MATH 311 Section 501
Exam 2
Spring 2013 P. Yasskin
1. (35 points 5 e.c.) Consider the vector space M 2, 2
of 2
1
/40
2
/30
3
/40
Total
/110
2 real matrices.
Consider the bases
1 0
E1
0 0
1 0
F1
0 1
E2
0 0
1 1
F2
0 0
Consider the function L : M 2, 2
1 0
1 1
F3
0 0
M 2, 2
0 0
E3
1 0
0 0
E4
0 1
1 1
F4
1 1
given by
2X X T
LX
a. (5 pts) Show L is linear.
Solution:
L aX bY
2 aX
bY
aX
bY
T
a 2X
X
T
b 2Y
b. (5 pts) Find the matrix of L relative to the E-basis. Call it
Y
T
aL X
bL Y
A.
E E
HINT: A is NOT a 2
2 matrix!
Solution:
L E1
L E2
L E3
L E4
A
E E
2 0
1 0
1 0
0 0
0 0
0 0
0 2
0 0
0
0 0
1 0
1 0
0 0
0 1
0
1
2 0
0 0
2
0
0 0
0 0
0 0
0 2
0 1
0 1
1
0
0
0
2
1 0
0
1
2
0
0
0
0
1
E1
2
2E 2
E2
E3
2E 3
E4
0
1
c. (5 pts) Find the change of basis matrix
from the F-basis to the E-basis.
C
E F
Solution:
F1
E1
F2
E1
E2
F3
E1
E2
E3
F4
E1
E2
E3
1 1 1 1
0 1 1 1
C
0 0 1 1
E F
E4
0 0 0 1
d. (5 pts) Find the change of basis matrix
from the E-basis to the F-basis.
C
F E
Solution:
1 1 1 1
1 0 0 0
R1
R4
1 1 1 0
1 0 0
1
R1
R3
0 1 1 1
0 1 0 0
R2
R4
0 1 1 0
0 1 0
1
R2
R3
0 0 1 1
0 0 1 0
R3
R4
0 0 1 0
0 0 1
1
0 0 0 1
0 0 0 1
0 0 0 1
0 0 0
1
1 1 0 0
1 0
1
0
0 1 0 0
0 1
1
0 0 1 0
0 0
0 0 0 1
0 0
C
F E
1
1
0
0
0
1
1
0
0
0
1
1
0
0
0
1
R1
R2
1 0 0 0
1
1
0
0
0
0 1 0 0
0
1
1
0
1
1
0 0 1 0
0
0
1
1
0
1
0 0 0 1
0
0
0
1
e. (10 pts) Find the matrix of L relative to the F-basis. Call it
B.
F F
Solution:
2
B
C
A
C
F F
F E
E E
E F
1
1
0
0
1
0
0
0
1 1 1 1
0
1
1
0
0
2
1 0
0 1 1 1
0
0
1
1
0
1
2
0
0 0 1 1
0
0
0
1
0
0
0
1
0 0 0 1
1
1
0
0
1
1
1 1
1
1 0 0
0
1
1
0
0
2
1 1
0
3
0
0
1
1
0
1 1 1
0
1 1 0
0
0
0
1
0
0
0
0
0 1
0 0
0 1
f. (5 pts) Find
B
by a second method.
F F
Solution:
L F1
L F2
d
1 0
0 0
0 0
0 0
2 2
1 0
1
0 0
1 0
1 0
bF 2
cF 3
c
1,
b
0,
L F4
F F
1 0
aF 1
L F3
B
2 0
dF 4
3,
a
a
b
c
d
1
1 1
1 1
2 0
1 0
1 0
2 2
1 1
1 1
2 2
1 1
1 1
1 0 0
0
3
0
1 1 0
0
0
2
c
d b
c
d
d
L F2
2 2
1
F1
F1
3F 2
F3
F3
F4
0 0
0 1
g. (5 pts Extra Credit) Looking at your solutions to (b), (e) and (f), find 3 linearly independent
eigenvectors of L. What are their eigenvalues?
Solution:
F1
L F3
F3
L F4
F4
L F1
F1, F3, F4
are linearly independent since they are part of a basis.
The eigenvalues are all 1.
3
2. (30 points) Consider the vector space M 2, 2
of 2
2 real matrices.
Consider the function of two matrices
tr XDY T
X, Y
2 0
where D
where tr means trace (sum of diagonal elements) and
a. (10 pts) Show X, Y is an inner product.
0 1
T
means transpose.
a b
HINT: First compute the inner product of X
e f
and Y
c d
.
g h
Solution:
tr
X, Y
tr
(1)
(2)
(3)
a b
2 0
e g
c d
0 1
f h
2ae
bf 2ag
bh
2ce
df 2cg
dh
2a 2
2ea
bZ
X, X
Y, X
X, aY
2ae
tr
bf
2cg
b tr XDZ T
a X, Y
f
2 1 1
0 0
2 0 0
1
3
arccos 1
3
h
0, i.e. X
d
0
b X, Z
Solution: P, Q
2 1 1 0 0 2 0 0 1
P, P
2 1 1 0 0 2 0 0 1 1
|P|
cos
c d
dh
1 0
b. (10 pts) Find the angle between the matrices P
Q, Q
P, Q
|P||Q|
2e 2g
b 2 2c 2 d 2 0 and
0 iff a b c
fb 2gc hd 2ae bf 2cg dh
X, Y
T
T
tr XD aY bZ
tr aXDY
bXDZ T
a tr XDY T
|Q|
a b
and Q
0 1
1
1 0
0
1
.
1
3
1
1
3
c. (10 pts) Let V
Find V
Solution: X
2a
d
So V
Span P, Q where P and Q are given in (b).
the orthogonal subspace to V within M 2, 2 .
a b
c d
0 and 2a
d
0 b
c 0
Find a basis for V
Solution: A basis is E 2
4
is in V
iff
X, P
0, or a
d
0, or Y
Span
0 1
0 0
,
0 and
X, Q
0 b
0, or
b
c 0
0 1
0 0
0 0
1 0
and its dimension.
0 1
0 0
,
E3
0 0
1 0
dim V
2
c
0 0
1 0
.
0 6
3. (35 points 5 e.c.) Consider the matrix A
.
1 5
a. (15 pts) Find the eigenvalues and corresponding eigenvectors of A.
Solution:
det A
1
6
det
5
1 5
2
6
5
6
2
3
0
2, 3
For
2:
For
3:
2 6
0
1
3
0
x
3r
1 3
0
0
0
0
y
r
3 6
0
1
2
0
x
2r
1 2
0
0
0
0
y
r
b. (10 pts) Find the diagonalizing matrix X so that A
are X
1
XDX
1
3
v2
1
2
v3
1
where D is diagonal. What
and D? Just state them. No derivation.
Solution:
3 2
X
X
1 1
1
1
2
1
3
2 0
D
0 3
c. (5 pts) Find A 4 .
Solution: A 4
XD 4 X
1
3 2
16
0
1
2
3 2
16
1 1
0
81
1
3
1 1
81 243
48
162
96
486
114 390
16
81
32
243
65
d. (5 pts) Find cos
Solution:
32
211
A .
cos A
X cos D X
1
3 2
cos 2
0
1
2
1 1
0
cos 3
1
3
3 2
1
0
1
2
3 2
1
2
5
12
1 1
0
1
1
3
1 1
1
3
2
5
e. (5 pts Extra Credit) What are the eigenvalues and corresponding eigenvectors of 5A?
Solution:
So
5A
A
3
1
10 and v 10
3
3
2
1
1
3
10
and
1
3
1
and
.
5A
2
A
1
2
1
15 and v 15
2
3
1
2
15
2
1
1
.
5