Name
Math 311
Exam 2 Version A
Section 502
Spring 2015
Solutions
P. Yasskin
Points indicated. Show all work.
1.
1
/15
4
/27
2
/38 5 E.C.
/10
3
/25
Total
/115
15 points
Let P 5 be the vector space of polynomials of degree less than 5.
Consider the subspace V Span v 1 , v 2 , v 3 , v 4 where
v 1 2 3x 2 , v 2 x 3x 3 , v 3 x 2 x 3 x 4 , v 4 2 x 3x 4
Find a basis for V What is dim V ?
Solution: To see if v 1 ,
av 1 bv 2 cv 3 dv 4
a 2 3x 2
b x 3x 3
2a 2d
b d x
b d
2a 2d 0
v 2 , v 3 , v 4 are linearly independent, we write:
0 and solve for a, b, c, d.
c x2 x3 x4
d 2 x 3x 4
0
2
3
3a c x
3b c x
c 3d x 4 0
0
3a c 0
3b d 0
c 3d 0
2
0
0
2
0
0
1
0
1
3
0
1
0
3
0
0
1
2
R1
1
0
0
1
0
0
0
1
0
1
0
0
0
3
0
1
0
0
1
0
0
0
3
1
0
0
1
3
0
0
0
1
3
0
R3
3R 1
1
0
0
1
0
1 0
0
1
0
0
1
0
1
0
0 1
0
1
0
0
0
1
3
0
0 0
1
3
0
0
3
1
0
0
0 0
1
3
0
R4
R3
0
0
1
3
0
0 0
1
3
0
R5
R3
1 0 0
1
0
0 1 0
1
0
0 0 1
3
0
0 0 0
0
0
0 0 0
0
0
R4
3R 2
1
3
R3
a
d
b d
c 3d
Pick d 1. So a
1
b 1
c 3
So they are not linearly independent. Further,
v 1 v 2 3v 3 v 4 0
or
v 4 v 1 v 2 3v 3
So v 1 , v 2 , v 3 also span and the above computation without column 4 shows they are linearly
independent. So v 1 , v 2 , v 3 is a basis and dim V 3.
1
2.
38 points
Consider the vector space V Span 1, sin 2x , cos 2x
with the usual addition and
scalar multiplication of functions. Two bases are:
e 1 1 e 2 sin 2x e 3 cos 2x
and
E 1 sin 2 x E 2 cos 2 x E 3 sin x cos x
Note: You do NOT need to prove they are bases.
Hints: Here are some useful identities:
1 cos 2x
1 cos 2x
2 sin x cos x , cos 2x
cos 2 x sin 2 x , sin 2 x
, cos 2 x
sin 2x
2
2
a. (5) Find the change of basis matrix
from the e basis to the E basis by using the
C
E e
identities.
e1
1
sin 2 x
e2
sin 2x
2 sin x cos x
e3
cos 2 x
cos x
E2
C
2E 3
2
2
cos 2x
E1
sin x
E1
b. (5) Find the change of basis matrix
E e
E2
1 0
1
1 0
1
0 2
0
from the E basis to the e basis by using the
C
e E
identities.
1
E1
sin 2 x
E2
cos 2 x
E3
cos 2x
2
1 cos 2x
2
sin 2x
2
sin x cos x
c. (2) Verify
C
and
E e
1 0
1
0 2
0
1 e3
2
1 e3
2
C
e E
1
2
0
1
2
0
1
2
1
2
0
1
2
0
are inverses.
C
e E
1
1 0
1 e1
2
1 e1
2
1 e2
2
1
2
0
1
2
0
1
2
1
2
d. (4) For the function f
1
2
1
2
0
1
2
0
sin 2x
1
2
1
2
1
2
1
2
0
1
2
1
2
0
0
1 0 0
0
2
0 1 0
0 0 1
1
2
4 sin 2 x , what are its components
f
e
and
f E?
2
f
sin 2x
2
4 sin x
sin 2x
21
cos 2x
2e 1
e2
2e 3
f
1
e
2
4
f
sin 2x
2
4 sin x
2 sin x cos x
2
4 sin x
4E 1
2E 3
f
E
0
2
2
e. (5) Find the matrix
d
dx
of the derivative operator D
A
e e
D e1
D1
0
0
D e2
D sin 2x
2 cos 2x
2e 3
D e3
D cos 2x
f. (5) Find the matrix
2 sin 2x
B
A
e e
2e 2
relative to the e basis.
0 0
0
0 0
2
0 2
0
d
dx
of the derivative operator D
E E
relative to the E basis.
Do NOT use the change of basis matrices.
D E1
D E2
D E3
D sin 2 x
2 sin x cos x
2
D cos x
2E 3
2 sin x cos x
cos 2 x
D sin x cos x
B
2E 3
E E
sin 2 x
E1
E2
B
C A C
So S
0
1
0
0
1
2
2
0
SAS 1 . What is S?
g. (2) A and B are related by a similarity transformation: B
E E E ee ee E
0
C
E e
h. (3) What is Im D ? Give a basis. What is dim Im D ?
HINT: Let f
a 1
c cos 2x .
b sin 2x
Df
2b cos 2x 2c sin 2x
Im D
Df
2b cos 2x 2c sin 2x
Basis: e 2 , e 3
dim Im D
2
Span cos 2x , sin 2x
Span e 2 , e 3
i. (3) What is Ker D ? Give a basis. What is dim Ker D ?
Ker D
f
Basis: e 1
Df
0
dim Ker D
C where C is a constant
f
Span 1
Span e 1
1
j. (2) Is D onto? Why or why not?
D is not onto because Im D
Span cos 2x , sin 2x
V
Span 1, sin 2x , cos 2x
k. (2) Is D 1-1? Why or why not?
D is not 1-1 because Ker D
Span e 1
0
3
3.
25 points
Consider the vector space S of symmetric 2 2 matrices. The following are
symmetric, biinear forms on S. Which one(s) are inner products? Why or why not? You do not need
to check they are symmetic or bilinear, just that they are positive definite.
a b
HINTS: Let A
a. (9)
A, B
A, A
tr
tr
If
So if A
b. (8)
A, A
A, B
tr
tr
A, A
A, B
tr
tr
1 1
1 1
a b
1 1
a b
b d
1 1
b d
a
b
b
d a
a
1
1
1
1
2
a
b b
b
b
b
2
b
then
d
2
1 1
b d
d a
b
3a
d b
b
3b
2
b b
2
b
b d
2 1
b d
b
2d a
2b
a
d b
d b
d
b d
2
b
d
2
0
b, but nothing says b
0.
b a
b
a b
3b
d b
d
b d
b d
d
d
3a
2a 2
2
2
0. So a
a
b
0, b
2
2b 2
0, d
b
2
d
b
0.
b
2d
0
2 1
a b
b2
b
1 2
1 2
4ab
a
2bd
2b 2
a b
a2
a b
0. So this is not an inner product.
tr
where G
tr AGB
b
1 1
b d
3b
b a
3 1
a b
b2
a
b and d
0 but A
A, A
3 1
2ab
b
0. So a
a b
3a 2
a
2
d
d
tr
where G
tr AGB
If A, A
0 then 2a 2 a
So this is an inner product.
c. (8)
A, A . Look for perfect squares or complete the squares.
where G
tr AGB
0 then
A, A
. Compute
b d
tr
2b b
b
2
This may not be positive, for example if a
A, A
3b 2 3d 2 0.
So this is not an inner product.
2a
4bd
d
a
2b 2a
b
a b
b
2d 2b
d
b d
b d
2
2b and b
a
2b
2
2d but b
3b 2
2
3d 2
0 then
4
4.
27 points
Consider the vector space S of symmetric 2
A, B
4 0
where G
tr AGB
tr
A, A
4 3
0 1
3 4
3 4
4 3
tr
3 4
4 0
3 4
4 3
0 1
4 3
4 3
4 0
4 3
3 4
0 1
3 4
73 60
tr
tr
120
2
125
120
125
24
25
U1
16 3
3 4
12 4
3 4
16 3
4 3
16 3
4 3
12 4
3 4
2 0
V1
V2
0 3
2 1
1 0
V3
1 3
0
1
.
basis to produce an orthogonal basis
2 0
V1
1 W1
5
4 3
arccos 24
25
Apply the Gram-Schmidt Procedure to the V 1 , V 2 , V 3
W 1 , W 2 , W 3 and an orthonormal basis U 1 , U 2 , U 3 .
tr
12 4
125
60 52
A, B
|A||B|
W1, W1
.
3 4
125
b. (19) A basis for S is
W1
4 3
and B
120
60 73
tr
cos
4 3
52 60
tr
tr
4 0
73 60
tr
B, B
3 4
60 52
tr
.
0 1
a. (8) Find the angle between the matrices A
A, B
2 matrices with the inner product
0 3
tr
16 0
0
9
1
5
2 0
4 0
2 0
0 3
0 1
0 3
25
tr
8 0
2 0
0 3
0 3
5
|W 1 |
2 0
2/5
0
0 3
0
3/5
5
V 2, W1
tr
8
W2, W2
0 1
0 3
1 3
25
25
0 1
4 0
0 1
1 0
0 1
1 0
5
0 4
1
5
0
tr
V3
W3, W3
0
4
1
0
1 tr
25
1 W3
2
0 3
1 0
tr
1 0
1/ 5
0
4 0
2 0
0
0 1
0 3
1 0
4 0
0 1
0
0 1
1 0
1
0 1
0 1
4 0
1 0
5
|W 2 |
1 0
tr
4 0
2 0
0
1
0 3
4 0
0 1
0
1 0
5
1
tr
1
0
1
5
4 0
0
0 1
0
0
64
1 0
V 3, W2
W2
W2, W2
3 0
36
1
2
4 3
0 3
1/ 5
V 3, W1
W1
W1, W1
tr
2 0
0 1
0
3
tr
8 1
2 0
0 1
8 0
tr
V 3, W2
2 1
1 0
tr
tr
25
9
1 W2
5
V 3, W1
U3
1 3
tr
tr
W3
2 0
V 2, W1
W1
W1, W1
V2
U2
4 0
16 3
tr
W2
2 1
8
100
25
4
0
1
5
5
25
1
3 0
0
8
2 0
3/5
0
0 3
0
8/5
1 tr
25
12 0
3 0
0
0
8
8
2
|W 3 |
3/5
0
3/10
0
0
8/5
0
4/5
6
5.
2
10 points EC
Consider the vector space V
x1, x2
x 1 0 and x 2 0
consisting of ordered pairs of positive numbers with addition and multiplication defined by
x1, x2
y1, y2
x 1 y 1 , x 2 y 2 and a
x1, x2
a
a
x1 , x2
So vector addition is real number multiplication of corresponding components and scalar
multiplication is real number exponentiation of each component. Note the zero vector is 0
a. (5) Is u 1
1, 2
and u 2
3, 1
1, 1 .
a basis? Why or why not?
Solution: To see if they span, we need to find a and b so that
x1, x2
a u1 b u2
But then
x1, x2
a
1, 2
b
3, 1
1a, 2a
3b, 1b
1a3b, 2a1b
3b, 2a
So b log 3 x 1 and a log 2 x 2 .
To see if they are linearly independent, we write
a u 1 b u 2 0 and must see if a b 0 is the only solution. But this says
3b, 2a
1, 1 So b log 3 1 0 and a log 2 1 0
So they span and are linearly independent and so are a basis.
b. (5) Is v 1
1, 1
and v 2
3, 1
a basis? Why or why not?
Solution: To see if they span, we need to find a and b so that
x1, x2
a v1 b v2 a
1, 1
b
3, 1
1a, 1a
3b, 1b
1a3b, 1a1b
So b log 3 x 1 and x 2 1. So there is no way to produce a vector with x 2 1.
They don’t span!
To see if they are linearly independent, we write
a v 1 b v 2 0 and must see if a b 0 is the only solution. But this says
3b, 1
1, 1 So b log 3 1 0 but a can be arbitrary.
They are not linearly independent!
Either one says they are not a basis.
3b, 1
7