1
Solutions to selected problems
Section 1.1, #11a,c,d.
a.
px = n − 1
for i = n − 2 : 0
px = xpx + i
end
b.
z = x1 , y = x 1
for i = 1 : n
y = y + xi
z = zy
end
c.
y = (t − x1 ), pt = a1
i=1:n−1
y = y(t − xi )
pt = pt + ai+1 y
for
end
Section 1.1, P
#12a,c,d
a. v = a + x a
c. v = P a x
d. v = P a x z = x
n
i=1
0
n
i=0
i
n
i=1
i
i
n−i
i
n+1
,
Section 1.2, # 13,14
13.
Recall
−1
ln(1.1) = ln(1 + 10 ) =
∞
X
−k
k−1 10
(−1)
k
k=1
≈ Pn (0.1) :=
n
X
k=1
(−1)k−1
10−k
.
k
By the alternating series theorem, the error in this approximation is bounded by
10−(n+1)
|ln(1.1) − Pn (0.1)| ≤
.
n+1
The RHS is less than
10−8
when
2
n ≥ 7.
Thus,
1
7
terms are needed.
14.
We compute
f (2) = 7, f 0 (2) = 8, f 00 (2) = 8
and
f 000 (2) = 6.
Thus,
f (2 + h) = 7h0 /0! + 8h1 /1! + 8h2 /2! + 6h3 /3! = 7 + 8h + 4h2 + h3
or equivalently,
f (x) = 7 + 8(x − 2) + 4(x − 2)2 + (x − 2)3 .
Section 2.1, #1a,b
a.
2
= (−1) 2
Since
−30
0 97−127
(1.00 . . .)2
and
97 = 26 + 25 + 20 = (01100001)2 ,
0|01100001|00000000000000000000000
is the
b.
32-bit
machine representation.
64.015625 = 26 (1 + 2−12 ) = (−1)0 2133−127 (1.000000000001)2
2 = (10000101)2 ,
0|10000101|00000000000100000000000
Since
and
133 = 27 + 22 +
0
is the
32-bit
machine representation.
Section 2.1, #3a-e
32-bit machine numbers must be in the range of the machine (roughly 2−126
they must have at most 24 signicant digits.
3a.
3b.
3c-d.
e.
to
2128 ), and
Not a machine number: too large to be in machine range
Not a machine number: binary expansion has more than
24
signicant digits
Not machine numbers: their binary expansions are nonterminating
1
= 2−8 is a machine number: it is in the range of the machine and its binary
256
expansion has 1 signicant digit.
Section 2.1, #5a-c, e-h
Conversions can be done as in Section 2.1 #1a,b above.
a. +0
b. −0
c. ∞
d. −∞
e. 2
f. 5.5
−126
2
g. 1
h.
1
10
Section 2.1, #24
x is positive and write x = 10m 1.a1 a2 . . .. After chopping to keep only 15
m
digits we get the machine representation x̂ = 10 1.a1 a2 . . . a14 . Observe that
WLOG assume
signicant
|x − x̂| = 10m 0.00 . . . a15 a16 . . . ≤ 10m 10−14 .
Thus,
10m 10−14
|x − x̂|
10−14
≤ m
≤ 10−14 .
=
|x|
10 1.a1 a2 . . .
1.a1 a2 . . .
Section 2.1, #29
Write
x = 2n (1.a1 a2 . . . a23 )2
and
y = 2m (1.b1 b2 . . . b23 )2 .
Write
x = 2m (2n−m (1.a1 a2 . . . a23 )2 ) = 2n (0.00 . . . a1 a2 . . .)2 .
x < 2−25 y , we have n − m ≤ −25, so in the last display a1
least 25 places to the rights of the decimal point. Thus,
Since
is at
appears in a spot which
x + y = 2m (1.b1 b2 . . . b23 0 . . . 0a1 a2 . . . a23 )2
where
a1
is
25
or more spots to the right of the decimal point.
After chopping (or
rounding), we get
(x
+ y) = 2m (1.b1 b2 . . . b23 )2 = y.
Section 2.2, #2
Using
ex = 1 + x + x2 /2! + x3 /3! + . . .,
f (x) =
write
f (x) = ex − x − 1
as
x2 x3 x4
+
+
+ ...
2
6
24
It's easy to check that the rst two terms above are enough for
5
signicant digits:
f (0.01) ≈ 0.000050167
Using
e0.01 ≈ 1.0101
and direct calculation gives
f (0.01) ≈ 1.0101 − 0.0100 − 1.0000 = 1.0001 − 1.0000 = 0.0001.
3
Section 2.2, #3
One option is to write
f (x) = (1 − e) + x + x2 /x + x3 /3 + . . ..
Section 2.2, #7
Note that if
x 6= 0,
1
sinh2 x − tanh2 x
tanh2 x sinh2 x
(sinh x − tanh x) =
=
,
x
x(sinh x + tanh x)
x(sinh x + tanh x)
where the last equality comes the hyperbolic trig identity
cosh2 x − 1 = sinh2 x
sinh2 x(cosh2 x − 1)
sinh x − tanh x =
= tanh2 x sinh2 x.
2
cosh x
2
2
The last expression in (1) avoids loss of signicance near
Section 2.2 , #14 √
By rewriting
f (x) =
x+2−
√
x
x = 0.
as
f (x) = √
2
√ .
x+2+ x
Section 2.2 #21
Using
sin x = x −
x3 x 5
+
− ...
3!
5!
and the alternating series theorem, we get
| sin x − x| ≤
For
|x| < 1/10
|x|3
.
3!
this becomes
| sin x − x| ≤
10−3
= 0.0001666...
6
Section 2.2 Computer Problems, #8
See the script section 2.2 #8.
4
(1)
and
Section 3.1 Computer Problems, #1
See the script section 3.1 #1.
Section 3.2 #1
Newton's method for nding roots of
xn+1
f (x) = x2 − R
is
f (xn )
x2 − R
2x2 − x2n + R
1
= xn − 0
= xn − n
= n
=
f (xn )
2xn
2xn
2
R
xn +
.
xn
Section 3.2 #2
Using #1 above,
x2n+1
Letting
2
R
1
R2
2
xn +
−R=
xn + 2R + 2 − R
xn
4
xn
2
2
2
xn − R
1
R
R2
1
2
=
=
xn − 2R + 2 =
xn −
.
4
xn
4
xn
2xn
1
−R=
4
en = x2n − R
be the error in the
nth
en+1 =
If
|en | < R
approximation,
1 2
e .
4x2n n
this means
|en+1 | ≤
In particular if
|e0 | ≤ R/2,
1
|en |2 .
4(R − |en |)
an induction argument shows that
|en+1 | ≤
1
|e |2 .
2R n
Section 3.2 #6
m = 8, and x0 = 1.1, we have x1 = 1.0875 and x2 = 1.0765625. When m = 12 and
x0 = 1.1, we have x1 = 1.091666... and x2 = 1.08402777... Note the slower convergence
to r = 1 when m = 12 compared to m = 8.
When
Section 3.2 #10
For
f (x) = tan x − R
we have
f (x)
tan x − R
=
= cos x sin x − R cos2 x.
f 0 (x)
sec2 x
5
Thus, the iteration formula
xn+1 = xn − cos xn sin xn + R cos2 xn
tan x = R.
can be used for nding solutions to
Section 3.2 Computer Problems, #7,19
See the scripts section 3.2 #7 and section 3.2 #19.
Section 4.1 #6
The polynomial, in Newton form, is
p(x) = 2 + x − 3x(x − 2) + 4x(x − 2)(x − 3).
Section 4.1 #7b
The columns can be lled top to bottom with
third column:
− 3, 25,
fourth column:
7, 9.5,
fth column:
0.5.
Thus, the interpolating polynomial is
1
p(x) = 2 − 3(x + 1) + 7(x + 1)(x − 1) + (x + 1)(x − 1)(x − 3).
2
Section 4.1 #10
Newton's interpolation polynomial is
p(x) = 7 + 2x + 5x(x − 2) + x(x − 2)(x − 3).
A nested form for ecient computation is
p(x) = 7 + x(2 + (x − 2)(5 + (x − 3))).
Section 4.1 #20
6
With
x = (−2, −1, 0, 1, 2), y = (2, 14, 4, 2, 2)
we have
P0 (x) = a0
P1 (x) = a0 + a1 (x + 2)
P2 (x) = a0 + a1 (x + 2) + a2 (x + 2)(x + 1)
P3 (x) = a0 + a1 (x + 2) + a2 (x + 2)(x + 1) + a3 (x + 2)(x + 1)x
P4 (x) = a0 + a1 (x + 2) + a2 (x + 2)(x + 1) + a3 (x + 2)(x + 1)x + a4 (x + 2)(x + 1)x(x − 1),
for our Newton polynomials. Using
Pi (xi ) = yi , i = 0, 1, 2, 3, 4,
we can solve to get
a0 = 2, a1 = 12, a2 = −11, a3 = 5, a4 = −1.5.
Q(x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4

   
1 −2 4 −8 16
b0
2
1 −1 1 −1 1  b1  14

   
1 0 0 0 0  b2  =  4 

   
1 1 1 1 1  b3   2 
1 2 4 8 16
b4
2
Alternatively, we may write
and solve
to get
b0 = 4, b1 = −8, b2 = 5.5, b3 = 2, b4 = −1.5.
Of course,
P4 (x) = Q(x).
Section 4.1 #21
To interpolate the additional point
(3, 10),
we look at
R(x) = Q(x) + c(x + 2)(x + 1)x(x − 1)(x − 2),
and plug in
10 = R(3) to get c = 2/5.
(3, 10).
Now
R(x) interpolates the points from #20, along
with the extra point
Section 4.1 Computer Problems #1,2,9
See scripts online.
Section 4.1 #43
The Newton form of the interpolating polynomial at
x0 , x1
is
P (x) = f [x0 ] + f [x0 , x1 ](x − x0 ).
7
Plugging in
x = x1
and solving for
f [x0 , x1 ] =
f [x0 , x1 ],
we nd
P (x1 ) − f [x0 ]
f (x1 ) − f (x0 )
=
.
x1 − x0
x1 − x0
By the mean value theorem, there is
f 0 (ξ) =
ξ ∈ (x0 , x1 )
such that
f (x1 ) − f (x0 )
= f [x0 , x1 ].
x1 − x0
Section 4.1 #45
Let
x − xi
[g(x) − h(x)].
xn − x0
i = 1, . . . , n − 1,
Q(x) = g(x) +
Since both
g
and
h
interpolate
Q(xi ) = g(xi ) +
for
i = 1, . . . , n − 1.
h
at
x0 − xi
x0 − xi
[g(xi ) − h(xi )] = f (xi ) +
[f (xi ) − f (xi )] = f (xi )
xn − x0
xn − x0
Since
g
interpolates
Q(x0 ) = g(x0 ) +
and since
f
interpolates
Q(xn ) = g(xn ) +
x0 ,
x0 − x0
[g(x0 ) − h(x0 )] = f (x0 ) + 0 = f (x0 ),
xn − x0
xn ,
x0 − xn
[g(xn ) − h(xn )] = g(xn ) − [g(xn ) − f (xn )] = f (xn ).
x n − x0
Note that this is the main step in proving Theorem 2 in Section 4.1.
Section 4.2 #2
Without loss of generality we can assume
xj = 0
h2
|(x − 0)(x − h)| ≤
4
and
xj+1 = h.
for all
We must show that
x ∈ [0, h].
Q(x) := (x − 0)(x − h) = x2 − hx. Since Q0 (x) = 2x − h = 0 when
x = h/2, the function Q has its smallest and largest values on [0, h] at h/2 or at the
2
endpoints 0, h. Since Q(0) = Q(h) = 0 we conclude max0≤x≤h |Q(x)| = |Q(h/2)| = h /4.
To do this, consider
Section 4.3 #3
8
Solution. We must show that
1
n
n!
=
≥ 1,
j = 0, . . . , n − 1.
(j + 1)!(n − j)!
j+1 j
1
When j = 1, . . . , n − 1, we have
≥ n1 and nj ≥ n so that
j+1
1
n
n
1 n
n
=
≥
≥ = 1.
j+1 j
j
n j
n
1 n
And when j = 0, we get
= 1.
j+1 j
Section 4.2 #4
Let
x ∈ [a, b]
and
a = x0 < x1 < . . . < xn = b
n
Y
|x − xi | ≤ hn+1 n!,
h := max (xi − xi−1 ).
1≤i≤n
i=0
Notice
h
be any choice of nodes. We must show
is the largest spacing between adjacent nodes. For
xj ≤ x ≤ xj+1 ,
1
1
|x − xj ||x − xj+1 | ≤ |xj+1 − xj |2 ≤ h2 .
4
4
where the rst inequality comes from Section 4.2, #2. Since
j−1
Y
j−1
j−1
j−1
Y
Y
Y
|x − xi | =
(x − xi ) ≤
(xj+1 − xi ) =
(j − i + 1)h = (j + 1)! hj .
i=0
i=0
Similarly, since
n
Y
x ≤ xj+1 ,
i=0
i=0
x ≥ xj−1 ,
n
Y
|x − xi | =
i=j+2
(xi − x) ≤
i=j+2
n
Y
(xi − xj ) =
i=j+2
n
Y
(i − j)h = (n − j)! hn−j−1 .
i=j+2
Combining the last three inequalities, we get
n
Y
1
1
|x − xi | ≤ h2 (j + 1)!hj (n − j)!hn−j−1 ≤ hn+1 n!
4
4
i=0
Section 4.2 #18
Let
P2 (x)
be the degree
2
interpolant of
f
at
x0 , x1 , x2 :
P2 (x) = f [x0 ] + f [x0 , x1 ](x − x0 ) + f [x0 , x1 , x2 ](x − x0 )(x − x1 ),
9
(2)
Q(x) = f (x) − P2 (x). Then Q has at least 3 roots namely x0 , x1 , x2 so,
0
00
due to Rolle's theorem, Q has at least 2 roots and Q has at least 1 root, call it ξ . Thus,
and dene
0 = Q00 (ξ) = f 00 (ξ) − P200 (ξ) = f 00 (ξ) − 2f [x0 , x1 , x2 ],
which shows
1
f [x0 , x1 , x2 ] = f 00 (ξ).
2
Section 4.2 Computer Problem #9
See script online.
Section 4.3 #3
Using Taylor's theorem,
1
1
f (x + h) = f (x) + f 0 (x)h + f 00 (x)h2 + f 000 (x)h3 + . . .
2
6
4
f (x + 2h) = f (x) + 2f 0 (x)h + 2f 00 (x)h2 + f 000 (x)h3 + . . .
3
Straightforward calculations give
1
1
[4f (x + h) − 3f (x) − f (x + 2h)] = f 0 (x) − f 000 (x)h2 + . . .
2h
3
1
Terminating this expansion at second order , we nd the error term
1
0
[4f (x + h) − 3f (x) − f (x + 2h)] − f (x) = 1 |f 000 (ξ)|h2 .
3
2h
Section 4.3 #5
We have
1
1
f (x + h) = f (x) + f 0 (x)h + f 00 (x)h2 + f 00 (α)h3 ,
2
6
1
1
f (x − h) = f (x) − f 0 (x)h + f 00 (x)h2 − f 00 (β)h3 .
2
6
Thus,
f (x + h) − f (x)
1
1
− f 0 (x) = f 00 (x)h + f 00 (α)h2 ,
h
2
6
f (x) − f (x − h)
1
1
− f 0 (x) = − f 00 (x)h + f 00 (β)h2 ,
h
2
6
1 and
using Taylor's theorem
10
so
1
2
f (x + h) − f (x) f (x) − f (x − h)
+
h
h
h2 00
− f (x) = [f (α) + f 00 (β)].
6
0
Section 4.3 #7
The problem with the analysis is that the expansions do not go to high enough order. If
we include one more term in the expansions for
2
nd the error is in fact O(h ).
f (x + h) − f (x)
and
f (x − h) − f (x),
we
Section 4.3 #10
a) We write the interpolating polynomial in Newton form:
P (x) = f (x0 ) + f [x0 , x1 ](x − x0 ) + f [x0 , x1 , x2 ](x − x0 )(x − x1 )
and compute
P 00 (x) = 2f [x0 , x1 , x2 ].
By the divided dierence recursion theorem,
f [x1 , x2 ] − f [x0 , x1 ]
f [x0 , x1 , x2 ] =
=
x2 − x0
f (x2 )−f (x1 )
x2 −x1
−
f (x1 )−f (x0 )
x1 −x0
− x0
(x0 )
− f (x1 )−f
h
=
h + αh
f (x2 )
2 f (x0 ) f (x1 )
−
+
.
= 2
h 1+α
α
α(α + 1)
b) Consider
P0 (x) = 1, P1 (x) = x − x1
and
x2
f (x2 )−f (x1 )
αh
P2 (x) = (x − x1 )2 .
Pi00 (x) = AP (x0 ) + BP (x1 ) + CP (x2 ),
for
A, B, C .
We will solve the equations
i = 0, 1, 2,
The equations are
0=A+B+C
0 = −hA + αhC
2 = h2 A + α2 h2 C.
It is straightforward to verify that
A=
2
,
h2 (1+α)
Section 4.3 #14
11
B = − h22α ,
and
C=
2
.
h2 α(1+α)
Write
1
1
1 (5)
1
f (x)h5 + . . .
f (x + h) = f (x) + f 0 (x)h + f 00 (x)h2 + f 000 (x)h3 + f (4) (x)h4 +
2
6
24
120
1
1
1
1 (5)
f (x − h) = f (x) − f 0 (x)h + f 00 (x)h2 − f 000 (x)h3 + f (4) (x)h4 −
f (x)h5 + . . .
2
6
24
120
and similarly
4
2
4
f (x + 2h) = f (x) + f 0 (x)2h + 2f 00 (x)h2 + f 000 (x)h3 + f (4) (x)h4 + f (5) (x)h5 + . . .
3
3
15
4
2
4
f (x − 2h) = f (x) − f 0 (x)2h + 2f 00 (x)h2 − f 000 (x)h3 + f (4) (x)h4 − f (5) (x)h5 + . . . .
3
3
15
Thus,
and
Notice
1
1
1 (5)
[f (x + h) − f (x − h)] = f 0 (x) + f 000 (x)h2 +
f (x)h4 + . . .
2h
6
120
(3)
1
1
2
2
[f (x + 2h) − f (x − 2h)] = f 0 (x) + f 000 (x)h2 + f (5) (x)h4 + . . . .
12h
3
9
45
that multiplying equation (3) by 4/3 and subtracting (4) gives
(4)
2
1
1
[f (x + h) − f (x − h)] −
[f (x + 2h) − f (x − 2h)] = f 0 (x) − f (5) (x)h4 + . . .
3
12h
30
Terminating this expansion at fourth order, we nd
2
1
1
0
[f (x + h) − f (x − h)] −
[f (x + 2h) − f (x − 2h)] − f (x) = |f (5) (ξ)|h4 .
3
12h
30
Section 4.3 Computer problems #4
See script online.
Section 7.1 #3 c,d
c.
In the rst step of naive Gaussian elimination, we divide
1 by 0, so the algorithm fails.
However, if we interchange rows 1 and 2, then we can do naive elimination to obtain
x2 = 2, x1 = 7.
d.
After doing naive elimination on the rst column, we nd a
0
in the
(2, 2)
and a
1
in
(3, 2) spot. Thus, continuing with naive elimination on the second column, we divide
1 by 0, and the algorithm fails.
the
Section 7.1 #7 d,e
12
d.
The Gaussian elimination sequence is, using

 
3 2 −1 7
3
2
−1
5 3 2


4
∼ 0 −1/3 11/3
−1 1 −3 −1
0 5/3 −10/3

 
3
2
0 68/15
3
0
0



∼ 0 −1/3 0 62/45 ∼ 0 −1/3 0
0
0
15 −37
0
0
15
Thus, to four signicant digits,
e.
Using
gauss_naive.m
gauss_naive.m,
 

7
3
2
−1
7
−23/3 ∼ 0 −1/3 11/3 −23/3
4/3
0
0
15
−37
 

64/5
1 0 0 64/15


62/45 ∼ 0 1 0 −62/15
0 0 1 −37/15
−37
x1 = 4.266, x2 = −4.133
again, we nd
and
x1 = 1, x2 = −1,
x3 = −2.466.
and
x3 = x4 = 0.
Section 7.1 Computer problem #3
See script online.
Section 7.3 #4
One example is

0
1

0
0
1
0
1
0
0
1
0
1
0
0
1
0

0
1
.
0
0
The system has a unique solution, but naive elimination fails at the very rst step.
Section 7.3 #6
Let A
j + 1.
be a diagonally dominant matrix such that the
Inductively, assume that after these eliminations row
ai,j
(i, j)th entry of A is 0 for i >
A up to and including row i.
Suppose naive elimination has been applied to the
be the
(i, j)th
entry of
A
(i + 1)st
is still diagonally dominant. Let
after these row operations. The next row operation is
ai+1,j = ai+1,j −
We check if the
i
row of
A
ai+1,i
ai,j ,
ai,i
(5)
is still diagonally dominant after this row operation.
Notice that the row operation makes the
the induction assumption that the
j = i, . . . , n.
ith
n
X
(i + 1, j) entries all 0 except for j ≥ i + 1.
A is diagonally dominant,
row of
|ai,j | ≤ |ai,i | − |ai,i+1 |.
j=i+2
13
By
Moreover, the
to begin with
(i + 1)th for of A is diagonally dominant, since A was diagonally dominant
and the row operations up to row i do not touch row (i + 1). Thus,
n
X
|ai+1,j | ≤ |ai+1,i+1 | − |ai+1,i |.
j=i+2
Combining the above displays,
n n X
X
ai,j ai+1,j − ai+1,i ai,j ≤
|ai+1,j | + |ai+1,i | a
a
i,i
i,i
j=i+2
j=i+2
ai,i+1
≤ (|ai+1,i+1 | − |ai+1,i |) + |ai+1,i | − ai,i+1 ai,i
ai,i+1
≤ ai+1,i+1 −
ai,i+1 .
ai,i
Thus, after the row operation (5), the
(i + 1)th
row of
A
is still diagonally dominant.
Section 8.1 #1b,c
Using
LU_fact.m,

1
0
a) L = 
3
0

the
LU
factorizations are
0
0
1
0
−3
1
2 −1/4

0
0
,
0
1
1
0
0
−1/20
1
0
b) L = 
 0
−4/3
1
0
0
−3/2


0 1/3
0
1 3
−1 

0 8
3 
0 0 −13/4


−20 −15 −10
−5
 0 −3/4 −1/2 −1/4
.
U =
 0
0
−2/3 −1/3
0
0
0
−1/2
1
0
U =
0
0

0
0
,
0
1
Section 8.1 #2b
The factorization is

1
0
L=
0
5
The
M
0
1
3
0
0
0
1
2

0
0
,
0
1

1
0
U =
0
0
from part a) is

1
0
0
0
1
0

 0 −3 1
−5 6 −2
14

0
0
.
0
1
0
3
0
0
0
0
4
0

2
0
.
0
0
It can be calculated by inverting the individual row operation matrices used to build
L,
then multiplying these matrices together.
Section 8.1 #4a
After the rst row operation, a zero appears in the
fails. Thus, the matrix has no
LU
(2, 2)
spot and so naive elimination
factorization.
Section 8.1 #11
a) The solution is
x1 = 2, x2 = −3, x3 = 0
and
x4 = −1.
b) In matrix notation the system is

6 0 0 0 12
3 6 0 0 −12


4 −2 7 0 14  .
5 −3 9 21 −2

Since the system is already in lower triangular form, one only needs to scale the columns
to get
L
(which must be unit lower triangular) and adjust

1
0
0
1/2
1
0
L=
2/3 −1/3 1
5/6 −1/2 9/7

0
0
,
0
1

6
0
U =
0
0
U
0
6
0
0
accordingly. Thus,

0 0
0 0
.
7 0
0 21
Section 8.1 Computer problems #11
See script online.
Section 8.2 #6,7,8
#6: c (Theorem 1, p. 329), #7: b (Theorem 2, p. 330) #8: a (Theorem 2, p. 330)
Section 8.2 #11
a) Since
A
is symmetric, its condition number can be computed from its eigenvalues.
15
Note that
−2 − λ
1
0
−2 − λ
1 det(A − λI) = 1
0
1
−2 − λ
= (−2 − λ)((2 + λ)2 − 1) + (2 + λ) = (2 + λ)(2 − (2 + λ)2 )
A
Thus, the eigenvalues of
in order of decreasing magnitude are
√
λ1 = − 2 − 2,
and so
λ2 = −2,
λ3 =
√
2−2
√
|λ1 |
2+2
√ ≈ 5.828.
κ(A) =
=
|λ3 |
2− 2
b) Since A is not symmetric, its condition number must be calculated from the eigenvalues
T
of A A (or singular values of A). Note that
1 − λ
1
1
T
2−λ
1 det(A A − λI) = 1
1
1
2 − λ
= (1 − λ)((2 − λ)2 − 1) − [(2 − λ) − 1] + [1 − (2 − λ)]
= (1 − λ)[(2 − λ)2 − 3].
Thus, the eigenvalues of
The singular values of
A
AT A
in order of decreasing magnitude are
λ1 =
√
3 + 2,
λ2 = 1,
in decreasing order are
√
λ3 = 3 − 2.
√
σi = λi , i = 1, 2, 3.
Thus,
p√
σ1
3+2
κ(A) =
= p√
≈ 3.732.
σ3
3−2
Section 8.2 Computer problems #2,7
See scripts online.
Section 8.3 #8,12,13
#8: d (Theorem 1, p.
345), #12: e (Gershgorin's theorem states that all eigenvalues
must be in the union of the intervals listed), #13: True (the
to
λ ∈ Ci (aii , ri ),
ith
inequality is equivalent
and every eigenvalue must be in at least one such disk).
16
Section 8.4 #2
Under appropriate assumptions on the inputs (including that
value zero), the output will be
of
A,
and
x
r ≈ 1/λn ,
with
λn
A does not have the eigen-
the smallest (in magnitude) eigenvalue
will be an approximation of the eigenvector corresponding to
λn .
Section 8.4 #4,5,6
We use the power method with
ϕ(x) = uT x
λ(5) = 17/5,
u = (1, 0, 0).


17/24
=  −1  .
17/24
and
x(5)
A
The aim is to estimate the dominant eigenvalue of
λ(5) ≈ λ = 2 +
√
x(5)
2,
In #4, ve iterations give
and the corresponding eigenvector,
 √ 
1/ 2
.

−1
≈x=
√
1/ 2
In #5, ve iterations give

λ(5) = −99/29,
The aim is to estimate the eigenvalue of
x(5)
A

99/140
=  1 .
99/140
furthest from
vector,
√
λ(5) + 4 ≈ λ = 2 − 2,
x(5)
4
and the corresponding eigen-
 √ 
1/ 2

1√  .
≈x=
1/ 2
In #6, ve iterations give
λ(5) = 99/58,
x(5)
The aim is to estimate smallest eigenvalue of
1/λ(5) ≈ λ = 2 −
√
A


99/140
=  1 .
99/140
and the corresponding eigenvector,
x(5)
2,
Section 8.4 Computer problems #4
See script online.
17
 √ 
1/ 2

1√  .
≈x=
1/ 2