MATH 260 – Final Exam
April 30, 2013
1. Let M pn,
Rq be the space of n-by-n matrices with real entries.
(a) We know that (with the operations of matrix addition and scalar multiplication), M pn,
is a vector space. What is the dimension of M pn, q? Exhibit a basis of M p2, q.
R
R
A : M p2, Rq Ñ M p2, Rq by
A pX q AX,
Suppose A is a fixed 2-by-2 matrix. Define the map
so the operation A is (left) multiplication by the fixed matrix A.
(b) Show that A is a linear map from M p2,
(c) Show that A B A
A and B and for any scalar c.
Rq to M p2, Rq.
c A , and that
B , that cA
Rq
AB
A B for 2-by-2 matrices
(d) What is the relationship of the determinant of A and the determinant of A ? What about
the trace of A and the trace of A ?
(e) Suppose A is diagonalizable. Does it follow that A is also diagonalizable?
(a) Since an n-by-n matrix has n2 independent entries, we have dim M pn,
M p2, q is
"
*
1 0
0 1
0 0
0 0
,
,
,
.
0 0
0 0
1 0
0 1
R
Rq n2. A basis for
(b) We have
A pαX
βY q ApαX
αAX βAY
α A pX q β A pY q
(and the latter is a 2-by-2 matrix) so A : M p2,
B pX q
(c) First, A
A
B
A
pA
B qX
βY q
Rq Ñ M p2, Rq is a linear map.
AX
BX
A pX q
B X for all X
P M p2, Rq, so
B.
Next, cA pX q pcAqX
cpAX q c A pX q for all X P M p2, Rq, so
Finally AB pX q pAB qX
ApBX q (d) If A a
c
b
d
and X
u
x
v
y
A
A pX q cA.
p B pX qq for all X P M p2, Rq, so
, then
cA
au
cu
bx
dx
av
cv
by
dy
,
AB
A B.
2
so the 4-by-4 matrix of A is
A
It’s easy to see that tr A
det A
a a
0
c
0 b
d 0
0 d
2a
b a
0
c
0
0
a
0
c
.
b 0
0 b
d 0
0 d
2d 2 tr A. Expanding along the first row gives us
0
c
0
a
0
c
b
0
d
2 2
a d abcd
b2 c2 abcd pad bcq2
pdet Aq2 .
(e) If v is an eigenvector of A corresponding to the eigenvalue λ, then the matrix X with v in its
first column and zeros in the second column is an eigenvector of A with eigenvalue λ. Likewise the
matrix X with v in the second column and zeros in the first column is an eigenvector of A with
eigenvalue λ. Thus, if A is diagonalizable, so is A and it has two eigenvectors for each eigenvector
of A — note that this gives another proof of the assertions about the trace and determinant of A .
2. Give examples of each of the following or explain why none can exist:
¾
(a) A 2-by-2 matrix A such that A8
I but Ak I for k P t1, 2, 3, 4, 5, 6, 7u.
(b) A 2-by-2 matrix A such that A4
I.
(c) A 3-by-3 matrix A such that A4
I
(d) A vector field F, defined and continuously differentiable for all px, y q
F dr π, where C is the unit circle x
2
y
2
1, traversed counterclockwise.
P R2 ,
such that
P R3 ,
such that
C
»»
S
(e) A vector field F, defined and continuously differentiable for all px, y, z q
F n dσ
4π, where S is the unit sphere x
2
y
2
z
2
1, n is the outward-pointing normal
vector to S, and dσ is the element of surface area.
(a) Rotation through the angle π {4 will do it, so the matrix is
b
b 1
1
b2 b2 .
1
2
1
2
(b) The same matrix as part (a) will do it.
(c) The determinant of the 3-by-3 identity matrix is
real matrix A. So there is no such example.
(d) The vector field x j will do, by Green’s theorem.
1, so this can never equal pdet Aq4 for a
3
(e) Either using the divergence theorem or the fact that n
the vector field r x i y j z k will do.
r on the unit sphere, we see that
3. Find and classify all of the critical points of the function f px, y q xy 2
We have
fx
y2
2xy
12
and fy
2xy
x2 y
12x.
x2 .
Since fy xp2y xq we must have either x 0 or x 2y at a critical point. If x 0, then from
fx 0 we get that y 2 12 0, which has no solutions. If x 2y then we have y 2 and so
x 4. So there are two critical points: p4, 2q and p4, 2q.
H
To classify the critical points, note that fxx 2y, fyy 2x and fxy
2y q2 4x2 4xy 4y 2 . Therefore:
fxx fyy pfxy q2 4xy p2x
Since H p4, 2q 64
32 16 0, this is a saddle point, as is
2y
2x. Therefore
p4, 2q.
Two critical points, both saddles.
4. Let L be a linear transformation from
Lpi
(a) Show that i
j
jq k ,
R3 to R3 that satisfies
Lpj
kq i ,
Lpk
iq j.
k is an eigenvector of L. What is its eigenvalue?
(b) Is L invertible?
(c) Find the matrix ML of L (with respect to the basis ti , j , ku.
(d) Find an orthogonal matrix R such that RT ML R is diagonal.
(a) Adding the three equations gives us 2Lpi
(b) Since the image of L is all of
pendent vectors), L is invertible.
1
2
kq i
j
j
k so the eigenvalue is 21 .
R3 (since we see that the image contains three linearly inde-
(c) We have Lpiq 12 Lppi jqpj kq pk iqq pk i jq and Lpkq 12 Lppi jq pj kq pk
1
2
pk i jq, Lpjq 12 Lppi jq pj kqpk iqq iqq 21 pk i jq. Therefore
1 1
2 2
ML 12 21
1
2
1
2
1
2
1
2
1
2
.
(d) We already know one eigenvector, namely r1, 1, 1s with eigenvalue 12 . Also, it’s easy to see
that ML I has all its entries equal to 21 , so it’s a matrix of rank 1. We can take any orthogonal pair
4
of vectors perpendicular to r1, 1, 1s as our eigenvectors for the eigenvalue
and r1, 1, 2s. Then we can take
R
1
3
1
?
3
1
?
3
?
R T ML R and then
1
2
1
?
2
?
0
1
2
0
0
1
6
1
?
6
2
?
6
?
0
1
0
0
0
1
1.
So we take r1, 1, 0s
.
R R
R
5. Suppose f :
Ñ is a differentiable function with f pxq ¥ 0 for all x. Let S be the surface
of revolution in 3 defined by revolving the graph of the function y f pxq (for a ¤ x ¤ b) in 2
around the x-axis.
R
(a) Using the parametrization x u, y
f puq cos v, z f puq sin v, derive the formula
»b
areapS q a
a
2πf pxq 1
pf pxqq2 dx.
1
(b) Apply this formula to calculate the (lateral) surface area of the cone obtained by rotating
the part of the line segment y 3 3x in the first quadrant around the x-axis.
(a) We have r u i f puq cospv q j f puq sin v k, therefore ru
and rv f puq sin v j f puq cos v k. Thus
i
f 1 puq cos v j
f 1 puq sin v k
}ru rv } du dv }f puqf puq i f puq cos v j f puq sin v k} du dv
a
f puq 1 pf puqq2 du dv
The surface S is parametrized by u (with a ¤ u ¤ b) and v (with 0 ¤ v ¤ 2π q. Therefore
1
dσ
1
areapS q »»
dσ
S
» b » 2π
a
0
a
pf puqq2 dv du f puq 1
»b
1
a
a
2πf puq 1
pf puqq2 du »b
1
a
a
2πf pxq 1
since u x.
(b) Since f pxq 3 3x (for 0 ¤ x ¤ 1) and f 1 pxq 3, we have
areapS q »1
0
y2
z2.
(a) Compute ∇ V.
? 2π p3 3xq 10 dx 2π 10
6. Let V be the vector field on
x2
?
R
3
3x 3x2
2
1
3π?10.
0
defined (except at the origin) by V ∇
1
, where ρ2
ρ
pf pxqq2 dx
1
5
»»
(b) Calculate
V n dσ where SR is the surface of the sphere x2
y2
z2
R ¡ 0 and n is the outward-pointing normal vector..
SR
R2 for some
(c) Why does the result of part (b) not contradict the divergence theorem?
(d) Use the divergence theorem to explain why the result of part (b) doesn’t depend on R.
»»
(e) Calculate
T
V n dσ where T is the surface of the sphere px 3q2
py 5q2 pz 7q2 4
(and n is the outward-pointing normal vector).
(a) We have
V∇
a
x2
1
y2
x
p x2
z2
Therefore
∇V y2
3
px2
z 2 q3{2
px2
y2
z 2 q3{2
3px2 y 2 z 2 q
px2 y2 z2 q5{2
z 2 q3{2
y2
,
y
,
px2
z
y2
z 2 q3{2
.
0.
R2 , we have
x , y , z ,
V
(b) From part (a), on the sphere x2
y2
z2
R3
R3
R3
and since n rx{R , y {R , z {Rs we have
Vn
which is a constant on SR , so
»»
SR
px2
V n dσ
y2
R4
z2q
R12
R12 areapSR q 4π.
(c) This doesn’t contradict the divergence theorem because the vector field V is not defined at
the origin, and it becomes singular there.
(d) The result doesn’t depend on R because if we consider the solid shell Ω between two spheres
SR1 and SR2 , we can apply the divergence theorem and conclude that the integral of V n is the
same over both spheres (and thus is independent of R).
(e) The sphere in question doesn’t surround the origin, so we can apply the divergence theorem
to it, and get that this integral is zero.
7. (a) Find the maximum value of the function f px, y, z q ln x ln y 3 ln z on the part of the
sphere x2 y 2 z 2 5r2 in the first octant (where x ¡ 0, y ¡ 0 and z ¡ 0). Be sure to explain
how you know that the critical point you find is the maximum.
6
(b) Your answer to part (a) should be in terms of r. Re-express r in terms of x, y and z using
the constraint equation, and write out the inequality that implied by your answer to part (a), which
will look like
ln x ln y 3 ln z ¤ an expression in x, y and z
(c) (Extra Credit:) Multiply both sides of the inequality in part (b) by 2, and let a x2 , b y 2
and c z 2 , to prove that
5
a b c
3
.
abc ¤ 27
5
(a) First off, as any of x, y or z approach zero, their logarithms approach 8, so f px, y, z q will
approach 8 at the edges of the part of the sphere. Therefore if we find only one critical point, it
will be the global maximum.
We set g px, y, z q x2 y 2
gives us the three equations
z 2 5r2 , and use Lagrange multipliers. The equation ∇f
1
x
2λx
2λy
1
y
3
z
λ∇g
2λz.
We multiply these equations by x, y and z respectively and add them together to get
5 2λpx2
y2
z 2 q 10λr2
using the constraint equation. Thus λ 1{p2r2 q, and we conclude that
1
x2
r12
r12
1
y2
3
z2
r12 .
Since we know x, y, and z are positive we have only the one critical point px, y, z q
Therefore, the maximum value of f on the domain is
ln r
(b) From (a), using that r2
z ¡ 0,
ln x
ln y
3 ln z
¤ 5 ln r
?
3 lnp 3 rq 5 ln r
ln r
51 px2
?
3 ln 3 5 ln
c
x2
y2
5
?
5 x2
ln
2
(c) Multiplying by 2, we have
ln x2
ln y 2
3 ln z 2
¤ 5 ln x
2
y2
5
z2
ln 27.
Now set a x2 , b y 2 and c z 2 to get
ln a
ln b
3 ln c ¤ 5 ln
a
b
5
3 r q.
?
3 ln 3 z2
?
3 ln 3.
z 2 q, we can conclude that for all x
y2
pr , r ,
c
ln 27
y2
5
¡ 0, y ¡ 0 and
z2
1
ln 27.
2
7
for all a ¡ 0, b ¡ 0 and c ¡ 0. Finally, exponentiate both sides and get
3
abc
¤ 27
a
b
5
c
5
(exponentiating both sides preserves the inequality because the exponential function is monotonically
increasing).
8. (a) Prove that for any (twice continuously differentiable) vector field V on
R3, we have
∇ p∇ Vq 0.
(b) Find a vector field V such that ∇ V 2x i
»»
(c) Evaluate
S
p2x i
above the xy-plane (so z
3y j 5z k.
3y j 5z kq n dσ where S is the part of the paraboloid z
4 x2 y 2
¡ 0), oriented so that the unit normal vector points away from the origin.
(a) Let Vpx, y, z q f px, y, z q i
g px, y, z q j
hpx, y, z q k. Then
∇ V phy gz q i
pfz hx q j pgx fy q k.
Therefore
∇ p∇ Vq phy gz qx
pfz hx qy pgx fy qz
fyz hxy gxz fyz
hxy gxz
0.
(b) First it’s an easy check that ∇ p2x i
3y j 5z k 2
3 5 0, so there’s hope.
Next, we need to solve hy gz 2x, fz hx 3y, gx fy 5z for f , g and h. Just to be
egalitarian, let’s tentatively set h xy and g xz, so that hy gz x pxq 2x, which makes
the first equation true. To make the second equation true, we’ll then need f 4yz so that fz hx 4y y 3y. And now we can check that the third equation becomes gx fy z 4z 5z
(that’s why we needed ∇ V 0, so that the third equation would be automatically satisfied once
we ran out of choices to make). So we’ll take
V 4yz i xz j
(c) By Stokes’s theorem, we have
»»
S
p2x i
3y j 5z kq n dσ
»»
S
xy k.
∇ V dS
»
BS
V dr.
Now B S is the circle of radius 2 centered at the origin in the xy-plane (where z 0). And when
z 0, we have V xy k. But k is perpendicular to dr, since dr is tangent to the xy-plane.
Therefore we end up integrating zero, and so we can conclude that
»»
S
p2x i
3y j 5z kq n dσ
0.
8
Just for the record, parametrize the paraboloid via x u cos v, y u sin v and z 4 u2 for
0 ¤ u ¤ 2 and 0 ¤ v ¤ 2π. Then ru cos v i sin v j 2u k and rv u sin v i u cos v j and so
n dσ
and so
»»
S
p2x i
ru rv du dv p2u2 cos v i
3y j 5z kq n dσ
» 2π » 2
0
16π
2u2 sin v j
4u3 cos2 v
0
24π 80π
u kq du dv
6u3 sin2 v 5p4 u2 qu du dv
40π
0.