + ex 2 + fy 2
for example giving the best fit to the given ordered triples. Is there any reason you have to use a
polynomial? Would similar approaches work for other combinations of functions just as well?
You do something similar to the above. You write an equation which would be satisfied if each
x i , y i , z i  were a real solution. Then you obtain the least squares solution.
132
Solution:
The two level surfaces, 2x + 3y − z + w = 0 and 3x − y + z + 2w = 0 intersect in a subspace of
R 4 , find a basis for this subspace. Next find an orthonormal basis for this subspace.
Using row reduced echelon form, it is easy to obtain
−2z − 7w
w + 5z
11z
11w
Thus a basis is
−2
−7
5
1
,
11
0
0
11
Then an orthonormal basis is
46
− 3135
6 209
1
− 15
6
1
6
6
11
30
6
1
1254
,
6 209
1
− 330
6 209
5
209
0
6 209
133
Solution:
−3
Find an orthonormal basis for the span of the vectors
1
0
3
− 10
10
1
10
10
0
134
Solution:
,
1
110
110
3
110
110
1
11
10 11
− 111 11
,
− 113 11
1
11
11
1
,
1
4
0
,
0
1
.
The set, V ≡ x, y, z : 2x + 3y − z = 0 is a subspace of R 3 . Find an orthonormal basis for
this subspace.
The subspace is of the form
x
y
2x + 3y
1
and a basis is
0
,
0
Therefore, an orthonormal basis is
1
2
3
1
5
− 353 5 14
5
,
0
2
5
5
1
14
5 14
3
70
5 14
135
Solution:
Find an orthonormal basis for the span of the vectors 3, −4, 0, 7, −1, 0, 1, 7, 1.
3/5
4/5
−4/5
,
0
,
3/5
0
0
0
1
136
Solution:
Find an orthonormal basis for the span of the vectors 3, 0, −4, 5, 0, 10, −7, 1, 1
3
5
0
−
4
5
,
0
,
0
4
5
3
5
1
0
137
Solution:
Using the Gram Schmidt process, find an orthonormal basis for the span of the vectors,
1, 2, 1, 0, 2, −1, 3, 1, and 1, 0, 0, 1. To do this, have the computer use the QR factorization.
1
2
1
2 −1 0
1
3
0
0
1
1
=
1
6
1
3
6
1
6
0
2 3
6
1
6
− 29
6
5
18
2 3
1
9
2 3
2 3
5
111
1
333
3 37
3 37
17
− 333
3 37
22
333
3 37
7
111
2
− 111
111
111
− 371 111
7
− 111
111
1
2
6
3
2
2 3
5
18
0
0
1
9
0
0
0
⋅
1
6
6
6
2 3
3 37
0
Then a solution is
1
6
1
3
6
1
6
6
1
6
6
2 3
− 29 2 3
,
5
18
1
9
0
,
2 3
2 3
5
111
3 37
1
333
17
− 333
22
333
3 37
3 37
3 37
138
Solution:
3
Find an orthonormal basis for the span of the vectors
1
0
3
10
1
10
1
− 410
410
10
3
410
,
10
2
41
0
,
410
10 41
2
41
− 416
41
1
41
41
1
,
1
4
0
,
0
.
1
41
139
Solution:
Find an orthonormal basis for the span of the vectors 3, 0, −4, 11, 0, 2, 1, 1, 7
3
11 1
0
0
−4
2
−
4
5
4
5
,
4
5
0
−
7
3
5
0
=
1
3
5
0
3
5
4
5
5
−5
0 1
0 10
5
3
5
0
1
0
0
5
0
0
,
1
0
140
Solution:
Let A, B be a m × n matrices. Define an inner product on the set of m × n matrices by
A, B F ≡ traceAB ∗ .
Show this is an inner product satisfying all the inner product axioms. Recall for M an n × n matrix,
n
traceM ≡ ∑ i=1 M ii . The resulting norm, ||⋅||F is called the Frobenius norm and it can be used to
measure the distance between two matrices.
It satisfies the properties of an inner product. Note that
traceAB ∗  =
∑ ∑ A ik B ik = ∑ ∑ A ik B ik = traceBa
k
i
k
i
so
A, B F = B, A F
The product is obviously linear in the first argument. If A, A F = 0, then
∑ ∑ A ik A ik = ∑|A ik |2 = 0
k
i
i,k
141
Solution:
Let A be an m × n matrix. Show
||A||2F ≡ A, A F =
∑ σ 2j
j
where the σ j are the singular values of A.
From the singular value decomposition,
σ 0
U ∗ AV =
0 0
σ 0
,A=U
V∗
0 0
Then
σ 0
traceAa = trace U
0 0
σ2 0
= trace U
0
0
V∗ V
U∗
σ 0
0 0
= trace
U∗
σ2 0
0
0
142
Solution:
Here is a matrix A
5
0
0
0
3
10
2 10 − 101 2 10
0
3
10
2 10 − 101 2 10
Find a singular value decomposition.
1
A=
0
0
0
1
2
1
2
0
5 0 0
1
2 − 12 2
0 2 0
0
1
2
0 0 0
0
2
143
Solution:
Here is a matrix A
2
0
3
10
1
10
0
10 − 101 10
10
3
10
10
=
∑ σ 2j
j
3 +
1
3
5
2
−
2
3
1
3
5
2
3
15
22
1
11
11 −
3 11 −
1
11
3 11 +
11 +
2 3 11 −
1
33
3
15
22
2
11
1
33
1
33
2 3 11
2 3 11
2 3 11 −
1
33
1
11
3 11 −
1
11
3 11
1
11
5
22
3 11 −
2 11 −
3 11 +
2
33
5
22
1
33
2 11
2 3 11
2 3 11
Find a singular value decomposition.
A=
1
2
2
1
6
1
2
2 − 16 2 3
1
3
0
2 3
2 3
− 13 3
1
3
1
3
5 0 0
1
2
2
3
0 2 0
1
2
2 − 223 2 11
3
0 0 1
0
3
22
1
11
2 11
2 11
− 111 11
1
11
3
11
11
11
144
Solution:
Here is a matrix.
7
26
− 653
10 13
10 13
3
26
21
65
2 5 13
2 5 13
Find a singular value decomposition.
2
13
13
3
13
3
13
2
13 − 13
13
13
4 0
1
10
10
3
10
0 3
3
10
10 − 101 2 5
2 5
145
Solution:
The trace of an n × n matrix M is defined as ∑ i M ii . In other words it is the sum of the entries
on the main diagonal. If A, B are n × n matrices, show traceAB = traceBA. Now explain why
if A = S −1 BS it follows traceA = traceB. Hint: For the first part, write these in terms of
components of the matrices and it just falls out.
traceAB = ∑ i ∑ k A ik B ki , traceBA = ∑ i ∑ k B ik A ki . These give the same thing. Now
traceA = traceS −1 BS = traceBSS −1  = traceB.
146
Solution:
Using the above problem and Schur’s theorem, show that the trace of an n × n matrix equals the
sum of the eigenvalues.
A = UTU ∗ for some unitary U. Then the trace of A equals the trace of T. However, the trace of
T is just the sum of the eigenvalues of A.
147
Solution:
If A is a general n × n matrix having possibly repeated eigenvalues, show there is a sequence
A k  of n × n matrices having distinct eigenvalues which has the property that the ij th entry of A k
converges to the ij th entry of A for all ij. Hint: Use Schur’s theorem.
A = U ∗ TU where T is upper triangular and U is unitary. Change the diagonal entries of T
slightly so that the resulting upper triangular matrix T k has all distinct diagonal entries and T k → T
in the sense that the ij th entry of T k converges to the ij th entry of T. Then let A k = U ∗ T k U. It
follows that A k → A in the sense that corresponding entries converge.
148
Solution:
Prove the Cayley Hamilton theorem as follows. First suppose A has a basis of eigenvectors
v k  nk=1 , Av k = λ k v k . Let pλ be the characteristic polynomial. Show pAv k = pλ k v k = 0.
Then since v k  is a basis, it follows pAx = 0 for all x and so pA = 0. Next in the general
case, use the above problem to obtain a sequence A k  of matrices whose entries converge to the
entries of A such that A k has n distinct eigenvalues. Explain why A k has a basis of eigenvectors.
Therefore, from the first part and for p k λ the characteristic polynomial for A k , it follows
p k A k  = 0. Now explain why and the sense in which
lim p k A k  = pA.
k→∞
First say A has a basis of eigenvectors v 1 , ⋯, v n . A k v j = λ kj v j . Then it follows that
n
pAv k = pλ k v k . Hence if x is any vector, let x = ∑ k=1 x k v k and it follows that
n
pAx = pA
∑ xkvk
n
=
k=1
=
∑ x k pAv k
k=1
n
n
k=1
k=1
∑ x k pλ k v k = ∑ x k 0v k = 0
Hence pA = 0. Now drop the assumption that A is nondefective. From the above, there exists a
sequence A k which is non defective which converges to A and also p k λ → pλ uniformly on
bounded intervals because these characteristic polynomials are defined in terms of determinants of
the corresponding matrix. Thus the coefficients of p k converge to the corresponding coefficients of
p. See the above construction of the A k . It is probably easiest to use the Frobinius norm for the last
part.
‖p k A k  − pA‖ F ≤ ‖p k A k  − pA k ‖ F + ‖pA k  − pA‖ F
The first term converges to 0 because the convergence of A k to A implies all entries of A k lie in a
bounded set. The second term converges to 0 also because the entries of A k converge to the
corresponding entries of A. Also, p k A k  = 0 from the first part.
149
Solution:
Show that the Moore Penrose inverse A + satisfies the following conditions.
AA + A = A, A + AA + = A + , A + A, AA + are Hermitian.
Next show that if A 0 satisfies the above conditions, then it must be the Moore Penrose inverse and
that if A is an n × n invertible matrix, then A −1 satisfies the above conditions. Thus the Moore
Penrose inverse generalizes the usual notion of inverse but does not contradict it. Hint: Let
U ∗ AV = Σ ≡
σ 0
0 0
and suppose
V+ A 0 U =
P Q
R S
where P is the same size as σ. Now use the conditions to identify P = σ, Q = 0 etc.
First recall what the Moore Penrose invers was in terms of the singular value decomposition. It
equals V
σ −1 0
0
σ 0
U ∗ where A = U
0
V ∗ with U, V unitary and of the right size.
0 0
Therefore,
σ 0
AA + A = U
0 0
σ 0
=U
σ −1 0
V∗ V
0
U∗U
0
σ 0
V∗
0 0
V∗ = A
0 0
Next
A + AA + = V
=V
σ −1 0
0
0
σ −1 0
0
σ 0
U∗U
V∗ V
0 0
σ −1 0
0
0
U∗
U∗ = A+
0
Next,
A+A = V
σ −1 0
0
σ 0
U∗U
0
I 0
V∗ = V
0 0
0 0
V∗
which is clearly Hermitian.
+
AA = U
σ 0
∗
V V
0 0
σ −1 0
0
0
I 0
U∗ = U
0 0
U∗
which is Hermitian.
Next suppose A 0 satisfies the above Penrose conditions. Then
A0 = V
P Q
R S
U∗, A = U
σ 0
V∗
0 0
and it is necessary to identify P, Q, R, S. Here P is the same size as σ.
A0A = V
=V
P Q
R S
σP ∗ σR ∗
0
σ 0
U∗U
0 0
V∗ = V
Pσ 0
Rσ 0
V∗
V∗
0
and so Rσ = 0 which implies R = 0.
AA 0 = U
=U
σ 0
0 0
P∗ σ 0
∗
Q σ 0
V∗ V
U∗
P Q
0 S
U∗ = U
σP σQ
0
0
U∗
Hence Q ∗ σ = 0 so Q = 0.
σ 0
A 0 AA 0 = V
P 0
=V
P 0
σ 0
P 0
0 S
0 0
0 S
=V
PσP 0
U∗ = A0 = V
U∗U
0 S
0
0
P 0
V∗ V
0 0
0 S
U∗
U∗
P 0
U∗
0 S
and so S = 0. Finally,
AA 0 A = U
=U
=U
σ 0
V∗ V
0 0
P 0
0 0
σ 0
P 0
σ 0
0 0
0 0
0 0
σPσ 0
0
0
V∗ = U
σ 0
U∗U
σ 0
0 0
V∗
V∗
V∗
0 0
Then σPσ = σ and so P = σ −1 . Hence A 0 = A + .
150
Solution:
Find the least squares solution to
1
1
1
1
x
y
1 1+
a
=
b
c
Next suppose  is so small that all  2 terms are ignored by the computer but the terms of order 
are not ignored. Show the least squares equations in this case reduce to
3+
x
3 +  3 + 2
y
3
=
a+b+c
a + b + 1 + c
.
Find the solution to this and compare the y values of the two solutions. Show that one of these is
−2 times the other. This illustrates a problem with the technique for finding least squares solutions
presented as the solutions to aAx = ay. One way of dealing with this problem is to use the QR
factorization. This is illustrated in the next problem. It turns out that this helps alleviate some of
the round off difficulties of the above.
The least squares problem is
1 1
1
1 1 +1
1
1
1
1
1 1+
x
y
=
1 1
1
1 1 +1
a
b
c
which reduces to
+3
3
x
 + 3  2 + 2 + 3
a+b+c
=
y
a + b + c + 1
and now, since the stupid computer doesn’t see  2 , what it sees is the following.
+3
x
 + 3 2 + 3
y
3
=
a+b+c
a + b + c + 1
This yields the solution
1
3 2
 2 a + b + c −  + 3a + b − 2c
1
3 2
3a + 3b − 6c
So what is the real least squares solution?
1
6 2
2 2 a + b + c +  + 3a + b − 2c
− 61 2 3a + 3b − 6c
The two y values are very different one is −2 times the other.
151
Solution:
Show that the equations aAx = ay can be written as R ∗ Rx = R ∗ Q ∗ y where R is upper
triangular and R ∗ is lower triangular. Explain how to solve this system efficiently. Hint: You first
find Rx and then you find x which will not be hard because R is upper triangular.
A = QR and so a = R ∗ Q ∗ . Hence the least squares problem reduces to
R ∗ Q ∗ QRx = R ∗ Q ∗ y
R ∗ Rx =R ∗ Q ∗ y
Then you can solve this by first solving R ∗ z = R ∗ Q ∗ y. Next, after finding z, you would solve for
x in Rx = z.
Numerical Methods For Linear Systems
152
Solution:
Solve the system
4 1 1
x
1 5 2
y
0 2 6
z
2
=
1
3
using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using
row operations.
4 1 1
x
1 5 2
y
0 2 6
z
2
=
0. 39
, Solution is:
1
3
−0. 09
0. 53
153
Solution:
Solve the system
6 1
1
x
1 13 2
y
0 2
z
5
1
=
2
3
using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using
row operations.
6
1 1
x
1 13 2
y
0
z
2 5
6. 060 6 × 10 −2
1
=
2
, Solution is:
6. 060 6 × 10 −2
3
0. 575 76
154
Solution:
Solve the system
10 1 1
x
1
8 2
y
0
2 6
z
1
=
2
3
using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using
row operations.
10 1 1
x
1
8 2
y
0
2 6
z
4. 128 4 × 10 −2
1
=
2
, Solution is:
0. 130 73
3
0. 456 42
155
Solution:
Solve the system
12 1 1
x
1
8 2
y
0
2 7
z
1
=
2
3
using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using
row operations.
12 1 1
x
1
8 2
y
0
2 7
z
3. 877 2 × 10 −2
1
=
, Solution is:
2
0. 148 63
3
0. 386 11
156
Solution:
Solve the system
13
1
1
x
1
13 2
y
0
2
z
5
1
=
2
3
using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using
row operations.
13
1
1
x
1
13 2
y
0
2
z
5
2. 784 8 × 10 −2
1
=
, Solution is:
2
3
6. 329 1 × 10 −2
0. 574 68
157
Solution:
If you are considering a system of the form Ax = b and A −1 does not exist, will either the Gauss
Seidel or Jacobi methods work? Explain. What does this indicate about using either of these
methods for finding eigenvectors for a given eigenvalue?
When this method works, there is a unique solution and so these methods are no good for the
eigenvalue problem.
Numerical Methods For Eigenvalues
158
Solution:
Using the power method, find the eigenvalue correct to one decimal place having largest
0 −4 −4
absolute value for the matrix A =
along with an eigenvector associated with
7 10 5
−2
0
6
this eigenvalue.
20
0
−4 −4
7
10
5
1
−2
0
6
1
−1729 412 493 480 034 304
1
=
1729 447 677 852 123 136
1729 348 721 805 623 296
−1729 412 493 480 034 304
1
1729 447 677 852 123 136
−0. 999 98
=
1729 447 677 852 123 136
1. 0
1729 348 721 805 623 296
0. 999 94
−1
It appears that an eigenvector is
. Does it work?
1
1
0
−4 −4
−1
7
10
5
1
−2
0
6
1
−8
=
−1
=8
8
8
.
1
1
159
Solution:
Using the power method, find the eigenvalue correct to one decimal place having largest
15 6 1
absolute value for the matrix A =
along with an eigenvector associated with this
−5 2 1
1
2 7
eigenvalue.
20
15 6 1
1
−5 2 1
1
1
1
2 7
1. 150 0 × 10 22
=
−5. 748 9 × 10 21
1. 729 4 × 10 18
1. 150 0 × 10 22
1
1. 150 0×10 22
−5. 748 9 × 10 21
1. 729 4 × 10
1. 0
=
−0. 499 9
1. 503 8 × 10 −4
18
1
−. 5
This is pretty close to
. How well does this work?
0
15 6 1
1
−5 2 1
−. 5
1
2 7
0
12. 0
=
1
= 12
−6. 0
0
−. 5
0
160
Solution:
Using the power method, find the eigenvalue correct to one decimal place having largest
10 4 2
absolute value for the matrix A =
along with an eigenvector associated with
−3 2 −1
0
this eigenvalue.
0
4
20
10 4
2
1
−3 2 −1
1
0
1
0
4
3. 458 8 × 10 18
=
−1. 729 4 × 10 18
1. 099 5 × 10 12
3. 458 8 × 10 18
1
3. 458 8×10 18
−1. 729 4 × 10
1. 0
=
18
−0. 5
3. 178 8 × 10 −7
1. 099 5 × 10 12
1
−. 5
This looks a lot like
. How well does this work?
0
10 4
2
1
−3 2 −1
0
0
8. 0
=
−. 5
4
1
=8
−4. 0
0
−. 5
0
0
161
Solution:
Using the power method, find the eigenvalue correct to one decimal place having largest
15 14 −3
absolute value for the matrix A =
along with an eigenvector associated
−13 −18 9
5
−1
10
with this eigenvalue.
20
14
−3
−13 −18
9
1
−1
1
15
5
10
−6. 044 6 × 10 23
1
=
1. 208 9 × 10 24
−6. 044 6 × 10 23
−6. 044 6 × 10 23
1
1. 208 9×10 24
1. 208 9 × 10
24
−0. 500 01
=
1. 0
−0. 500 01
−6. 044 6 × 10 23
−. 5
Looks like
1
−. 5
14
−3
−. 5
−13 −18
9
1
−1
−. 5
15
5
10
−. 5
8. 0
=
−16. 0
= −16
8. 0
162
Solution:
Find the eigenvalues and eigenvectors for the matrix
1
−. 5
1 2 3
2 2 1
3 1 4
First of all, note that the matrix is symmetric and so all the eigenvalues are real. You can get
fairly close to them by just graphing.
1 2 3
, characteristic polynomial: X 3 − 7X 2 + 15
2 2 1
3 1 4
X − 7X 2 + 15
3
y
10
-2 -10
-20
-30
2
4
6
x
You can see that the eigenvalues are close to −1. 5, 1. 8, 6. 5. Use the shifted inverse power
method.
−1
1 2 3
1 0 0
+ 1. 5
2 2 1
=
0 1 0
3 1 4
0 0 1
−1. 939 4 −2. 060 6
−1. 939 4
1. 151 5
0. 848 48
1
−2. 060 6 0. 848 48
1. 151 5
1
−1. 939 4
1. 151 5
0. 848 48
−2. 060 6 0. 848 48
1. 151 5
4. 432 0 × 10 14
1
4. 432 0 × 10 14
−2. 021 3 × 10 14
−1. 939 4 −2. 060 6
20
4. 424 2
1
4. 432 0×10 14
4. 424 2
=
−2. 021 3 × 10 14
−2. 111 0 × 10 14
1. 0
=
−2. 111 0 × 10 14
−0. 456 07
Thus we would expect
−0. 476 31
1. 0
−0. 456 07
to be close to an eigenvector for A. Then
−0. 476 31
1 2 3
1. 0
2 2 1
−0. 456 07
3 1 4
−0. 476 31
−1. 341 1
=
0. 611 55
and so the eigenvalue should be close to
0. 638 69
−1. 341 1.
−1. 341 1
1. 0
Check: −1. 341 1
−0. 456 07
−0. 476 31
=
0. 611 64
0. 638 78
so it works very well, to three decimal
places.
Next find the one close to 1. 8
−1
1 2 3
−0. 303 03 −0. 757 58 0. 757 58
1 0 0
− 1. 8
2 2 1
3 1 4
=
0 1 0
0 0 1
−0. 303 03 −0. 757 58 0. 757 58
1
0. 757 58
−2. 251 1
1
1
−5. 129 7×10 17
8. 122 6 × 10
=
2 2 1
−1. 583 4
3 1 4
1. 0
1. 261 1 × 10 17
=
8. 122 6 × 10 17
−5. 129 7 × 10 17
−1. 583 4
−5. 129 7 × 10 17
−0. 245 84
−2. 251 1
3. 679 7
−0. 245 84
17
1 2 3
0. 757 58
1
3. 679 7
1. 261 1 × 10 17
3. 679 7
20
−0. 757 58 −5. 822 5
3. 679 7
−0. 757 58 −5. 822 5
This ought to be close.
1. 0
−0. 412 64
=
−2. 658 5
1. 679 1
An eigenvalue ought to be clost to 1. 679 1.
−0. 245 84
Check: 1. 679 1
−0. 412 79
=
−1. 583 4
−2. 658 7
1. 0
It worked well. One could iterate some
1. 679 1
more if desired.
Next find the one close to 6. 5
−1
1 2 3
2 2 1
1 0 0
− 6. 5
3 1 4
1. 673 5 1. 306 1 2. 530 6
=
0 1 0
1. 306 1 0. 775 51 1. 877 6
0 0 1
2. 530 6 1. 877 6 3. 387 8
20
1. 673 5 1. 306 1 2. 530 6
1
1. 306 1 0. 775 51 1. 877 6
1
2. 530 6 1. 877 6 3. 387 8
1
5. 758 6 × 10 15
1
8. 067 9×10 15
4. 201 0 × 10
15
0. 713 77
2 2 1
0. 520 71
3 1 4
1. 0
=
4. 201 0 × 10 15
8. 067 9 × 10 15
0. 713 77
=
8. 067 9 × 10 15
1 2 3
5. 758 6 × 10 15
0. 520 71
1. 0
4. 755 2
=
3. 469 0
6. 662
eigenvalue is about 6. 662 and an
0. 713 77
eigenvector ought to be about
.
0. 520 71
1. 0
0. 713 77
Check: 6. 662
4. 755 1
=
0. 520 71
, very close to the matrix times this
3. 469 0
1. 0
6. 662
eigenvector.
−0. 245 84
0. 713 77
6. 662,
,
0. 520 71
1. 0
−1. 583 4
1. 679 1,
1. 0
,
−1. 341 1,
−0. 456 07
−0. 476 31
1. 0
163
Solution:
3 2 1
Find the eigenvalues and eigenvectors of the matrix A =
2 1 3
numerically. In this
1 3 2
case the exact eigenvalues are ± 3 , 6. Compare with the exact answers.
Graphing the characteristic equation, you see that the eigenvectors are close to 1. 7, −1. 7, 6. You
see this from graphing. In fact 6 is an eigenvalue and then you find the eigenvector the usual way.
1
. You need to find the others.
1
1
−1
3 2 1
1 0 0
− 1. 7
2 1 3
=
0 1 0
1 3 2
19. 471
19. 471
0 0 1
1. 289 6
4. 016 9
1
−14. 165
4. 016 9
10. 381
1
−1. 216 9 × 10 25
3. 260 8 × 10
4. 016 9
−14. 165
4. 016 9
10. 381
−1. 216 9 × 10 25
=
3. 260 8 × 10 24
8. 908 7 × 10 24
1. 0
1
−1. 216 9×10 25
8. 908 7 × 10 24
732.
1. 289 6
1
−5. 074 0
24
−5. 074 0
20
−5. 074 0 −14. 165
−5. 074 0 −14. 165
=
−0. 267 96
−0. 732 08
3 2 1
1. 0
2 1 3
−0. 267 96
1 3 2
−0. 732 08
1. 732
=
−0. 464 2
−1. 268
so the eigenvalue ought to be about 1.
1. 0
1. 732
=
−0. 267 96
Check: 1. 732
−0. 464 11
−0. 732 08
−1. 268 0
−1
3 2 1
−1. 168 8 5. 194 8 −3. 896 1
1 0 0
+ 1. 7
2 1 3
=
0 1 0
1 3 2
0 0 1
−19. 351
14. 286
−3. 896 1
14. 286
−10. 260
20
−1. 168 8 5. 194 8 −3. 896 1
−4. 066 7 × 10 23
1
5. 194 8
−19. 351
14. 286
1
−3. 896 1
14. 286
−10. 260
1
−4. 066 7 × 10 23
=
1. 517 7 × 10 24
−1. 111 1 × 10 24
−0. 267 95
1
1. 517 7×10 24
1. 517 7 × 10 24
−1. 111 1 × 10
5. 194 8
=
1. 0
−0. 732 09
24
3 2 1
−0. 267 95
2 1 3
1. 0
1 3 2
−0. 732 09
0. 464 06
=
−1. 732 2
1. 267 9
It appears the eigenvalue is −1. 732 2 or close to it. How well does it work?
−0. 267 95
−1. 732 2
0. 464 14
=
1. 0
−0. 732 09
−1. 732 2
of course 3 = 1. 732 1
1. 268 1
164
Solution:
3 2 1
Find the eigenvalues and eigenvectors of the matrix A =
2 5 3
numerically. The exact
1 3 2
eigenvalues are 2, 4 + 15 , 4 − 15 . Compare your numerical results with the exact values. Is it
much fun to compute the exact eigenvectors?
This time, I shall just raise to a high power and take a QR factorization. The first column of Q
will then be close to an eigenvector unless the first column of A is an eigenvector.
20
1. 448 7 × 10 17 2. 713 5 × 10 17 1. 632 8 × 10 17
3 2 1
2 5 3
1 3 2
0. 415 99
=
2. 713 5 × 10 17 5. 082 3 × 10 17 3. 058 1 × 10 17
1. 632 8 × 10 17 3. 058 1 × 10 17 1. 840 1 × 10 17
0. 806 58
0. 779 18 −7. 803 9 × 10
0. 468 86
−0. 585 95
0. 419 97
−2
−0. 621 92
0. 660 94
=
3. 482 5 × 10 17 6. 522 6 × 10 17 3. 924 8 × 10 17
0
1. 539 × 10 13
1. 324 1 × 10 13
0
0
1. 399 9 × 10 12
T
0. 415 99
0. 806 58
0. 419 97
3 2 1
0. 415 99
0. 779 18 −7. 803 9 × 10 −2 −0. 621 92
2 5 3
0. 779 18 −7. 803 9 × 10 −2 −0. 621 92
−0. 585 95
1 3 2
0. 468 86
0. 468 86
0. 660 94
0. 806 58
−0. 585 95
8. 768 9 × 10 −5 3. 295 6 × 10 −5
7. 873 0
8. 768 9 × 10 −5
1. 746 2
0. 641 05
−5
0. 641 05
0. 380 82
3. 295 6 × 10
Now it appears that one eigenvalue is close to 7. 873 0 and the others are the eigenvalues for the
1. 746 2 0. 641 05
lower right corner.
, eigenvalues: 2. 000 0, 0. 127 02. Note that 2 is an
0. 641 05 0. 380 82
−1
3 2 1
eigenvalue and you will see this if you try to take
2 5 3
1 0 0
−2
1 3 2
1 2 1 0
1 0
0 0 1
0
0 1 −1 0
, row echelon form:
2 3 3 0
3
0 1 0
1 3 0 0
0 0
0
0
−3
an eigenvector is
1
1
Check:
3 2 1
−3
2 5 3
1
1 3 2
1
−6
=
2
2
next, look for the one close to . 127
−1
3 2 1
2 5 3
1 0 0
−. 127
1 3 2
526. 16
526. 16
=
0 1 0
0 0 1
−3087. 5 4664. 4
10
18133.
−27395.
1
4664. 4
−27395.
41388.
1
−3087. 5
18133.
−27395.
4664. 4
−27395.
41388.
2. 133 9 × 10 46
1
−3087. 5
−3087. 5 4664. 4
=
−1. 253 3 × 10 47
1. 893 4 × 10 47
.
0. 419 97
0. 660 94
2. 133 9 × 10 46
1
1. 893 4×10 47
0. 112 7
=
−1. 253 3 × 10 47
1. 893 4 × 10
3 2 1
0. 112 7
2 5 3
−0. 661 93
1 3 2
1. 0
1. 0
0. 014 24
=
−0. 084 25
0. 126 91
1. 430 3 × 10 −2
0. 112 7
0. 126 91
−0. 661 93
47
−0. 661 93
=
−8. 400 6 × 10 −2
1. 0
so this worked well.
0. 126 91
Next look for the other one.
−1
3 2 1
2 5 3
1 0 0
− 7. 87
=
0 1 0
1 3 2
0 0 1
57. 869
108. 7
65. 414
1
108. 7
203. 45 122. 50
1
65. 414 122. 50 73. 578
1
1
4. 156 9×10 50
4. 156 9 × 10
=
2. 501 4 × 10 50
0. 533 88
2 5 3
1. 0
1 3 2
0. 601 75
1. 0
0. 601 75
108. 7
203. 45 122. 50
2. 219 3 × 10 50
=
4. 156 9 × 10 50
2. 501 4 × 10 50
1. 0
0. 601 75
4. 203 4
=
7. 873
so the eigenvalue is about 7. 873
4. 737 4
0. 533 88
Check: 7. 873
65. 414
0. 533 88
50
3 2 1
108. 7
65. 414 122. 50 73. 578
20
2. 219 3 × 10 50
57. 869
4. 203 2
=
7. 873
. Worked well.
4. 737 6
4 + 15 = 7. 873 0
165
Solution:
0 2 1
Find the eigenvalues and eigenvectors of the matrix A =
2 5 3
numerically. We don’t
1 3 2
know the exact eigenvalues in this case. Check your answers by multiplying your numerically
computed eigenvectors by the matrix.
The characteristic polynomial is x 3 − 7x 2 − 4x + 1.
x 3 − 7x 2 − 4x + 1
y
-2
2
-20
4
6
x
-40
-60
The eigenvalues are close to −. 7, . 4, 7. 5. Now we find these more closely and also the
eigenvectors.
−1
0 2 1
1 0 0
+. 7
2 5 3
=
0 1 0
1 3 2
88. 889
−11. 111
88. 889
−32. 963 3. 703 7
−11. 111 3. 703 7 0. 370 37
0 0 1
20
−236. 67
88. 889
88. 889
−32. 963 3. 703 7
1
−11. 111 3. 703 7 0. 370 37
1
−11. 111
2. 585 6 × 10 48
1
2. 585 6 × 10 48
−9. 693 9 × 10
−236. 67
=
−9. 693 9 × 10 47
1. 193 1 × 10 47
1. 0
1
2. 585 6×10 48
47
=
−0. 374 92
4. 614 4 × 10 −2
1. 193 1 × 10 47
−0. 703 70
0 2 1
1. 0
2 5 3
−0. 374 92
1 3 2
4. 614 4 × 10 −2
=
0. 263 83
−3. 247 2 × 10 −2
λ 1 = −0. 703 70
−0. 703 7
1. 0
Check: −0. 703 70
=
−0. 374 92
4. 614 4 × 10
0. 263 83
−2
−3. 247 2 × 10 −2
−1
0 2 1
−0. 990 34 −0. 120 77 0. 845 41
1 0 0
−. 4
2 5 3
=
0 1 0
1 3 2
0 0 1
−0. 990 34 −0. 120 77 0. 845 41
1
0. 845 41
−3. 526 6
1
−3. 045 5 × 10
12
5. 824 3 × 10 12
0. 845 41
−3. 526 6
−0. 242 19
1
5. 824 3×10 12
=
−0. 522 90
1. 0
1. 932 4
−1. 410 6 × 10 12
1
1. 932 4
−1. 410 6 × 10 12
1. 932 4
20
−0. 120 77 −0. 990 34
1. 932 4
−0. 120 77 −0. 990 34
=
−3. 045 5 × 10 12
5. 824 3 × 10 12
0 2 1
−0. 242 19
2 5 3
−0. 522 90
1 3 2
1. 0
−0. 045 8
=
−0. 098 88
0. 189 11
λ 2 = 0. 189 11
−4. 580 1 × 10 −2
−0. 242 19
=
−0. 522 90
check: 0. 189 11
−9. 888 6 × 10 −2
1. 0
worked well.
0. 189 11
−1
0 2 1
1 0 0
− 7. 5
2 5 3
1 3 2
5. 428 6 16. 0 9. 714 3
=
0 1 0
0 0 1
16. 0
46. 0
1
9. 714 3 28. 0 16. 857
6. 878 8 × 10
36
4. 174 9 × 10
36
=
6. 878 8 × 10 36
4. 174 9 × 10 36
1
2. 386 4 × 10 36
28. 0
2. 386 4 × 10 36
1
28. 0
46. 0
9. 714 3 28. 0 16. 857
20
5. 428 6 16. 0 9. 714 3
16. 0
0. 346 92
1
6. 878 8×10 36
=
1. 0
0. 606 92
0 2 1
0. 346 92
2. 606 9
2 5 3
1. 0
1 3 2
0. 606 92
4. 560 8
0. 346 92
2. 607 0
=
7. 514 6
λ 3 = 7. 514 6
Check: 7. 514 6
1. 0
0. 606 92
=
7. 514 6
4. 560 8
166
Solution:
0 2 1
Find the eigenvalues and eigenvectors of the matrix A =
2 0 3
numerically. We don’t
1 3 2
know the exact eigenvalues in this case. Check your answers by multiplying your numerically
computed eigenvectors by the matrix.
0 2 1. 0
2 0
3
1 3
2
0. 379 2
, eigenvectors:
0. 584 81
0. 717 08
↔ 4. 975 4,
0. 814 41
0. 439 25
↔ −0. 300 56,
0. 156 94
↔ −2. 674 9
−0. 795 85
−0. 558 66
0. 416 76
167
Solution:
3 2 3
Consider the matrix A =
2 1 4
and the vector 1, 1, 1 T . Estimate the distance
3 4 0
between the Rayleigh quotient determined by this vector and some eigenvalue of A.
Recall that there was a formula for this. Let
1 1 1
q=
3 2 3
1
2 1 4
1
3 4 0
1
= 22
3
3
Then there is an eigenvalue λ such that
3 2 3
λ − 22
3
≤
1
2 1 4
1
3 4 0
1
1
−
22
3
1
1
= 1 2
3
3
168
Solution:
1 2 1
Consider the matrix A =
2 1 4
and the vector 1, 1, 1 T . Estimate the distance
1 4 5
between the Rayleigh quotient determined by this vector and some eigenvalue of A.
1 1 1
q=
1 2 1
1
2 1 4
1
1 4 5
1
3
=7
Then
|λ − 7| ≤
1 2 1
1
2 1 4
1
1 4 5
1
3
1
−7
1
1
=
6
169
Solution:
Using Gerschgorin’s theorem, find upper and lower bounds for the eigenvalues of
3 2 3
A=
.
2 6 4
3 4 −3
From the bottom line, a lower bound is −10 From the second line, an upper bound is 12.
170
Solution:
The QR algorithm works very well on general matrices. Try the QR algorithm on the following
matrix which happens to have some complex eigenvalues.
A=
1
2
3
1
2
−1
−1 −1
1
Use the QR algorithm to get approximate eigenvalues and then use the shifted inverse power
method on one of these to get an approximate eigenvector for one of the complex eigenvalues.
The real parts of the eigenvalues are larger than −4. I will consider the matrix which comes from
adding 5I to the given matrix. Thus, consider
1
2
1
2
1
0
0
0
1
2
1
2
6
=
2
0
15
2
1
2
1
7
−1
−1 −1
6
1
2
−1
2 − 12 2
− 12 2
3
1
2
1
2
2
2 − 12 2
0
2
6
2
5
2
2
0
6
2
3
1
1
7
−1
0
−1 −1
6
0
2
1
2
11
2
2
=
6. 0
=
0
1
2
− 12
0
2
2
1
2
1
2
2
2
−0. 707 11 3. 535 5
1. 414 2
7. 5
0. 5
0
0. 5
5. 5
This is the upper Hessenberg matrix which goes with A.
35
−0. 707 11 3. 535 5
6. 0
1. 414 2
7. 5
0. 5
0
0. 5
5. 5
=
3. 582 3 × 10 29 4. 290 6 × 10 29 6. 739 6 × 10 29
6. 190 5 × 10 30 7. 422 7 × 10 30 1. 165 9 × 10 31
1. 409 7 × 10 30 1. 690 3 × 10 30 2. 655 3 × 10 30
This has a QR factorization with Q =
5. 633 4 × 10 −2
−0. 998 39
7. 287 3 × 10 −3
0. 973 49
5. 330 5 × 10 −2
−0. 222 43
0. 221 68
1. 962 4 × 10 −2
0. 974 92
Then
−0. 707 11 3. 535 5
6. 0
Q
T
1. 414 2
7. 5
0. 5
0
0. 5
5. 5
−1. 281 6 0. 230 07
7. 695 7
1. 310 2 × 10
Q=
−4
5. 898 3
−3. 601 3
−1. 781 5 × 10 −5 0. 310 73
5. 406 1
It is now clear that the eigenvalues are approximately those of
5. 898 3 −3. 601 3
0. 310 73
5. 406 1
and 7. 695 7. That is,
5. 652 2 + 1. 028 8i, 5. 652 2 − 1. 028 8i, 7. 695 7
Subtracting 5 gives the approximate eigenvalues for the original matrix,
0. 652 2 + 1. 028 8i, 0. 652 2 − 1. 028 8i, 2. 695 7
Lets find the eigenvector which goes with the first of the above.
−1
1
2
3
1
2
−1
−1 −1
1
1 0 0
− 0. 652 2 + 1. 028 8i
0 1 0
0 0 1
=
−4251. 5 − 8395. 3i
−16391. +3972. 6i
−26508. +5426. 7i
1411. 0 + 4647. 5i
8687. 2 − 554. 36i
13959. −389. 59i
2431. 6 − 3583. 0i
−5253. 6 − 5712. 0i −8094. 1 − 9459. 8i
−4251. 5 − 8395. 3i
−16391. +3972. 6i
−26508. +5426. 7i
1411. 0 + 4647. 5i
8687. 2 − 554. 36i
13959. −389. 59i
2431. 6 − 3583. 0i
−5253. 6 − 5712. 0i −8094. 1 − 9459. 8i
12
1
=
1
1
−3. 942 8 × 10 51 + 2. 747 1 × 10 51 i
2. 249 7 × 10 51 − 1. 044 0 × 10 51 i
−1. 984 8 × 10 51 − 9. 746 7 × 10 50 i
Then the first approximation will be
−3. 942 8 × 10 51 + 2. 747 1 × 10 51 i
1
−3. 942 8 × 10 51 + 2. 747 1 × 10 51 i
2. 249 7 × 10 51 − 1. 044 0 × 10 51 i
−1. 984 8 × 10 51 − 9. 746 7 × 10 50 i
1. 0
=
−0. 508 31 − 8. 937 5 × 10 −2 i
0. 222 94 + 0. 402 53i
−4251. 5 − 8395. 3i
−16391. +3972. 6i
−26508. +5426. 7i
1411. 0 + 4647. 5i
8687. 2 − 554. 36i
13959. −389. 59i
2431. 6 − 3583. 0i
−5253. 6 − 5712. 0i −8094. 1 − 9459. 8i
−3658. 8 − 18410. 0i
1. 0
−2
−0. 508 31 − 8. 937 5 × 10 i
=
214. 5 + 9684. 9i
6594. 9 − 5577. 1i
0. 222 94 + 0. 402 53i
The next approximate eigenvector is
−3658. 8 − 18410. 0i
214. 5 + 9684. 9i
6594. 9 − 5577. 1i
1
−3658. 8 − 18410. 0i
1. 0
=
−0. 508 31 − 8. 936 9 × 10 −2 i
0. 222 94 + 0. 402 53i
⋅
This didn’t change by much. In fact, it didn’t change at all. Thus the approximate eigenvalue is
obtained by solving
1
= −3658. 8 − 18410. 0i
λ − 0. 652 2 + 1. 028 8i
Thus
λ = 0. 652 19 + 1. 028 9i
and the approximate eigenvector is
1. 0
−0. 508 31 − 8. 936 9 × 10 −2 i
0. 222 94 + 0. 402 53i
Lets check it.
1
2
3
0. 652 2 + 1. 028 9i
1. 0
−2
2
−1
−0. 508 31 − 8. 936 9 × 10 i
−1 −1
1
0. 222 94 + 0. 402 53i
1
=
−0. 268 75 + 0. 491 90i
0. 652 19 + 1. 028 9i
1. 0
0. 652 19 + 1. 028 9i
−0. 239 56 − 0. 581 27i
=
−2
−0. 508 31 − 8. 936 9 × 10 i
0. 222 94 + 0. 402 53i
−0. 239 56 − 0. 581 29i
−0. 268 76 + 0. 491 91i
It worked very well. Thus you know the other eigenvalue for the other complex eigenvalue is
1. 0
−0. 508 31 + 8. 936 9 × 10 −2 i
0. 222 94 − 0. 402 53i
171
Solution:
Use the QR algorithm to approximate the eigenvalues of the symmetric matrix
1
2
3
2 −8 1
3
20
1
2
3
2 −8 1
3
1
0
=
1
0
1. 250 3 × 10 17
−6. 367 1 × 10 17
3. 099 4 × 10 16
−6. 367 1 × 10 17
3. 242 5 × 10 18
−1. 578 3 × 10 17
3. 099 4 × 10
−1. 578 3 × 10
7. 683 2 × 10
16
0. 192 47
0. 835 59
−0. 980 14
0. 189 19 5. 939 7 × 10 −2
4. 771 2 × 10 −2 0. 515 74
0. 514 53
−0. 855 41
17
15
=
6. 496 1 × 10 17 −3. 308 2 × 10 18 1. 610 3 × 10 17
0
1. 826 7 × 10 13
1. 076 7 × 10 12
0
0
3. 180 1 × 10 11
T
0. 192 47
0. 835 59
−0. 980 14
0. 189 19 5. 939 7 × 10 −2
4. 771 2 × 10
−2
−7. 008 4 × 10
−6
3
0. 192 47
2 −8 1
−0. 980 14
3
3. 825
−0. 845 44
−0. 845 44
−2. 383 7
20
−0. 845 44
=
−0. 845 44 −2. 383 7
1
0
4. 771 2 × 10
−0. 132 54 0. 991 18
7. 906 × 10 11
−1. 057 2 × 10 11
−1. 057 2 × 10 11
1. 422 7 × 10 10
0. 132 54
−0. 845 44
3. 825
−0. 132 54 0. 991 18
0. 991 18
−0. 845 44 −2. 383 7
3. 938 1
−9. 737 8 × 10 −5
−9. 737 8 × 10 −5
−2. 496 8
−0. 132 54 0. 991 18
Now the eigenvalues are close to −8. 441 4, 3. 938 1, −2. 496 8. To find eigenvectors,
−1
2
3
1 0 0
+ 8. 441 4
2 −8 1
3
1
=
0 1 0
0
0 0 1
−4375. 9
22285.
22285.
−1. 134 9 × 10
−1084. 8
5524. 2
−1084. 8
5
5524. 2
−268. 77
20
−4375. 9
22285.
−1084. 8
22285.
−1. 134 9 × 10 5
5524. 2
1
−1084. 8
5524. 2
−268. 77
1
−9. 894 1 × 10
2
3
2 −8 1
3
1
0
=
−0. 196 36
1
2. 032 6×10 101
2. 032 6 × 10 101
−3. 991 3 × 10 100
1
−3. 991 3 × 10 100
1
0. 515 74
=
0. 132 54
=
1. 0
−4. 867 7 × 10 −2
99
−0. 196 36
1. 0
−4. 867 7 × 10 −2
1. 657 6
=
−8. 441 4
0. 410 92
2. 032 6 × 10 101
−9. 894 1 × 10 99
0. 514 53
0. 189 19 5. 939 7 × 1
−2
8. 919 8 × 10 7
0
T
1
0. 835 59
7. 976 4 × 10 11 −1. 066 7 × 10 11
0. 132 54
0. 991 18
2
4. 960 7 × 10 −5 −7. 008 4 × 10 −6
4. 960 7 × 10 −5
0. 991 18
1
−0. 855 41
0. 515 74
−8. 441 4
3. 825
0. 514 53
=
−0. 855 41
−0. 196 36
8. 441 4
−1. 657 6
=
1. 0
−4. 867 7 × 10
,
8. 441 4
−2
−0. 410 9
−1
1
2
3
1 0 0
− 3. 938 1
2 −8 1
3
1
14217. 3360. 4 11683.
=
0 1 0
0
0 0 1
11683. 2761. 6 9601. 3
8
14217. 3360. 4 11683.
1
3360. 4 794. 21 2761. 6
1
11683. 2761. 6 9601. 3
1
1. 600 8 × 10 35
1. 315 5 × 10 35
1. 0
2
0. 236 37
35
0. 821 78
3
1. 0
2 −8 1
0. 236 37
3
0. 821 78
1
=
1
1. 600 8×10 35
3. 783 8 × 10 34
1
0
1. 0
3. 938 1
=
0. 930 82
3. 236 4
3. 938 1
3. 938 1 =
0. 236 37
0. 821 78
0. 930 85
3. 236 3
λ 2 = 3. 938 1
1
2
−1
3
1 0 0
+ 2. 496 8
2 −8 1
3
1
0
=
0 1 0
0 0 1
8
5680. 5
768. 27
−7133. 0
768. 27
103. 74
−964. 66
1
−7133. 0 −964. 66
8957. 4
1
−1. 400 5 × 10
31
=
1. 300 4 × 10 32
−0. 796 37
2 −8 1
−0. 107 70
3
1
0
−0. 107 70
1. 0
1. 988 2
=
768. 27
−7133. 0
768. 27
103. 74
−964. 66
−7133. 0 −964. 66
8957. 4
−1. 400 5 × 10 31
1. 300 4 × 10 32
1. 0
3
2
=
−0. 796 37
1
1. 300 4×10 32
5680. 5
−1. 035 6 × 10 32
1
−1. 035 6 × 10 32
1
3. 783 8 × 10 34
:=
1. 600 8 × 10 35
1. 315 5 × 10
3360. 4 794. 21 2761. 6
0. 268 86
−2. 496 8
−0. 796 37
1. 988 4
−2. 496 8 =
−0. 107 70
0. 268 91
−2. 496 8
1. 0
−0. 196 36
1. 0
3. 938 2,
,
0. 236 38
8. 441 4,
,
1. 0
−4. 867 7 × 10 −2
0. 821 82
−0. 796 37
−2. 496 8,
−0. 107 70
1. 0
172
Solution:
3
Try to find the eigenvalues of the matrix
3
1
−2 −2 −1
using the QR algorithm. It has
0 1 0
eigenvalues 1, i, −i. You will see the algorithm won’t work well.
To see that the method will not work well, consider the powers of this matrix.
2
3
3
1
3
=
−2 −2 −1
0
1
0
3
3
1
4
0
−2 −3
0
−2 −2 −1
3
=
−2 −2 −1
0
1
0
3
3
1
1
1
−1
0
0
1
−2 −3
0
4
−2 −2 −1
0
1
0
1 0 0
=
0 1 0
0 0 1
Thus the powers of the matrix will repeat forever. Therefore, you cannot expect things to get
small and be able to find the eigenvalues by looking at those of a block upper triangular matrix
which has very small entries below the block diagonal. Recall that, in terms of size of the entries,
you can at least theoretically consider the QR factorization of A k and then look at Q ∗ AQ however,
not much can happen given the fact that there are only finitely many A k . The problem here is that
there is no gap between the size of the eigenvalues. The matrix has eigenvalues equal to i, −i and
1. To produce a situation in which this unfortunate situation will not occur, simply add a multiple
of the identity to the matrix and use the modified one. This is always a reasonable idea. You can
identify an interval, using Gerschgorin’s theorem which will contain all the eigenvalues of the
matrix. Then when you add a large enough multiple of the identity, the result will have all positive
real parts for the eigenvalues. If you do this to this matrix, the problem goes away. The idea is to
produce gaps between the absolute values of the eigenvalues.
Fields
173
Solution:
Prove the Euclidean algorithm: If m, n are positive integers, then there exist integers q, r ≥ 0
such that r < m and
n = qm + r
Hint: You might try considering
S ≡ n − km : k ∈ N and n − km < 0
and picking the smallest integer in S or something like this.
The hint is a good suggestion. Pick the first thing in S. By the Archimedean property, S ≠ ∅.
That is km > n for all k sufficiently large. Call this first thing q + 1. Thus n − q + 1m < 0 but
n − qm ≥ 0. Then
n − qm < m
and so
0 ≤ r ≡ n − qm < m.
174
Solution:
The greatest common divisor of two positive integers m, n, denoted as q is a positive number
which divides both m and n and if p is any other positive number which divides both m, n, then p
divides q. Recall what it means for p to divide q. It means that q = pk for some integer k. Show
that the greatest common divisor of m, n is the smallest positive integer in the set S
S ≡ xm + yn : x, y ∈ Z and xm + yn > 0
Two positive integers are called relatively prime if their greatest common divisor is 1.
First note that either m or −m is in S so S is a nonempty set of positive integers. By well
ordering, there is a smallest element of S, called p = x 0 m + y 0 n. Either p divides m or it does not.
If p does not divide m, then by the above problem,
m = pq + r
where 0 < r < p. Thus m = x 0 m + y 0 nq + r and so, solving for r,
r = m1 − x 0  + −y 0 qn ∈ S.
However, this is a contradiction because p was the smallest element of S. Thus p|m. Similarly
p|n. Now suppose q divides both m and n. Then m = qx and n = qy for integers, x and y.
Therefore,
p = mx 0 + ny 0 = x 0 qx + y 0 qy = qx 0 x + y 0 y
showing q|p. Therefore, p = m, n.
175
Solution:
A positive integer larger than 1 is called a prime number if the only positive numbers which
divide it are 1 and itself. Thus 2,3,5,7, etc. are prime numbers. If m is a positive integer and p
does not divide m where p is a prime number, show that p and m are relatively prime.
Suppose r is the greatest common divisor of p and m. Then if r ≠ 1, it must equal p because it
must divide p. Hence there exist integers x, y such that
p = xp + ym
which requires that p must divide m which is assumed not to happen. Hence r = 1 and so the two
numbers are relatively prime.
176
Solution:
There are lots of fields. This will give an example of a finite field. Let Z denote the set of
integers. Thus Z = ⋯, −3, −2, −1, 0, 1, 2, 3, ⋯. Also let p be a prime number. We will say that
two integers, a, b are equivalent and write a ∼ b if a − b is divisible by p. Thus they are equivalent
if a − b = px for some integer x. First show that a ∼ a. Next show that if a ∼ b then b ∼ a.
Finally show that if a ∼ b and b ∼ c then a ∼ c. For a an integer, denote by a the set of all
integers which is equivalent to a, the equivalence class of a. Show first that is suffices to consider
only a for a = 0, 1, 2, ⋯, p − 1 and that for 0 ≤ a < b ≤ p − 1, a ≠ b. That is, a = r
where r ∈ 0, 1, 2, ⋯, p − 1. Thus there are exactly p of these equivalence classes. Hint: Recall
the Euclidean algorithm. For a > 0, a = mp + r where r < p. Next define the following
operations.
a + b ≡ a + b
ab ≡ ab
Show these operations are well defined. That is, if a = a ′  and b = b ′ , then
a + b = a ′  + b ′  with a similar conclusion holding for multiplication. Thus for addition you
need to verify a + b = a ′ + b ′  and for multiplication you need to verify ab = a ′ b ′ . For
example, if p = 5 you have 3 = 8 and 2 = 7. Is 2 × 3 = 8 × 7? Is 2 + 3 = 8 + 7?
Clearly so in this example because when you subtract, the result is divisible by 5. So why is this so
in general? Now verify that 0, 1, ⋯, p − 1 with these operations is a Field. This is called
the integers modulo a prime and is written Z p . Since there are infinitely many primes p, it follows
there are infinitely many of these finite fields. Hint: Most of the axioms are easy once you have
shown the operations are well defined. The only two which are tricky are the ones which give the
existence of the additive inverse and the multiplicative inverse. Of these, the first is not hard.
−x = −x. Since p is prime, there exist integers x, y such that 1 = px + ky and so 1 − ky = px
which says 1 ∼ ky and so 1 = ky. Now you finish the argument. What is the multiplicative
identity in this collection of equivalence classes?
The only substantive issue is why Z p is a field. Let x ∈ Z p where x ≠ 0. Thus x is not a
multiple of p. Then from the above problem, x and p are relatively prime. Hence from another of
the above problems, there exist integers a, b such that
1 = ap + bx
Then
1 − bx = ap = 0
and it follows that
bx = 1
−1
so b = x .
Abstract Vector Spaces
177
Solution:
Let M = u = u 1 , u 2 , u 3 , u 4  ∈ R 4 : |u 1 | ≤ 4. Is M a subspace? Explain.
No. 1, 0, 0, 0 ∈ M but 101, 0, 0, 0 ∉ M.
178
Solution:
Let M = u = u 1 , u 2 , u 3 , u 4  ∈ R 4 : sinu 1  = 1. Is M a subspace? Explain.
Absolutely not. Let u =  π2 , 0, 0, 0. Then 2u ∉ M although u ∈ M.
179
Solution:
If you have 5 vectors in F 5 and the vectors are linearly independent, can it always be concluded
they span F 5 ? Explain.
If not, you could add in a vector not in their span and obtain 6 vectors which are linearly
independent. This cannot occur thanks to the exchange theorem.
180
Solution:
If you have 6 vectors in F 5 , is it possible they are linearly independent? Explain.
No because there exists a spanning set of 5 vectors. Note that it doesn’t matter what F is here.
181
Solution:
Show in any vector space, 0 is unique.
Say 0 ′ is another one. Then 0 = 0 + 0 ′ = 0 ′ . This happens from the definition of what it means
for a vector to be an additive identity.
182
Solution:
↑In any vector space, show that if x + y = 0, then y = −x.
Say x + y = 0. Then −x = −x +x + y = −x + x + y = 0 + y = y. If it acts like the additive
inverse, it is the additive inverse.
183
Solution:
↑Show that in any vector space, 0x = 0. That is, the scalar 0 times the vector x gives the vector
0.
0x = 0 + 0x = 0x + 0x. Now add −0x to both sides to conclude that 0x = 0.
184
Solution:
↑Show that in any vector space, −1x = −x.
Lets show −1x acts like the additive inverse. x +−1x =1 + −1x = 0x = 0 from one of
the above problems. Hence −1x = − x.
185
Solution:
Let X be a vector space and suppose x 1 , ⋯, x k  is a set of vectors from X. Show that 0 is in
spanx 1 , ⋯, x k .
Pick any vector, x i . Then 0x i is in the span, but this was shown to be 0 above.
186
Solution:
Let X consist of the real valued functions which are defined on an interval a, b. For
f, g ∈ X, f + g is the name of the function which satisfies f + gx = fx + gx and for α a real
number, αfx ≡ αfx. Show this is a vector space with field of scalars equal to R. Also
explain why it cannot possibly be finite dimensional.
It doesn’t matter where the functions have their values provided it is some real vector space. The
axioms of a vector space are all routine because they hold for a vector space. The only thing
maybe not completely obvious is the assertions about the things which are supposed to exist. 0
would be the zero function which sends everything to 0. This is an additive identity. Now if f is a
function, −fx ≡ −fx. Then
f + −fx ≡ fx + −fx ≡ fx + −fx = 0
Hence f + −f = 0.
For each x ∈ a, b, let f x x = 1 and f x y = 0 if y ≠ x. Then these vectors are obviously
linearly independent.
187
Solution:
Let S be a nonempty set and let V denote the set of all functions which are defined on S and have
values in W a vector space having field of scalars F. Also define vector addition according to the
usual rule, f + gs ≡ fs + gs and scalar multiplication by αfs ≡ αfs. Show that V is a
vector space with field of scalars F.
This is no different than the above. You simply understand that the vector space has the same
field of scalars as the W.
188
Solution:
Verify that any field F is a vector space with field of scalars F. However, show that R is a vector
space with field of scalars Q.
A field also has multiplication. However, you can consider the elements of the field as vectors
and then it satisfies all the vector space axioms. When you multiply a number (vector) in R by a
scalar in Q you get something in R. All the axioms for a vector space are now obvious. For
example, if α ∈ Q and x, y ∈ R,
αx + y = αx + αy
from the distributive law on R.
189
Solution:
Let F be a field and consider functions defined on 1, 2, ⋯, n having values in F. Explain how,
if V is the set of all such functions, V can be considered as F n .
Simply let fi be the i th component of a vector x ∈ F n . Thus a typical thing in F n is
f1, ⋯, fn.
190
Solution:
Let V be the set of all functions defined on N ≡1, 2, ⋯ having values in a field F such that
vector addition and scalar multiplication are defined by f + gs ≡ fs + gs and
αfs ≡ αfs respectively, for f, g ∈ V and α ∈ F. Explain how this is a vector space and show
that for e i given by
e i k ≡
1 if i = k
0 if i ≠ k
,
the vectors e k  ∞k=1 are linearly independent.
n
Say for some n, ∑ k=1 c k e k = 0, the zero function. Then pick i,
n
0=
∑ c k e k i = c i e i i = c i
k=1
Since i was arbitrary, this shows these vectors are linearly independent.
191
Solution:
Suppose you have smooth functions y 1 , y 2 , ⋯, y n  (all derivatives exist) defined on an interval
a, b. Then the Wronskian of these functions is the determinant
Wy 1 , ⋯, y n x = det
y 1 x
⋯
y n x
y ′1 x
⋯
y ′n x
⋮
⋮
y n−1
x
1
⋯
y n−1
x
n
Show that if Wy 1 , ⋯, y n x ≠ 0 for some x, then the functions are linearly independent.
Say
n
∑ ckyk = 0
k=1
Then taking derivatives you have
n
∑ c k y jk = 0,
j = 0, 1, 2⋯, n − 1
k=1
This must hold when each equation is evaluated at x where you can pick the x at which the above
determinant is nonzero. Therefore, this is a system of n equations in n variables, the c i and the
coefficient matrix is invertible. Therefore, each c i = 0.
192
Solution:
Give an example of two functions, y 1 , y 2 defined on −1, 1 such that Wy 1 , y 2 x = 0 for all
x ∈ −1, 1 and yet y 1 , y 2  is linearly independent.
Let y 1 x = x 2 and let y 2 x = x|x|. Suppose c 1 y 1 + c 2 y 2 = 0 Then you could consider
evaluating at 1 and get
c1 + c2 = 0
and then at −1 and get
c1 − c2 = 0
then it follows that c 1 = c 2 = 0. Thus, these are linearly independent. However, if x > 0,
W=
x2 x2
2x 2x
=0
and if x < 0
W=
x 2 −x 2
2x −2x
Also, W = 0 if x = 0 because at this point, W =
0 0
0 0
=0
= 0.
193
Solution:
Let the vectors be polynomials of degree no more than 3. Show that with the usual definitions of
scalar multiplication and addition wherein, for px a polynomial, αpx = αpx and for p, q
polynomials p + qx ≡ px + qx, this is a vector space.
This is just a subspace of the vector space of functions because it is closed with respect to vector
addition and scalar multiplication. Hence this is a vector space.
194
Solution:
In the previous problem show that a basis for the vector space is 1, x, x 2 , x 3 .
Using one of the above problems, this is really easy if you take the Wronskian of these functions.
1 x x2
det
x3
0 1 2x 3x 2
0 0 2
6x
0 0 0
6
≠ 0.
195
Solution:
In the previous problem show that a basis for the vector space is 1, x, x 2 , x 3 .
a. x + 1, x 3 + x 2 + 2x, x 2 + x, x 3 + x 2 + x
i. Lets look at the Wronskian of these functions. If it is nonzero somewhere, then these are
linearly independent and are therefore a basis because there are four of them.
1 + x x 3 + x 2 + 2x x 2 + x
det
1
x3 + x2 + x
3x 2 + 2x + 2 2x + 1 3x 2 + 2x + 1
0
6x + 2
2
6x + 2
0
6
0
6
Lets plug in x = 0 and see what happens. You only need to have it be nonzero at one
point.
1 0 0 0
det
1 2 1 1
0 2 2 2
= 12 ≠ 0
0 6 0 6
so these are linearly independent.
b. x + 1, x 2 + x, 2x 3 + x 2 , 2x 3 − x 2 − 3x + 1
i. Suppose
c 1 x 3 + 1 + c 2 x 2 + x + c 3 2x 3 + x 2  + c 4 2x 3 − x 2 − 3x + 1 = 0
Then combine the terms according to power of x.
c 1 + 2c 3 + 2c 4 x 3 + c 2 + c 3 − c 4 x 2 + c 2 − 3c 4 x + c 1 + c 4  = 0
Is there a non zero solution to the system
3
c 1 + 2c 3 + 2c 4 = 0
c2 + c3 − c4 = 0
c 2 − 3c 4 = 0
, Solution is: c 1 = 0, c 2 = 0, c 3 = 0, c 4 = 0 Therefore, these are
c1 + c4 = 0
linearly independent.
196
Solution:
Let V be the space of polynomials of degree three or less. Consider a collection of polynomials in
this vector space which are of the form
a i x 3 + b i x 2 + c i x + d i , i = 1, 2, 3, 4
Show that this collection of polynomials is linearly independent on an interval a, b if and only if
a1 b1 c1 d1
a2 b2 c2 d2
a3 b3 c3 d3
a4 b4 c4 d4
is an invertible matrix.
Let p i x denote the i th of these polynomials. Suppose ∑ i C i p i x = 0. Then collecting terms
according to the exponent of x, you need to have
C1a1 + C2a2 + C3a3 + C4a4 = 0
C1b1 + C2b2 + C3b3 + C4b4 = 0
C1c1 + C2c2 + C3c3 + C4c4 = 0
C1d1 + C2d2 + C3d3 + C4d4 = 0
The matrix of coefficients is just the transpose of the above matrix. There exists a non trivial
solution if and only if the determinant of this matrix equals 0.
197
Solution:
Let the field of scalars be Q, the rational numbers and let the vectors be of the form a + b 2
where a, b are rational numbers. Show that this collection of vectors is a vector space with field of
scalars Q and give a basis for this vector space.
This is obvious because when you add two of these you get one and when you multiply one of
these by a scalar, you get another one. A basis is 1, 2 . By definition, the span of these gives
the collection of vectors. Are they independent? Say a + b 2 = 0 where a, b are rational numbers.
If a ≠ 0, then b 2 = −a which can’t happen since a is rational. If b ≠ 0, then −a = b 2 which
again can’t happen because on the left is a rational number and on the right is an irrational. Hence
both a, b = 0 and so this is a basis.
198
Solution:
Suppose V is a finite dimensional vector space. Based on the exchange theorem above, it was
shown that any two bases have the same number of vectors in them. Give a different proof of this
fact using the earlier material in the book. Hint: Suppose x 1 , ⋯, x n  and y 1 , ⋯, y m  are two
bases with m < n. Then define
φ : F n → V, ψ : F m → V
by
n
φa ≡
∑ akxk,
m
ψb ≡
k=1
∑ b jyj
j=1
−1
Consider the linear transformation, ψ ∘ φ. Argue it is a one to one and onto mapping from F n to
F m . Now consider a matrix of this linear transformation and its row reduced echelon form.
Since both of these are bases, both ψ and φ are one to one and onto. Therefore, the given linear
transformation is also one to one and onto. It has a matrix A which has more columns than rows
such that multiplication by A has the same effect as doing ψ −1 ∘ φ. However, the augmented
matrix A|0 has free variables because there are more columns than rows for A. Hence A cannot
be one to one. Therefore, ψ −1 ∘ φ also must fail to be one to one. This is a contradiction.
199
Solution:
This and the following problems will present most of a differential equations course. To begin
with, consider the scalar initial value problem
y ′ = ay, yt 0  = y 0
When a is real, show the unique solution to this problem is y = y 0 e at−t 0  . Next suppose
y ′ = a + iby, yt 0  = y 0
where yt = ut + ivt. Show there exists a unique solution and it is
yt = y 0 e at−t 0  cos bt − t 0  + i sin bt − t 0 
≡ e a+ibt−t 0  y 0 .
Next show that for a real or complex there exists a unique solution to the initial value problem
y ′ = ay + f, yt 0  = y 0
and it is given by
yt = e at−t 0  y 0 + e at ∫ e −as fsds.
t
t0
Consider the first part. First you can verify that yt = y 0 e at−t 0  works using elementary
calculus. Why is it the only one which does so? Suppose y 1 t is a solution. Then
yt − y 1 t = zt solves the initial value problem
z ′ t = azt, zt 0  = 0.
Thus z ′ − az = 0. Multiply by e −at on both sides. By the chain rule,
e −at z ′ − az = d e −at zt = 0,
dt
−at
and so there is a constant C such that e zt = C. Since zt 0  = 0, this constant is 0, and so
zt = 0.
Next consider the second part involving the complex stuff. You can verify that
y 0 e at−t 0  cos bt − t 0  + i sin bt − t 0  = yt
does indeed solve the initial value problem from using elementary calculus. Now suppose you
have another solution y 1 t. Then let zt = yt − y 1 t. It solves
z ′ = a + ibz, zt 0  = 0
Now zt = ut + ivt. It is also clear that z t ≡ ut − ivt solves the equation
z ′ = a − ib z , z t 0  = 0
Thus from the product rule which you can easily verify from the usual proof of the product rule to
be also true for complex valued functions,
d |zt|2 = d z z  = z ′ z + z z ′ = a + ibz z + za − ib z
dt
dt
= 2a|z|2 , |z|t 0  = 0
Therefore, from the first part, |z| = 0 and so y = y 1 .
′
Note that this implies that e ia+ibt  = a + ibe ia+ibt where e ia+ibt is given above. Now
consider the last part. Sove y ′ = ay + f, yt 0  = y 0 .
y ′ − ay = f, yt 0  = y 0
Multiply both sides by e −at−t 0 
d e −at−t 0  y = e −at−t 0  ft
dt
Now integrate from t 0 to t. Then
e −at−t 0  yt − y 0 =
∫t
t
e −as−t 0  fsds
0
Hence
yt = e at−t 0  y 0 + e at−t 0  ∫ e −as−t 0  fsds
t
t0
= e at−t 0  y 0 + e at ∫ e −as fsds
t
t0
200
Solution:
Now consider A an n × n matrix. By Schur’s theorem there exists unitary Q such that
Q −1 AQ = T
where T is upper triangular. Now consider the first order initial value problem
x ′ = Ax, xt 0  = x 0 .
Show there exists a unique solution to this first order system. Hint: Let y = Q −1 x and so the
system becomes
y ′ = Ty, yt 0  = Q −1 x 0
Now letting y =y 1 , ⋯, y n  T , the bottom equation becomes
y ′n = t nn y n , y n t 0  = Q −1 x 0  n .
*
Then use the solution you get in this to get the solution to the initial value problem which occurs
one level up, namely
y ′n−1 = t n−1n−1 y n−1 + t n−1n y n , y n−1 t 0  = Q −1 x 0  n−1
Continue doing this to obtain a unique solution to *.
There isn’t much to do. Just see that you understand it. In the above problem, equations of the
above form are shown to have a unique solution and there is even a formula given. Thus there is a
unique solution to
y ′ = Ty, yt 0  = Q −1 x 0
Then let y = Q −1 x and plug this in. Thus
Q −1 x ′ = TQ −1 x, Q −1 xt 0  = Q −1 x 0
Then
x ′ = QTQ −1 x = Ax, xt 0  = x 0 .
The solution is unique because of uniqueness of the solution for y. You can go backwards, starting
with x and defining y ≡ Q −1 x and then there is only one thing y can be and so x is also unique.
201
Solution:
Now suppose Φt is an n × n matrix of the form
Φt =
x 1 t ⋯ x n t
where
x ′k t = Ax k t.
Explain why
Φ ′ t = AΦt
if and only if Φt is given in the form of ♠. Also explain why if c ∈ F n ,
yt ≡ Φtc
solves the equation
y ′ t = Ayt.
Suppose Φ ′ t = AΦt where Φt is as described. This happens if and only if
Φ ′ t ≡
=
=A
x ′1 t ⋯ x ′n t
x 1 t ⋯ x n t
Ax 1 t ⋯ Ax n t
from the way we multiply matrices. Which happens if and only if
x ′k = Ax k
Say Φ ′ t = AΦt. Then consider
yt =
∑ c k x k t.
k
♠
Then
y ′ t =
∑ c k x ′k t = ∑ c k Ax k t = A ∑ c k x k t
k
k
= AΦtc = Ayt
k
202
Solution:
In the above problem, consider the question whether all solutions to
x ′ = Ax
♣
n
are obtained in the form Φtc for some choice of c ∈ F . In other words, is the general solution
to this equation Φtc for c ∈ F n ? Prove the following theorem using linear algebra.
Theorem: Suppose Φt is an n × n matrix which satisfies
Φ ′ t = AΦt.
Then the general solution to ♣ is Φtc if and only if Φt −1 exists for some t. Furthermore, if
Φ ′ t = AΦt, then either Φt −1 exists for all t or Φt −1 never exists for any t.
(detΦt is called the Wronskian and this theorem is sometimes called the Wronskian
alternative.)
Suppose first that Φtc is the general solution. Why does Φt −1 necessarily exist? If for some
t 0 , Φt 0  fails to have an inverse, then there exists c ∉ Φt 0 R n . However, there exists a unique
a solution to the system
x ′ = Ax, xt 0  = c
from one of the above problems. It follows that xt ≠ Φtv for any choice of v because for any
v,
c = xt 0  ≠ Φt 0 v
Thus Φtc for arbitrary choice of c fails to deliver the general solution as assumed.
Next suppose Φt 0  −1 exists for some t 0 and let xt be any solution to the system of differential
equations. Then let yt = ΦtΦt 0  −1 xt 0 . It follows y ′ = Ay and x ′ = Ax and also that
xt 0  = yt 0  so by the uniqueness part of the above problem, x = y. Hence Φtc for arbitrary c
is the general solution. Consider det Φt if it equals zero for any t 0 then Φtc does not deliver the
general solutions. If det Φt is nonzero for any t then Φtc delivers the general solution. Either
Φtc does deliver the general solution or it does not. Therefore, you cannot have two different
values of t, t 1 , t 2 such that det Φt 1  = 0 but det Φt 2  ≠ 0. In other words det Φt either
vanishes for all t or it vanishes for no t. Thus the inverse of Φt either exists for all t or for no t.
203
Solution:
Let Φ ′ t = AΦt. Then Φt is called a fundamental matrix if Φt −1 exists for all t. Show
there exists a unique solution to the equation
x ′ = Ax + f, xt 0  = x 0
*
and it is given by the formula
xt = ΦtΦt 0  −1 x 0 + Φt ∫ Φs −1 fsds
t
t0
Now these few problems have done virtually everything of significance in an entire undergraduate
differential equations course, illustrating the superiority of linear algebra. The above formula is
called the variation of constants formula.
It is easy to see that x given by the formula does solve the initial value problem. This is just
calculus.
x ′ t = Φ ′ tΦt 0  −1 x 0 + Φ ′ t ∫ Φs −1 fsds + ΦtΦt −1 ft
t
t0
= A ΦtΦt 0  −1 x 0 + Φt ∫ Φs −1 fsds
t
+ ft
t0
= Axt + ft
As for the initial condition,
xt 0  = Φt 0 Φt 0  −1 x 0 + Φt 0  ∫ Φs −1 fsds = x 0
t0
t0
It is also easy to see that this must be the solution. If you have two solutions, then let
ut = x 1 t − x 2 t and observe that
u ′ t = Aut, ut 0  = 0.
From the above problem, there is at most one solution to this initial value problem and it is u = 0.
204
Solution:
Show there exists a special Φ such that Φ ′ t = AΦt, Φ0 = I, and suppose Φt −1 exists
for all t. Show using uniqueness that
Φ−t = Φt −1
and that for all t, s ∈ R
Φt + s = ΦtΦs
Explain why with this special Φ, the solution to the differential equation
x ′ = Ax + f, xt 0  = x 0
can be written as
xt = Φt − t 0 x 0 + ∫ Φt − sfsds.
t
t0
Hint: Let Φt be such that the j th column is x j t where
x ′j = Ax j , x j 0 = e j .
Use uniqueness as required.
You simply let Φt =
x 1 t ⋯ x n t
where
x ′k t = Ax k t, x k 0 = e k
Then det Φ0 = 1 and so det Φt ≠ 0 for any t. Furthermore, there is only one solution to
Φ ′ t = AΦt along with the initial condition Φ0 = I and it is the one just described. This
follows from the uniqueness of the intial value problem x ′ = Ax, x0 = x 0 discussed earlier.
Now it is interesting to note that A and Φt commute. To see this, consider
yt ≡ AΦtc − ΦtAc
When t = 0 y0 = Ac − Ac = 0. What about its derivative?
y ′ t = AΦ ′ tc − Φ ′ tAc = A 2 Φtc − AΦtAc
= AAΦtc − ΦtAc = Ayt
By uniqueness, it follows that yt = 0. Thus these commute as claimed, since c is arbitrary. Now
from the product rule (the usual product rule holds when the functions are matrices by the same
proof given in calculus. )
ΦtΦ−t ′ = Φ ′ tΦ−t − ΦtΦ ′ −t
= AΦtΦ−t − ΦtAΦ−t
= ΦtAΦ−t − ΦtAΦ−t = 0
Hence ΦtΦ−t must be a constant matrix since the derivative of each component of this
product equals 0. However, this constant can only equal I because when t = 0 the product is I.
Therefore, Φt −1 = Φ−t. Next consider the claim that Φt + s = ΦtΦs. Fix s and let t
vary.
Φ ′ t + s − Φ ′ tΦs = AΦt + s − AΦtΦs = AΦt + s − ΦtΦs
Now Φ0 + s − Φ0Φs = Φs − Φs = 0 and so, by uniqueness, t → Φt + s − ΦtΦs
equals 0. The rest follows from the variation of constants formula derived earlier.
xt = ΦtΦ0 −1 x 0 + Φt ∫ Φs −1 fsds
t
0
= Φtx 0 + Φt ∫ Φ−sfs
t
0
= Φtx 0 + ∫ ΦtΦ−sfs
t
0
t
= Φtx 0 + ∫ Φt − sfs
0
The reason you can take Φt inside the integral is that this is constant with respect to the variable
of integration. The j th entry of the product is
Φt ∫ Φ−sfs
t
0
=
j
∑ Φt jk ∫ 0 ∑ Φ−s ki f i sds
t
k
=
i
∑ ∫ 0 Φt jk Φ−s ki f i ds
t
k,i
=
∑ ∫ 0 ΦtΦ−s ji f i ds
t
i
t
=
∫ 0 ∑ΦtΦ−s ji f i ds
=
∫ 0 ΦtΦ−sfsds
i
t
j
You could also simply verify directly that this formula works much as was done earlier.
205
Solution:
∗
Using the Lindemann Weierstrass theorem show that if σ is an algebraic number
sin σ, cos σ, ln σ, and e are all transcendental. Hint: Observe, that
ee −1 + −1e 0 = 0, 1e lnσ + −1σe 0 = 0,
1 e iσ − 1 e −iσ + −1 sinσe 0 = 0.
2i
2i
The hint gives it away. Consider the claim about ln σ. The equation shown does hold from the
definition of ln σ. However, if ln σ were algebraic, then e lnσ , e 0 would be linearly dependent with
field of scalars equal to the algebraic numbers, contrary to the Lindemann Weierstrass theorem.
The other instances are similar. In the case of cos σ, you could use the identity
1 e iσ + 1 e −iσ − e 0 cos σ = 0
2
2
iσ −iσ 0
contradicting independence of e , e , e .
Inner Product Spaces
206
Solution:
Let V be the continuous complex valued functions defined on a finite closed interval I. Define an
inner product as follows.
⟨f, g⟩ ≡
∫I fxgxpxdx
where px some function which is strictly positive on the closed interval I. It is understood in
writing this that
∫I fx + igxdx ≡ ∫I fxdx + i ∫I gxdx
Then with this convention, the usual calculus theorems hold about evaluating integrals using the
fundamental theorem of calculus and so forth. You simply apply these theorems to the real and
imaginary parts of a complex valued function. Show V is an inner product space with the given
inner product.
All of the axioms of the inner product are obvious except one, the one which says that if
⟨f, f⟩ = 0 then f = 0. This one depends on continuity of the functions. Suppose then that it is not
true. In other words, ⟨f, f⟩ = 0 and yet f ≠ 0. Then for some x ∈ I, fx ≠ 0. By continuity, there
exists δ > 0 such that if y ∈ I ∩ x − δ, x + δ ≡ I δ , then
|fy − fx| < |fx|/2
It follows that for y ∈ I δ ,
|fy| > |fx| − |fx/2| = |fx|/2.
Hence
⟨f, f⟩ ≥
∫I |fy|2 pxdy ≥
|fx|2 /2
length of I δ minp > 0,
δ
a contradiction. Note that min p > 0 because p is a continuous function defined on a closed and
bounded interval and so it achieves its minimum by the extreme value theorem of calculus.
207
Solution:
Let V be the polynomials of degree at most n which are defined on a closed interval I and let
x 0 , x 1 , ⋯, x n  be n + 1 distinct points in I. Then define
n
⟨f, g⟩ ≡
∑ fx k gx k 
k=0
Show that V is an inner product space with respect to the given inner product.
All of the axioms of the inner product are obvious for this one also, except for the one which
says that if ⟨f, f⟩ = 0, then f = 0. Suppose than that ⟨f, f⟩ = 0. Then f equals 0 at n + 1 points of
the interval and yet f is a polynomial of degree at most n. Therefore, this would be a contradiction
unless f is identically equal to 0 for all x. This is because a polynomial of degree n has at most n
zeros.
208
Solution:
Let V be any complex vector space and let v 1 , ⋯, v n  be a basis. Decree that
⟨v i , v j ⟩ = δ ij .
Then define
n
n
j=1
k=1
n
∑ cjv j, ∑ d k v k
≡
∑ c jd k ⟨v j, v k ⟩ = ∑ c k d k
j,k
k=1
This makes the complex vector space into an inner product space.
n
A generic vector of V will be denoted by w = ∑ i=1 wi v i . Now
⟨w, u⟩ ≡
n
n
k=1
k=1
∑ wk u k = ∑ wk u k ≡ ⟨u, w⟩
Letting a, b be scalars,
⟨aw + bz, u⟩ =
∑awk + bz k u k = a ∑ wk u k + b ∑ z k u k
k
k
k
≡ a⟨w, u⟩ + b⟨z, u⟩
⟨w, w⟩ ≡
n
n
k=1
k=1
∑ wk wk = ∑|wk |2 ≥ 0
The only way this can equal zero is to have each wk = 0. Since v k  is a basis, this happens if
and only if w = 0. Thus the inner product by decree is an inner product.
209
Solution:
Let V consist of sequences a = a k  ∞k=1 , a k ∈ C, with the property that
∞
∑|a k |2 < ∞
k=1
and the inner product is then defined as
∞
⟨a, b⟩ ≡
∑ akbk
k=1
The last example is obviously an inner product as noted earlier except for needing to verify that
the inner product makes sense. However, from the Cauchy Schwarz inequality,
n
∑ akbk
1/2
n
≤
k=1
∑|a k |
2
k=1
∑|a k |
k=1
∑|b k |
2
k=1
1/2
∞
≤
1/2
n
2
1/2
∞
∑|b k |
2
<∞
k=1
210
Solution:
In each of the examples of the above problems, write the Cauchy Schwarz inequality.
In the first example, the Cauchy Schwarz inequality says
∫I fxgxpxdx
1/2
∫I |fx|2 pxdx
≤
∫I |gx|2 pxdx
1/2
In the next example, the Cauchy Schwarz inequality says
n
∑ fx k gx k 
1/2
n
≤
k=0
∑|fx k |
1/2
n
∑|gx k |
2
k=0
2
k=0
In the third example, the Cauchy Schwarz inequality says
n
∑ u k wk
1/2
n
≤
k=1
∑|u k |
2
k=1
1/2
n
∑|wk |
2
k=1
where u = ∑ k u k v k and w = ∑ k wk v k .
In the last example,
∞
∑ akbk
k=1
1/2
∞
≤
∑|a k |
2
k=1
1/2
∞
∑|b k |
2
k=1
211
Solution:
Verify that in an inner product space,
|αz| = |α||z|
and that |z| ≥ 0 and equals zero if and only if z = 0.
|αz|2 ≡ ⟨αz, αz⟩ = α⟨z, αz⟩ = α⟨αz, z⟩ = α α ⟨z, z⟩ ≡ |α|2 |z|2
and so
|α||z| = |αz|
It is clear |z| ≥ 0 because |z| = ⟨z, z⟩ . Suppose |z| = 0. Then ⟨z, z⟩ = 0 and so, by the axioms of
the inner product, z = 0.
212
Solution:
Consider the Cauchy Schwarz inequality. Show that it still holds under the assumptions
⟨u, v⟩ = ⟨v, u⟩, ⟨au + bv, z⟩ = a⟨u, z⟩ + b⟨v, z⟩, and ⟨u, u⟩ ≥ 0. Thus it is not necessary to say
that ⟨u, u⟩ = 0 only if u = 0. It is enough to simply state that ⟨u, u⟩ ≥ 0.
This was all that was used in the proof. Look at the proof carefully and you will see this is the
case.
213
Solution:
Consider the integers modulo a prime, Z p . This is a field of scalars. Now let the vector space be
Z p  n where n ≥ p. Define now
1/2
n
⟨z, w⟩ ≡
∑ z i wi
i=1
Does this satisfy the axioms of an inner product? Does the Cauchy Schwarz inequality hold for
this ⟨ ⟩? Does the Cauchy Schwarz inequality even make any sense?
It might be the case that ⟨z, z⟩ = 0 and yet z ≠ 0. Just let z =z 1 , ⋯, z n  where exactly p of the
z i equal 1 but the remaining are equal to 0. Then ⟨z, z⟩ would reduce to 0 in the integers mod p.
Another problem is the failure to have an order on Z p . Consider first Z 2 . Is 1 positive or negative?
If it is positive, then 1 + 1 would need to be positive. But 1 + 1 = 0 in this case. If 1 is negative,
then −1 is positive, but −1 is equal to 1. Thus 1 would be both positive and negative. You can
consider the general case where p > 2 also. Simply take a ≠ 1. If a is positive, then consider
a, a 2 , a 3 ⋯. These would all have to be positive. However, eventually a repeat will take place.
Thus a n = a m m < n, and so a m a k − 1 = 0 where k = n − m. Since a m ≠ 0, it follows that
a k = 1 for a suitable k. It follows that the sequence of powers of a must include each of
1, 2, ⋯, p − 1 and all these would therefore, be positive. However, 1 + p − 1 = 0
contradicting the assertion that Z p can be ordered. So what would you mean by saying ⟨z, z⟩ ≥ 0?
The Cauchy Schwarz inequality would not even apply.
214
Solution:
If you only know that ⟨u, u⟩ ≥ 0 along with the other axioms of the inner product and if you
define |z| the same way, which properties of a norm are lost?
You lose the one which says that if |z| = 0 then z = 0. However, all the others are retained.
215
Solution:
In an inner product space, an open ball is the set
Bx, r ≡ y : |y − x| < r.
If z ∈ Bx, r, show there exists δ > 0 such that Bz, δ ⊆ Bx, r. In words, this says that an
open ball is open. Hint: This depends on the triangle inequality.
Let δ = r − |z − x|. Then if y ∈ Bz, δ,
|y − x| ≤ |y − z| + |z − x| < δ + |z − x| = r − |z − x| + |z − x| = r
and so Bz, δ ⊆ Bx, r.
216
Solution:
Let V be the real inner product space consisting of continuous functions defined on −1, 1 with
the inner product given by
∫ −1 fxgxdx
1
Show that 1, x, x 2  are linearly independent and find an orthonormal basis for the span of these
vectors.
Lets first find the Grammian.
2
G=
1
1
∫ −1 xdx
1
∫ −1 xdx ∫ −1 x 2 dx
1
∫ −1 x 2
1
∫ −1 x 3
1
∫ −1 x 2
1
∫ −1 x 3
=
1
∫ −1 x 4 dx
Then the inverse is
G −1 =
9
8
0 − 158
0
3
2
0
− 158
0
45
8
2
0
2
3
0
2
3
0
2
3
0
2
5
a b c
Now let R =
0 d h
. Solve
0 0 f
T
a b c
a b c
0 d h
0 d h
0 0 f
0 0 f
9
8
0 − 158
0
3
2
0
− 158
0
45
8
=
Thus
9
8
0 − 158
0
3
2
0
− 158
0
45
8
a 2 + b 2 + c 2 bd + ch cf
bd + ch
=
d 2 + h 2 hf
cf
f2
hf
Hence a solution is
f = 3 2 5 , h = 0, c = − 1 2 5 , d = 1 2 3 , b = 0, a = 1 2
4
4
2
2
and so the orthonormal basis is
1
2
1 x x2
2
0
1
2
0
0
1
2
2
1
2
2 3x
− 14 2 5
2 3
0
3
4
2 5 x2 −
0
3
4
2 5
1
4
2 5
217
Solution:
A regular Sturm Liouville problem involves the differential equation for an unknown function
of x which is denoted here by y,
′
pxy ′  + λqx + rxy = 0, x ∈ a, b
and it is assumed that pt, qt > 0 for any t along with boundary conditions,
C 1 ya + C 2 y ′ a = 0
C 3 yb + C 4 y ′ b = 0
where
C 21 + C 22 > 0, and C 23 + C 24 > 0.
There is an immense theory connected to these important problems. The constant λ is called an
eigenvalue. Show that if y is a solution to the above problem corresponding to λ = λ 1 and if z is a
solution corresponding to λ = λ 2 ≠ λ 1 , then
∫ a qxyxzxdx = 0.
b
Let y go with λ and z go with μ.
zpxy ′  ′ + λqx + rxyz = 0
′
ypxz ′  + μqx + rxzy = 0
Subtract.
zpxy ′  ′ − ypxz ′  ′ + λ − μqxyz = 0
Now integrate from a to b. First note that
zpxy ′  ′ − ypxz ′  ′ = d pxy ′ z − pxz ′ y
dx
and so what you get is
pby ′ bzb − pbz ′ byb − pay ′ aza − paz ′ aya
+ λ − μ ∫ qxyxzxdx = 0
b
a
Look at the stuff on the top line. From the assumptions on the boundary conditions,
C 1 ya + C 2 y ′ a = 0
C 1 za + C 2 z ′ a = 0
and so
yaz ′ a − y ′ aza = 0
Similarly,
ybz ′ b − y ′ bzb = 0
Hence, that stuff on the top line equals zero and so the orthogonality condition holds.
218
Solution:
Using the above problem or standard techniques of calculus, show that
2
sinnx
π
∞
n=1
are orthonormal with respect to the inner product
⟨f, g⟩ =
π
∫ 0 fxgxdx
Hint: If you want to use the above problem, show that sinnx is a solution to the boundary value
problem
y ′′ + n 2 y = 0, y0 = yπ = 0
It is obvious that sinnx solves this boundary value problem. Those boundary conditions in the
above problem are implied by these. In fact,
1 sin 0 + 0 cos0 = 0
1 sin π + 0 cos π = 0
Then it follows that
π
∫ 0 sinnx sinmxdx = 0 unless n = m.
π
Now also ∫ sin 2 nxdx =
0
219
Solution:
1
2
π and so
2
π
sinnx
are orthonormal.
Find S 5 fx where fx = x on −π, π. Then graph both S 5 fx and fx if you have access to a
system which will do a good job of it.
5
Recall that this is the partial sum of the Fourier series. ∑ k=−5 b k e ikx where
bk = 1
2π
−1 k
i
k
π
∫ −π xe −ikx dx =
Thus
5
S 5 fx =
∑
k=−5
−1 k
i e ikx =
k
5
=
∑ −1k
k+1
∑ −1
k
k
icoskx + i sinkx
k=−5
5
sinkx =
k=−5
∑ 2−1
k
k+1
sinkx
k=1
y
-3
5
-2
2
-1
1
2
3
x
-2
220
Solution:
Find S 5 fx where fx = |x| on −π, π. Then graph both S 5 fx and fx if you have access to a
system which will do a good job of it.
a.
π
b k = 1 ∫ |x|e −ikx dx = − 22
2π −π
k π
if k is odd and 0 if k is even. However, b 0 = 12 π Thus the sum desired is
−1
S 5 fx =
5
∑ − k12 π 1 + −1 k+1 coskx + ∑ − k12 π 1 + −1 k+1 coskx
k=−5
k=1
+ 1π
2
since the sin terms cancel due to the fact that sin is odd. The above equals
2
1π−∑
4
cos2k + 1x
2
+
1 2 π
2k
k=0
y3
2
1
-3
-2
-1
0
1
2
3
x
221
Solution:
Find S 5 fx where fx = x 2 on −π, π. Then graph both S 5 fx and fx if you have access to a
system which will do a good job of it.
bk = 1
2π
b0 = 1
2π
π
∫ −π x 2 e −kix dx =
π
∫ −π x 2 dx =
2 −1 k
k2
1 π2
3
Then the series is
5
S 5 fx =
∑
k=1
−1
2 −1 k e ikx + ∑ 2 −1 k e ikx + π 2
3
k2
k2
k=−5
5
=
∑ k42 −1 k coskx +
k=1
∑
5
4
k=1 k 2
−1 coskx +
k
π2
3
π2
3
y
10
8
6
4
2
-3
-2
-1
0
1
2
3
x
222
Solution:
Let V be the set of real polynomials defined on 0, 1 which have degree at most 2. Make this
into a real inner product space by defining
⟨f, g⟩ ≡ f0g0 + f1/2g1/2 + f1g1
Find an orthonormal basis and explain why this is an inner product.
It was described more generally why this was an inner product in an earlier problem. To find an
orthonormal basis, consider a basis 1, x, x 2 . Then the Grammian is
3
0 + 1/2 + 1
0 + 1/4 + 1
0 + 1/2 + 1
0 + 1/4 + 1
0 + 1/21/4 + 1
0 + 1/4 + 1 0 + 1/21/4 + 1
3
2
5
4
9
8
3
=
3
2
5
4
0 + 1/4 2 + 1
5
4
9
8
17
16
Now G −1 equals
1
−3
2
−3
26
−24
2
−24
24
a b c
Let R =
0 d h
0 0 f
T
a b c
a b c
0 d h
0 d h
0 0 f
0 0 f
a 2 + b 2 + c 2 bd + ch cf
=
bd + ch
d 2 + h 2 hf
cf
hf
f2
So you need to solve the following .
a 2 + b 2 + c 2 bd + ch cf
bd + ch
d +h
cf
hf
2
2
hf
f2
f = 2 6 , h = −2 6 , c = 1 6 , d =
6
So the orthonormal basis is
1
3
3
2x−
1
2
223
Solution:
Consider R n with the following definition.
−3
2
−3
26
−24
2
−24
24
2 ,b = − 1 2 ,a = 1 3
2
3
3 − 12 2
1
3
1 x x2
=
1
1
6
6
0
2
−2 6
0
0
2 6
2 2 6 x2 − 2 6 x +
1
6
6
n
⟨x, y⟩ ≡
∑ xiyii
i=1
Does this define an inner product? If so, explain why and state the Cauchy Schwarz inequality in
terms of sums.
It obviously defines an inner product because it satisfies all the axioms of one. The Cauchy
Schwarz inequality says
n
∑ xiyii
1/2
n
∑
≤
i=1
x 2i i
i=1
1/2
n
∑
y 2i i
i=1
This is a fairly surprising inequality it seems to me.
224
Solution:
For f a piecewise continuous function,
n
S n fx = 1
2π
π
∑ e ikx ∫ −π fye −iky dy
.
k=−n
Show this can be written in the form
S n fx =
π
∫ −π fyD n x − ydy
where
n
D n t = 1
2π
∑ e ikt
k=−n
This is called the Dirichlet kernel. Show that
sinn + 1/2t
D n t = 1
2π
sint/2
For V the vector space of piecewise continuous functions, define S n : V → V by
S n fx =
π
∫ −π fyD n x − ydy.
Show that S n is a linear transformation. (In fact, S n f is not just piecewise continuous but infinitely
π
differentiable. Why?) Explain why ∫ D n tdt = 1. Hint: To obtain the formula, do the
−π
following.
n
e it/2 D n t = 1
2π
∑ e ik+1/2t
e i−t/2 D n t = 1
2π
∑ e ik−1/2t
k=−n
n
k=−n
Change the variable of summation in the bottom sum and then subtract and solve for D n t.
n
S n fx =
1
2π
∑ e ikx
k=−n
π
∫ −π fye −iky dy
= 1
2π
Let D n t be as described. Then the above equals
π
∫ −π D n x − yfydy.
π
n
∫ −π fy ∑ e ikx−y dy.
k=−n
So now you want to find a formula for D n t. The hint is really good.
n
e
it/2
D n t = 1
2π
∑ e ik+1/2t
k=−n
n
e i−t/2 D n t = 1
2π
n−1
∑ e ik−1/2t =
1
2π
k=−n
∑
e ik+1/2t
k=−n+1
D n te it/2 − e −it/2  = 1 e in+1/2t − e −in+1/2t 
2π
D n t2i sint/2 = 1 2i sin n + 1 t
2
2π
1
sintn + 2 
D n t = 1
2π
sin 12 t
You know that t → D n t is periodic of period 2π. Therefore, if fy = 1,
S n fx =
π
π
∫ −π D n x − ydy = ∫ −π D n tdt
However, it follows directly from computation that S n fx = 1.
225
Solution:
Let V be an inner product space and let U be a finite dimensional subspace with an orthonormal
basis u i  ni=1 . If y ∈ V, show
n
|y|2 ≥
∑|⟨y, u k ⟩|2
k=1
Now suppose that
u k  ∞k=1
is an orthonormal set of vectors of V. Explain why
lim⟨y, u k ⟩ = 0.
k→∞
When applied to functions, this is a special case of the Riemann Lebesgue lemma.
n
y − ∑⟨y, u k ⟩u k , w
=0
k=1
for all w ∈ spanu i  ni=1 . Therefore,
n
n
k=1
k=1
|y| = y − ∑⟨y, u k ⟩u k + ∑⟨y, u k ⟩u k
2
2
Now if ⟨u, v⟩ = 0, then you can see right away from the definition that
|u + v|2 = |u|2 + |v|2
n
n
Applying this to u = y − ∑ k=1 ⟨y, u k ⟩u k , v = ∑ k=1 ⟨y, u k ⟩u k , the above equals
n
= y − ∑⟨y, u k ⟩u k
2
k=1
n
= y − ∑⟨y, u k ⟩u k
k=1
n
+
∑⟨y, u k ⟩u k
2
k=1
2
n
+ ∑|⟨y, u k ⟩|2 ,
k=1
the last step following because of similar reasoning to the above and the assumption that the u k
∞
are orthonormal. It follows the sum ∑ k=1 |⟨y, u k ⟩|2 converges and so lim k→∞ ⟨y, u k ⟩ = 0 because if
a series converges, then the k th term must converge to 0.
226
Solution:
Let f be any piecewise continuous function which is bounded on −π, π. Show, using the above
problem, that
π
lim
n→∞
π
∫ −π ft sinntdt = lim
∫ ft cosntdt = 0
n→∞ −π
Let the inner product space consist of piecewise continuous bounded functions with the inner
product defined by
⟨f, g⟩ ≡
π
∫ −π fxgxdx
Then, from the above problem and the fact shown earlier that
1
2π
e ikx
form an orthonormal
k∈Z
set of vectors in this inner product space, it follows that
lim
⟨f, e inx ⟩ = 0
n→∞
without loss of generality, assume that f has real values. Then the above limit reduces to having
both the real and imaginary parts converge to 0. This implies the thing which was desired. Note
also that if α ∈ −1, 1, then
π
lim
n→∞
π
∫ −π ft sinn + αtdt = lim
∫ ftsinnt cos α + cosnt sin αdt = 0
n→∞ −π
227
Solution:
∗
Let f be a function which is defined on −π, π. The 2π periodic extension is given by the
formula fx + 2π = fx. In the rest of this problem, f will refer to this 2π periodic extension.
Assume that f is piecewise continuous, bounded, and also that the following limits exist
fx + y − fx +
fx − y − fx +
, lim
lim
y
y
y→0+
y→0+
Here it is assumed that
fx + ≡ lim fx + h, fx − ≡ lim fx − h
h→0+
h→0+
both exist at every point. The above conditions rule out functions where the slope taken from
either side becomes infinite. Justify the following assertions and eventually conclude that under
these very reasonable conditions
lim S fx = fx + + fx −/2
n→∞ n
the mid point of the jump. In words, the Fourier series converges to the midpoint of the jump of
the function.
S n fx =
S n fx −
fx + + fx −
2
=
π
∫ −π fx − yD n ydy
π
∫ −π
fx − y −
fx + + fx −
D n ydy
2
=
π
π
∫ 0 fx − yD n ydy + ∫ 0 fx + yD n ydy
π
− ∫ fx + + fx −D n ydy
0
≤
π
∫ 0 fx − y − fx −D n ydy
+
π
∫ 0 fx + y − fx +D n ydy
Now apply some trig. identities and use the result of the above problems to conclude that both of
these terms must converge to 0.
From the definition of D n , the top formula holds. Now observe that D n is an even function.
Therefore, the formula equals
S n fx =
=
π
0
π
π
∫ 0 fx − yD n ydy + ∫ −π fx − yD n ydy
∫ 0 fx − yD n ydy + ∫ 0 fx + yD n ydy
fx + y + fx − y
2D n ydy
2
π
π
Now note that ∫ 2D n y = 1 because ∫ D n ydy = 1 and D n is even. Therefore,
=
0
π
∫0
−π
fx + + fx −
S n fx −
2
π fx + y − fx + + fx − y − fx −
= ∫
2D n ydy
2
0
From the formula for D n y given earlier, this is dominated by an expression of the form
π fx + y − fx + + fx − y − fx −
C ∫
sinn + 1/2ydy
siny/2
0
for a suitable constant C. The above is equal to
π
fx + y − fx + + fx − y − fx −
y
C ∫
sinn + 1/2ydy
y
0 siny/2
y
and the expression siny/2 equals a bounded continuous function on 0, π except at 0 where it is
undefined. This follows from elementary calculus. Therefore, changing the function at this single
point does not change the integral and so we can consider this as a continuous bounded function
defined on 0, π. Also, from the assumptions on f,
fx + y − fx + + fx − y − fx −
y→
y
is equal to a piecewise continuous function on 0, π except at the point 0. Therefore, the above
integral converges to 0 by the previous problem. This shows that the Fourier series generally tries
to converge to the midpoint of the jump.
228
Solution:
Using the Fourier series obtained above for fx = x and the result that the Fourier series
converges to the midpoint of the jump obtained above to find an interesting formula by examining
where the Fourier series converges when x = π/2. Of course you can get many other interesting
formulas in the same way. Hint: You should get
n
S n fx =
∑ 2−1
k
k=1
k+1
sinkx
Let x = π/2. Then you must have
π = lim
n→∞
2
= lim
n→∞
n
2−1 k+1
sin k π
k
2
∑
k=1
n
n
= lim
n→∞
∑ 2k2− 1 sin
k=1
2k − 1 π
2
∑ 2k2− 1 −1 k+1 .
k=1
Then, dividing by 2 you get
π = lim
n→∞
4
n
k+1
∑ −1
2k − 1
k=1
You could also consider the Fourier series for fx = x 2 and get
n
lim
n→∞
∑ k42 −1 k coskx +
k=1
π2 = x2
3
because the periodic extension of this function is continuous. Let x = 0
n
lim
n→∞
∑ k42 −1 k +
k=1
π2 = 0
3
and so
π 2 = lim
n→∞
3
n
∑
k=1
4 −1 k+1 ≡
k2
∞
∑ k42 −1 k+1
k=1
This is one of those calculus problems where you show it converges absolutely by the comparison
test with a p series. However, here is what it converges to.
229
Solution:
Let V be an inner product space and let K be a convex subset of V. This means that if x, z ∈ K,
then the line segment x + tz − x = 1 − tx + tz is contained in K for all t ∈ 0, 1. Note that
every subspace is a convex set. Let y ∈ V and let x ∈ K. Show that x is the closest point to y out
of all points in K if and only if for all w ∈ K,
Re⟨y − x, w − x⟩ ≤ 0.
n
In R , a picture of the above situation where x is the closest point to y is as follows.
The condition of the above variational inequality is that the angle θ shown in the picture is
larger than 90 degrees. Recall the geometric description of the dot product in terms of an included
angle.
Consider for t ∈ 0, 1 the following.
|y − x + tw − x|2
where w ∈ K and x ∈ K. It equals
ft = |y − x|2 + t 2 |w − x|2 − 2t Re⟨y − x, w − x⟩
Suppose x is the point of K which is closest to y. Then f ′ 0 ≥ 0. However,
f ′ 0 = −2 Re⟨y − x, w − x⟩. Therefore, if x is closest to y,
Re⟨y − x, w − x⟩ ≤ 0.
Next suppose this condition holds. Then you have
|y − x + tw − x|2 ≥ |y − x|2 + t 2 |w − x|2 ≥ |y − x|2
By convexity of K, a generic point of K is of the form x + tw − x for w ∈ K. Hence x is the
closest point.
230
Solution:
Show that in any inner product space the parallelogram identity holds.
|x + y|2 + |x − y|2 = 2|x|2 + 2|y|2
Next show that in a real inner product space, the polarization identity holds.
⟨x, y⟩ = 1 |x + y|2 − |x − y|2 .
4
This follows right away from the axioms of the inner product.
|x + y|2 + |x − y|2 = |x|2 + |y|2 + 2 Re⟨x, y⟩
+ |x|2 + |y|2 − 2 Re⟨x, y⟩
= 2|x|2 + 2|y|2
Of course the same reasoning yields
1 |x + y|2 − |x − y|2  = 1 |x|2 + |y|2 + 2⟨x, y⟩ − |x|2 + |y|2 − 2⟨x, y⟩
4
4
= ⟨x, y⟩
231
Solution:
∗
This problem is for those who know about Cauchy sequences and completeness of F p for F
either R or C, and about closed sets. Suppose K is a closed nonempty convex subset of a finite
dimensional subspace U of an inner product space V. Let y ∈ V. Then show there exists a unique
point x ∈ K which is closest to y. Hint: Let
λ = inf|y − z| : z ∈ K
Let x n  be a minimizing sequence,
|y − x n | → λ.
Use the parallelogram identity in the above problem to show that x n  is a Cauchy sequence.
Now let u k  pk=1 be an orthonormal basis for U. Say
p
xn =
∑ c nk u k
k=1
Verify that for c n ≡ c n1 , ⋯, c np  ∈ F p
|x n − x m | = |c n − c m |F p .
Now use completeness of F p and the assumption that K is closed to get the existence of x ∈ K
such that |x − y| = λ.
The hint is pretty good. Let x k be a minimizing sequence. The connection between x k and
c k ∈ F k is obvious because the u k  are orthonormal. That is,
|x n − x m | = |c n − c m |F p .
Use the parallelogram identity.
y − x k − y − x m 
2
2
+
y − x k + y − x m 
2
2
=2
y − xk
2
2
+2
y − xm
2
Hence
1 |x − x |2 = 1 |y − x |2 + 1 |y − x |2 − y − x k + x m 2
k
k
m
4 m
2
2
2
2
2
≤ 1 |y − x k | + 1 |y − x m | − λ 2
2
2
Now the right hand side converges to 0 since x k  is a minimizing sequence. Therefore, x k  is a
Cauchy sequence in U. Hence the sequence of component vectors c k  is a Cauchy sequence in
F n and so it converges thanks to completeness of F. It follows that x k  also must converge to
some x. Then since K is closed, it follows that x ∈ K. Hence
λ = |x − y|.
232
Solution:
∗
Let K be a closed nonempty convex subset of a finite dimensional subspace U of a real inner
product space V. (It is true for complex ones also.) For x ∈ V, denote by Px the unique closest
point to x in K. Verify that P is Lipschitz continuous with Lipschitz constant 1,
|Px − Py| ≤ |x − y|.
Hint: Use the above problem which involves characterizing the closest point with a variational
inequality.
From the problem,
⟨Px − Py, y − Py⟩ ≤ 0
⟨Py − Px, x − Px⟩ ≤ 0
Thus
⟨Px − Py, x − Px⟩ ≥ 0
Hence
⟨Px − Py, x − Px⟩ − ⟨Px − Py, y − Py⟩ ≥ 0
and so
⟨Px − Py, x − y − Px − Py⟩ ≥ 0
|x − y||Px − Py| ≥ ⟨Px − Py, Px − Py⟩ = |Px − Py|2
233
Solution:
∗
This problem is for people who know about compactness. It is an analysis problem. If you
have only had the usual undergraduate calculus course, don’t waste your time with this problem.
Suppose V is a finite dimensional normed linear space. Recall this means that there exists a norm
‖⋅‖ defined on V as described above,
‖v‖ ≥ 0 equals 0 if and only if v = 0
‖v + u‖ ≤ ‖u‖ + ‖v‖, ‖αv‖ = |α|‖v‖.
Let |⋅| denote the norm which is defined by
|v|2 ≡
n
n
j=1
j=1
∑|v j |2 where v = ∑ v ju j
u j  a basis, the norm which comes from the inner product by decree. Show |⋅| and ‖⋅‖ are
equivalent. That is, there exist constants δ, Δ > 0 such that for all x ∈ V,
δ|x| ≤ ‖x‖ ≤ Δ|x|.
Next explain why every two norms on a finite dimensional vector space must be equivalent in the
above sense. Let u k  nk=1 be a basis for V and if x ∈ V, let x i be the components of x relative to
this basis. Thus the x i are defined according to
∑ xi ui = x
i
Then since u i  is an orthonormal basis by decree, it follows that
∑|x i |2 .
|x|2 =
i
Now letting x k  be a sequence of vectors of V let x k  denote the sequence of component
vectors in F n . One direction is easy, saying that ‖x‖ ≤ Δ|x|. If this is not so, then there exists a
sequence of vectors x k  such that
‖x k ‖ > k|x k |
dividing both sides by ‖x k ‖ it can be assumed that 1 > k|x k | = |x k |. Hence x k → 0 in F k . But
from the triangle inequality,
n
‖x k ‖ ≤
∑ x ki ‖u i ‖
i=1
= 0, this is a contradiction to each ‖x k ‖ = 1. It follows that there
Therefore, since
exists Δ such that for all x,
‖x‖ ≤ Δ|x|
lim k→∞ x ki
Now consider the other direction. If it is not true, then there exists a sequence x k  such that
1 |x k | > ‖x k ‖
k
Dividing both sides by |x k |, it can be assumed that |x k | = |x k | = 1. Hence, by compactness of the
closed unit ball in F n , there exists a further subsequence, still denoted by k such that x k → a ∈ F n
and it also follows that |a|F n = 1. Also the above inequality implies lim k→∞ ‖x k ‖ = 0. Therefore,
n
n
j=1
j=1
xk = 0
∑ x kj u j = lim
∑ a ju j = lim
k→∞
k→∞
which is a contradiction to the u j being linearly independent. Therefore, there exists δ > 0 such
that for all x,
δ|x| ≤ ‖x‖.
Now if you have any other norm on this finite dimensional vector space, say |||⋅|||, then from what
was just shown, there exist scalars δ i and Δ i all positive, such that
δ 1 |x| ≤ ‖x‖ ≤ Δ 1 |x|
δ 2 |x| ≤ |||x||| ≤ Δ 2 |x|
It follows that
Δ
Δ Δ
Δ Δ
|||x||| ≤ 2 ‖x‖ ≤ 2 1 |x| ≤ 2 1 |||x|||.
δ1
δ1
δ1δ2
Hence
δ1
Δ
|||x||| ≤ ‖x‖ ≤ 1 |||x|||.
Δ2
δ2
In other words, any two norms on a finite dimensional vector space are equivalent norms. What
this means is that every consideration which depends on analysis or topology is exactly the same
for any two norms. What might change are geometric properties of the norms.
Linear Transformations
234
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of − 14 π.
1
2
− 12
2
2
1
2
1
2
2
2
235
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 34 π.
− 12 2 − 12 2
1
2
2
− 12 2
236
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 43 π.
− 12
− 12 3
1
2
3
− 12
237
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 14 π. and then reflects through the y axis.
1
0
0 −1
1
2
1
2
2 − 12 2
2
1
2
2
=
1
2
− 12
2
− 12 2
2 − 12 2
238
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4.
cos 2π
− sin 2π
3 
3 
cos −π
− sin −π
4 
4 
sin 2π
3 
sin −π
4 
cos 2π
3 
1
4
2 3 −
1
4
2 − 14 2 3 −
1
4
2 3 +
1
4
2
2 3 −
1
4
=
cos −π
4 
1
4
1
4
2
2
239
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 13 π. and then reflects through the y axis.
−1 0
0
− 12 3
1
2
1
2
1
1
2
3
− 12
=
1
2
1
2
3
3
1
2
240
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 16 π.
1
2
3
1
2
− 12
1
2
3
241
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of 76 π.
− 12 3
1
2
− 12
− 12 3
242
Solution:
Find the matrix for the linear transformation which rotates every vector in R 2 through an angle
of π/12. Hint: Note that π/12 = π/3 − π/4.
− 12 3
1
2
1
2
3
1
2
1
2
−
1
2
2
2
− 12 2
1
2
2
=
1
4
2 3 +
1
4
1
4
2 3 −
1
4
2 − 14 2 3 −
2
1
4
2 −
1
4
1
4
2
2 3
243
Solution:
Let V be a finite dimensional inner product space, the field of scalars equal to either R or C.
Verify that f given by fv ≡ ⟨v, z⟩ is in LV, F, the space of linear transformations from V to the
field F. Next suppose f is an arbitrary element of LV, F. Show the following.
a. If f = 0, the zero mapping, then fv = ⟨v, 0⟩ for all v ∈ V.
This is obvious from the properties of the inner product.
⟨v, 0⟩ = ⟨v, 0 + 0⟩ = ⟨v, 0⟩ + ⟨v, 0⟩
so ⟨v, 0⟩ = 0.
b. If f ≠ 0 then there exists z ≠ 0 satisfying ⟨u, z⟩ = 0 for all u ∈ kerf.
kerf is a subspace and so there exists z 1 ∉ kerf. Then there exists a closest point of
kerf to z 1 called x. Then let z = z 1 − x. By the theorems on minimization, ⟨u, z⟩ = 0 for all
u ∈ kerf.
c. Explain why fyz − fzy ∈ kerf.
ffyz − fzy = fyfz − fzfy = 0.
d. Use the above to show that there exists w such that fy = ⟨y, w⟩ for all y ∈ V.
0 = ⟨fyz − fzy, z⟩ = fy|z|2 − fz⟨y, z⟩
and so
fz
fy = y, 2 z
|z|
so w =
fz
|z|2
z appears to work.
e. Show there is at most one such w.
If w 1 , w 2 both work, then for every y,
0 = fy − fy = ⟨y, w 1 ⟩ − ⟨y, w 2 ⟩ = ⟨y, w 1 − w 2 ⟩
In particular, this is true for y = w 1 − w 2 and so w 1 = w 2 .
You have now proved the Riesz representation theorem which states that every f ∈ LV, F is of
the form
fy = ⟨y, w⟩
for a unique w ∈ V.
244
Solution:
↑Let A ∈ LV, W where V, W are two finite dimensional inner product spaces, both having field
of scalars equal to F which is either R or C. Let f ∈ LV, F be given by
fy ≡ ⟨Ay, z⟩
where ⟨ ⟩ now refers to the inner product in W. Use the above problem to verify that there exists
a unique w ∈ V such that fy = ⟨y, w⟩, the inner product here being the one on V. Let A ∗ z ≡ w.
Show that A ∗ ∈ LW, V and by construction,
⟨Ay, z⟩ = ⟨y, A ∗ z⟩.
In the case that V = F n and W = F m and A consists of multiplication on the left by an m × n
matrix, give a description of A ∗ .
It is required to show that A ∗ is linear.
⟨y, A ∗ αz + βw⟩ ≡ ⟨Ay, αz + βw⟩ = α ⟨Ay, z⟩ + β ⟨Ay, w⟩
≡ α ⟨y, A ∗ z⟩ + β ⟨y, A ∗ w⟩
= ⟨y, αA ∗ z⟩ + ⟨y, βA ∗ w⟩ = ⟨y, αA ∗ z + βA ∗ w⟩
Since y is arbitrary, this shows that A ∗ is linear. In case A is an m × n matrix as described,
A ∗ = A T 
245
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 + 2D + 1f. Find kerA.
First solve D + 1z = 0. This is easy to do and gives zt = C 1 e −t . Now solve
D + 1y = C 1 e −t . To do this, you multiply by an integrating factor. Thus
d e t y = C
1
dt
Take an antiderivative and obtain
ety = C1t + C2
Thus the desired kernel consists of all functions y which are of the form
yt = C 1 te t + C 2 e t
where C i is a constant.
246
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 + 5D + 4f. Find kerA. Note that you could first find
kerD + 4 where D is the differentiation operator and then consider kerD + 1D + 4 = kerA
and consider Sylvester’s theorem.
In this case, the two operators D + 1 and D + 4 commute and are each one to one on the kernel
of the other. Also, it is obvious that kerD + a consists of functions of the form Ce −at . Therefore,
kerD + 1D + 4 consists of functions of the form
y = C 1 e −t + C 2 e −4t
where C 1 , C 2 are arbitrary constants. In other words, a basis for kerD + 1D + 4 is e −t , e −4t .
247
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 + 6D + 8f. Find kerA. Note that you could first find
kerD + 2 where D is the differentiation operator and then consider kerD + 2D + 4 = kerA
and consider Sylvester’s theorem.
In this case, the two operators D + 2 and D + 4 commute and are each one to one on the
kernel of the other. Also, it is obvious that kera + D consists of functions of the form Ce −at .
Therefore, kerD + 2D + 4 consists of functions of the form
y = C 1 e −2t + C 2 e −4t
where C 1 , C 2 are arbitrary constants. In other words, a basis for kerD + 2D + 4 is e −2t , e −4t .
248
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 + 3D − 4f. Find kerA. Note that you could first find
kerD − 1 where D is the differentiation operator and then consider kerD − 1D + 4 = kerA
and consider Sylvester’s theorem.
In this case, the two operators D − 1 and D + 4 commute and are each one to one on the
kernel of the other. Also, it is obvious that kera + D consists of functions of the form Ce −at .
Therefore, kerD − 1D + 4 consists of functions of the form
y = C 1 e t + C 2 e −4t
where C 1 , C 2 are arbitrary constants. In other words, a basis for kerD − 1D + 4 is e t , e −4t .
249
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 − 7D + 12f. Find kerA. Note that you could first find
kerD − 4 where D is the differentiation operator and then consider kerD − 4D − 3 = kerA
and consider Sylvester’s theorem.
In this case, the two operators D − 4 and D − 3 commute and are each one to one on the
kernel of the other. Also, it is obvious that kera + D consists of functions of the form Ce −at .
Therefore, kerD − 4D − 3 consists of functions of the form
y = C 1 e 4t + C 2 e 3t
where C 1 , C 2 are arbitrary constants. In other words, a basis for kerD − 4D − 3 is e 4t , e 3t .
250
Solution:
Let A be the linear transformation defined on the vector space of smooth functions (Those which
have all derivatives) given by Af = D 2 + 4D + 4f. Find kerA.
First, kerD + a = C 1 e −2t . Therefore, D + ay = C 1 e −2t and so, solving this linear first order
equation,
y = C 1 te −2t + C 2 e −2t
251
Solution:
Suppose Ax = b has a solution where A is a linear transformation. Explain why the solution is
unique precisely when Ax = 0 has only the trivial (zero) solution.
Suppose first there is a unique solution z. Then there can only be one solution to Ax = 0 because
if this last equation had more than one solution, say y i , i = 1, 2, then y i + z, i = 1, 2 would each be
solutions to Ay = b because A is linear. Recall also that the general solution to Ax = b consists of
the general solution to Ax = 0 added to any solution to Ax = b. Therefore, if there is only one
solution to Ax = 0, then there is only one solution to Ax = b.
252
Solution:
Verify the linear transformation determined by the matrix
A=
1 0 2
0 1 4
maps R 3 onto R 2 but the linear transformation determined by this matrix is not one to one.
This is an old problem. The rank is obviously 2 so the matrix maps onto R 2 . It is not one to one
because the system Ax = 0 must have a free variable.
253
Solution:
Let L be the linear transformation taking polynomials of degree at most three to polynomials of
degree at most three given by
D 2 + 4D + 3
where D is the differentiation operator. Find the matrix of this linear transformation relative to the
basis 1, x, x 2 , x 3 . Also find the matrices with of D + 3 and D + 1 and multiply these. You should
get the same thing. Why?
3 3x + 4 3x 2 + 8x + 2 3x 3 + 12x 2 + 6x
3 4 2 0
=
0 3 8 6
1 x x2 x3
0 0 3 12
0 0 0 3
Done the other way, the matrix of D + 3, and D + 1 are respectively
3 1 0 0
0 3 2 0
0 0 3 3
0 0 0 3
1 1 0 0
,
0 1 2 0
0 0 1 3
0 0 0 1
Then since composition of transformation corresponds to multiplication of the matrices, the
matrix of the full operator is
3 1 0 0
1 1 0 0
0 3 2 0
0 1 2 0
0 0 3 3
0 0 1 3
0 0 0 3
0 0 0 1
=
3 4 2
0
0 3 8
6
0 0 3 12
0 0 0
3
254
Solution:
Let L be the linear transformation taking polynomials of degree at most three to polynomials of
degree at most three given by
D 2 − 16
where D is the differentiation operator. Find the matrix of this linear transformation relative to the
basis 1, x, x 2 , x 3 . Also find the matrices with of D + a and D + b and multiply these. You should
get the same thing. Why?
−16 −16x 2 − 16x 2 6x − 16x 3
=
1 x x2 x3
−16
0
2
0
0
−16
0
6
0
0
−16
0
0
0
0
−16
Done the other way, the matrix of D − 4, and D + 4 are respectively
−4
1
0
0
0
−4
2
0
0
0
−4
3
0
0
0
−4
4 1 0 0
,
0 4 2 0
0 0 4 3
0 0 0 4
Then since composition of transformation corresponds to multiplication of the matrices, the
matrix of the full operator is
−4
1
0
0
4 1 0 0
0
−4
2
0
0 4 2 0
0
0
−4
3
0 0 4 3
0
0
0
−4
0 0 0 4
=
−16
0
2
0
0
−16
0
6
0
0
−16
0
0
0
0
−16
255
Solution:
Let L be the linear transformation taking polynomials of degree at most three to polynomials of
degree at most three given by
D 2 + 2D + 1
where D is the differentiation operator. Find the matrix of this linear transformation relative to the
basis 1, x, x 2 , x 3 . Find the matrix directly and then find the matrix with respect to the differential
operator D + 1 and multiply this matrix by itself. You should get the same thing. Why?
You should get the same thing because the multiplication of matrices corresponds to
composition of linear transformations. Lets find the matrix for D + 1 first.
1 x + 1 2x + x 2 3x 2 + x 3
1 1 0 0
=
1 x x2 x3
0 1 2 0
0 0 1 3
0 0 0 1
so the matrix is
1 1 0 0
0 1 2 0
0 0 1 3
0 0 0 1
Then the matrix of the desired transformation is just this one squared.
1 2 2 0
0 1 4 6
0 0 1 6
0 0 0 1
256
Solution:
Show that if L ∈ LV, W (linear transformation) where V and W are vector spaces, then if
Ly p = f for some y p ∈ V, then the general solution of Ly = f is of the form
kerL + y p .
This is really old stuff. However, here it applies to an arbitrary linear transformation. Suppose
Lz = f. Then L z − y p = Lz − L y p = f − f = 0. Therefore, letting y = z − y p , it follows
that y is a solution to Ly = 0 and z = y p + y. Thus every solution to Ly = f is of the form y p + y
for some y which solves Ly = 0.
257
Solution:
Let L ∈ LV, W where V, W are vector spaces, finite or infinite dimensional, and define x ∼ y if
x − y ∈ kerL. Show that ∼ is an equivalence relation. Next define addition and scalar
multiplication on the space of equivalence classes as follows.
x + y ≡ x + y
αx = αx
Show that these are well defined definitions and that the set of equivalence classes is a vector
space with respect to these operations. The zero is ker L. Denote the resulting vector space by
V/ kerL. Now suppose L is onto W. Define a mapping A : V/ kerK → W as follows.
Ax ≡ Lx
Show that A is well defined, linear, one to one and onto.
Now suppose L is onto W. Define a mapping A : V/ kerK → W as follows.
Ax ≡ Lx
Show that A is well defined, one to one and onto.
It is obvious that x ∼ x. If x ∼ y, then y ∼ x is also clear. If x ∼ y and y ∼ z, then
z−x = z−y+y−x
and by assumption, both z − y and y − x ∈ kerL which is a subspace. Therefore, z − x ∈ kerL
also and so ∼ is an equivalence relation. Are the operations well defined? If x = x ′ , y = y ′ ,
is it true that x + y = y ′ + x ′ ? Of course. x ′ + y ′ − x + y = x ′ − x + y ′ − y ∈ kerL
because kerL is a subspace. Similar reasoning applies to the case of scalar multiplication. Now
why is A well defined? If x = x ′ , is Lx = Lx ′ ? Of course this is so. x − x ′ ∈ kerL by
assumption. Therefore, Lx = Lx ′ . It is clear also that A is linear.
Aax + by = Aax + by = aLx + bLy = aAx + bAy
If Ax = 0, then Lx = 0 and so x ∈ kerL and so x = 0. Therefore, A is one to one. It is
obviously onto LV = W.
258
Solution:
If V is a finite dimensional vector space and L ∈ LV, V, show that the minimal polynomial for
L equals the minimal polynomial of A where A is the n × n matrix of L with respect to some basis.
This is really easy because of the definition of what you mean by the matrix of a linear
transformation.
L = qAq −1
where q, q −1 are one to one and onto linear transformations from V to F n or from F n to V. Thus if
pλ is a polynomial,
pL = qpAq −1
Thus the polynomials which send L to 0 are the same as those which send A to 0.
259
Solution:
Let A be an n × n matrix. Describe a fairly simple method based on row operations for
computing the minimal polynomial of A. Recall, that this is a monic polynomial pλ such that
pA = 0 and it has smallest degree of all such monic polynomials. Hint: Consider I, A 2 , ⋯.
2
Regard each as a vector in F n and consider taking the row reduced echelon form or something
like this. You might also use the Cayley Hamilton theorem to note that you can stop the above
sequence at A n .
2
An easy way to do this is to “unravel" the powers of the matrix making vectors in F n and then
making these the columns of a n 2 × n matrix. Look for linear relationships between the columns
by obtaining the row reduced echelon form. As an example, consider the following matrix.
1
1
0
−1 0 −1
2
1
3
Lets find its minimal polynomial. We have the following powers
1 0 0
0 1 0
0 0 1
1
,
1
0
0
−1 0 −1
2
1
,
1
−1
−3 −2 −3
3
7
5
8
−3 −1 −4
,
−7 −6 −7
18 15 19
By the Cayley Hamilton theorem, I won’t need to consider any higher powers than this. Now I will
unravel each and make them the columns of a matrix.
1
1
0
−3
0
1
1
−1
0
0
−1 −4
0 −1 −3 −7
1
0
−2 −6
0 −1 −3 −7
0
2
7
18
0
1
5
15
1
3
8
19
Next you can do row operations and obtain the row reduced echelon form for this matrix and then
look for linear relationships.
1 0 0
2
0 1 0 −5
0 0 1
4
0 0 0
0
0 0 0
0
0 0 0
0
0 0 0
0
0 0 0
0
0 0 0
0
From this you see that for A denoting the matrix,
A 3 = 4A 2 − 5A + 2I
and so the minimal polynomial is
λ 3 − 4λ 2 + 5λ − 2
No smaller degree polynomial can work either. Since it is of degree 3, this is also the characteristic
polynomial. Note how we got this without expanding any determinants or solving any polynomial
equations. If you factor this polynomial, you get λ 3 − 4λ 2 + 5λ − 2 = λ − 2λ − 1 2 so this is an
easy problem, but you see that this procedure for finding the minimal polynomial will work even
when you can’t factor the characteristic polynomial. If you want to work with smaller matrices,
you could also look at A k e i for e 1 , e 2 , etc., use similar techniques on each of these and then find
the least common multiple of the resulting polynomials.
260
Solution:
Let A be an n × n matrix which is non defective. That is, there exists a basis of eigenvectors.
Show that if pλ is the minimal polynomial, then pλ has no repeated roots. Hint: First show
that the minimal polynomial of A is the same as the minimal polynomial of the diagonal matrix
Dλ 1 
D=
⋱
Dλ r 
Where Dλ is a diagonal matrix having λ down the main diagonal and in the above, the λ i are
r
distinct. Show that the minimal polynomial is ∏ i=1 λ − λ i .
If two matrices are similar, then they must have the same minimal polynomial. This is obvious
from the fact that for pλ any polynomial and A = S −1 BS,
pA = S −1 pBS
r
So what is the minimal polynomial of the above D? It is obviously ∏ i=1 λ − λ i . Thus there are
no repeated roots.
261
Solution:
Show that if A is an n × n matrix and the minimal polynomial has no repeated roots, then A is
non defective and there exists a basis of eigenvectors. Thus, from the above problem, a matrix
may be diagonalized if and only if its minimal polynomial has no repeated roots. It turns out this
condition is something which is relatively easy to determine.
If A has a minimal polynomial which has no repeated roots, say
m
pλ =
∏λ − λ i ,
j=1
then from the material on decomposing into direct sums of generalized eigenspaces, you have
F n = kerA − λ 1 I ⊕ kerA − λ 2 I ⊕ ⋯ ⊕ kerA − λ m I
and by definition, the basis vectors for kerA − λ 2 I are all eigenvectors. Thus F n has a basis of
eigenvectors and is therefore diagonalizable or non defective.
Conversely, if F n has a basis of eigenvectors, then A is similar to a diagonal matrix. Say the
distinct entries on the diagonal are λ 1 , ⋯, λ m , m ≤ n. Then it is easy to check that the minimal
m
polynomial is ∏ j=1 λ − λ i  because this polynomial will send the diagonal matrix to zero.
262
Solution:
Recall the linearization of the Lotka Volterra equations used to model the interaction between
predators and prey. It was shown earlier that if x, y are the deviations from the equilibrium point,
then
x ′ = −bxy − b c y
d
a
′
y = dxy + dx
b
If one is interested only in small values of x, y, that is, in behavior near the equilibrium point, these
are approximately equal to the system
x ′ = −b c y
d
a
′
y = dx
b
Written more simply, for α, β > 0,
x ′ = −αy
y ′ = βx
Find the solution to the initial value problem
x′
=
y′
0 −α
x
β
y
0
x0
,
y0
x0
=
y0
where x 0 , y 0 are both real. Also have a computer graph the vector fields −xy − y, xy + x and
−y, x which result from setting all the parameters equal to 1.
From the discussion in the book, the eigenvalues are ± αβ i. The eigenpairs are
−i α
β
i α
↔ −i α β ,
β
↔i α β
Then the fundamental matrix is
−i α i α
β
β
exp −t αβ i
0
0
exp t αβ i
1
2
− 12
i
α
i
α
1
2 β
1
2 β
Then the solution desired is
xt
yt
=
exp −t αβ i
−i α i α
β
β
1
2
0
−
exp t αβ i
0
1
2
i
α
i
α
1
2 β
x0
1
2 β
y0
That is,
xt
yt
e −it
αβ
β e −it
αβ
1
2
=
1
2
i
α
+
1
2
−
1
2
αβ
e it
i
α
1
2
β e it
α
i
β
αβ
1
2
e it
e −it
αβ
−
αβ
+
1
2
i
1
2
α
β
e it
e −it
αβ
αβ
x0
y0
Since the solution must be real, this is of the form
xt
yt
αβ t
cos
=
β
α
sin
αβ t
α
β
sin
cos
αβ t
=
α
β
y0
αβ t
Thus both x and y oscillate about the origin. For example, letting ω =
xt = x 0 cosωt + y 0
x0
αβ ,
sinωt
x 20 + y 20 α cosωt − φ
β
where φ is a phase shift. A similar formula holds for y. First is the field for the full Lotka Volterra
equations and then is the one for the linearized Lotka Volterra equations.
These suggest that the behavior of the linearized system and the nonlinear system is similar only
for small x, y.