Name
Math 311
Exam 2
Section 502
Spring 2010
Solutions
P. Yasskin
1
/45
3
/30
2
/40 Total
/115
1. (45 points) Consider the vector space of functions V = Span(1, cos x, cos 2 x) with basis
v1 = 1
v 2 = cos x
v 3 = cos 2 x
a. (5 pts) Show cos 2x ∈ V.
HINT: Use a trig identity.
cos 2x = 2 cos 2 x − 1 = −1 ⋅ v 1 + 2 ⋅ v 3 ∈ V.
b. (10 pts) Show another basis for V is
p1 = 1
p 2 = cos x
p 3 = cos 2x
Since there are 3 p’s and dim V = 3, we only need to show the p’s span OR are linearly
independent.
To show they span, let f ∈ V. Then
f = a ⋅ 1 + b ⋅ cos x + c ⋅ cos 2 x = a ⋅ 1 + b ⋅ cos x + c ⋅ 1 + cos 2x
2
c
c
c
= a+
⋅ 1 + b ⋅ cos x +
⋅ cos 2x = a +
p + bp 2 + c p 3
2
2
2 1
2
So {p 1 , p 2 , p 3 } span V.
To show they are linearly independent, suppose a ⋅ 1 + b ⋅ cos x + c ⋅ cos 2x = 0. Then
x = 0 : a+b+c = 0
x= π :
a−c = 0
2
x = π : a−b+c = 0
(1)
(2) 
(3)
(1) − (3) 
b=0
(4)
(1) + (3)  a + c = 0
(5)
So {p 1 , p 2 , p 3 } are linearly independent.

(2) + (5)  a = 0
(2) − (5)  c = 0
(OR expand in v’s.)
c. (6 pts) Find the change of basis matrix from the p basis to the v basis. Call it
p1 = 1 = v1
1 0 −1
p 2 = cos x = v 2

p 3 = cos 2x = −1 ⋅ v 1 + 2 ⋅ v 3
C
v↼p
=
0 1
0
0 0
2
d. (6 pts) Find the change of basis matrix from the
v basis to the p basis. Call it
v1 = 1 = p1
1 0
v 2 = cos x = p 2

1
+
cos
2x
1
1
2
v 3 = cos x =
= p1 + p3
2
2
2
(OR find the inverse of
C.
v↼p
C
p↼v
=
0 1
0 0
C.
p↼v
1
2
0
1
2
C .)
v↼p
1
e. (6 pts) Consider the linear function L : V  ℝ 2 given by L(f) =
f(0)
f π
2
Find the matrix of L relative to the v basis for V and the basis e 1 =
.
1
0
, e2 =
0
1
for ℝ . Call it A .
2
e↼v
1
L(v 1 ) = L(1) =
= e1 + e2
1
1
L(v 2 ) = L(cos x) =
= e1
0
1
L(v 3 ) = L(cos 2 x) =

A
e↼v
1 1 1
=
1 0 0
= e1
0
f. (6 pts) Find the matrix of L relative to the p basis for V and the e basis for ℝ 2 .
Call it B .
e↼p
1
L(p 1 ) = L(1) =
= e1 + e2
1
L(p 2 ) = L(cos x) =
1
0
1
L(p 3 ) = L(cos 2x) =
g. (6 pts) Recompute
= e1
B
B
e↼p
1 1
=
1
1 0 −1
= e1
−1
e↼p

by a second method.
(One method should use the change of basis matrix, the other should not.)
B
e↼p
=
A
e↼v
C
v↼p
=
1 1 1
1 0 0
1 0 −1
0 1
0
0 0
2
=
1 1
1
1 0 −1
2
2. (40 points) Consider a linear map L : R n  R p whose matrix relative to the standard bases is
A=
1
−2 0 −4
−1
2
1
0
0
1 −2
2
.
a. (4 pts) What are n and p?
n=4
p=3
b. (9 pts) Identify the kernel of L, a basis for the kernel, and the dimension of the kernel.
1
−2 0 −4
0
−1
2
1
2
0
0
0
1 −2
0
2s + 4t
w
x
y
=s
2t
t
0
0
0
0
1 −2
0
0
0
1 −2
0
1
0
0
2
1
0
2
s
=
z
basis =

1 −2 0 −4

1 −2 0 −4
0
0
0
1 −2
0
0
0
0
0
4
+t
0
2
1
0
2
 Ker(L) = Span
1
0
0
4
,
0
2
1
4
,
0
2
dim Ker(L) = 2
1
3
c. (9 pts) Identify the image of L, a basis for the image, and the dimension of the image.
−2
1
Im(L) = Span
−1
,
= Span
,
1
0
a
+b
−1
0
1
−1 1
0
+c
1
0
2
0
1 −2
0

2
Are they independent?
2
−2
0
=
0
−2
0
1 0 −4
0
0 1 −2
0
0 1 −2
0

1 0 −4
0
0 1 −2
0
0 0
0
0
a

b
4t
=
c
2t
t
Throw away the 3 rd vector because c is the free variable.
Not independent.
1
Im(L) = Span
,
−4
1
0 −4
−4
1
0
2
−2
1
0
−1
,
1
0
1
1
,
2
0
−4
0
−1
0
0
,
1
1
basis =
1
−1
0
0
,
1
dim Im(L) = 2
1
d. (6 pts) Is the function L one-to-one? Why?
L is not 1 − 1 because Ker(L) ≠ {0}
e. (6 pts) Is the function L onto? Why?
L is not onto because dim Im(L) = 2 but dim Codom(L) = 3
f. (6 pts) Verify the dimensions in a, b and c agree with the Nullity-Rank Theorem.
dim Ker(L) + dim Im(L) = 2 + 2 = 4 = dim Dom(L)
4
3. (30 points) Consider the vector space of functions V = Span(1, cos x, cos 2 x)
with the inner product ⟨f, g⟩ =
π
∫ 0 f(x)g(x) sin x dx.
Here is a table of integrals:
π
π
∫ 0 sin x dx = 2
π
∫ 0 cos x sin x dx = 0
π
∫ 0 cos 2 x sin x dx =
π
∫ 0 cos 3 x sin x dx = 0
π
2
5
∫ 0 cos 4 x sin x dx =
2
3
∫ 0 cos 5 x sin x dx = 0
a. (10 pts) Find the angle between the functions p = 1 and q = cos 2 x.
π
|p| =
⟨p, p⟩ =
⟨1, 1⟩ =
|q| =
⟨q, q⟩ =
⟨cos 2 x, cos 2 x⟩ =
⟨p, q⟩ = ⟨1, cos 2 x⟩ =
=
π
π
5 =
2
2
∫ 0 cos 4 x sin x dx
∫ 0 cos 2 x sin x dx =
⟨p, q⟩
= 2
|p||q|
3 2
cos(θ) =
∫ 0 sin x dx
2
5
=
2
3
5
3
θ = arccos
5
≈ 0. 73 rad ≈ 41. 8°
3
b. (20 pts) Apply the Gram-Schmidt procedure to the basis
v1 = 1
v 2 = cos x
v 3 = cos 2 x
to produce an orthogonal basis w 1 , w 2 , w 3 and an orthonormal basis u 1 , u 2 , u 3 .
w1 = v1 = 1
⟨w 1 , w 1 ⟩ = ⟨1, 1⟩ =
π
⟨v 2 , w 1 ⟩ = ⟨cos x, 1⟩ =
w2 = v2 −
|w 1 | =
∫ 0 sin x dx = 2
π
∫ 0 cos x sin x dx = 0
⟨v 2 , w 1 ⟩
w = v 2 = cos x
⟨w 1 , w 1 ⟩ 1
π
⟨w 2 , w 2 ⟩ = ⟨cos x, cos x⟩ =
∫ 0 cos 2 x sin x dx =
π
2
3
⟨v 3 , w 1 ⟩ = ⟨cos 2 x, 1⟩ =
w3 = v3 −
∫ 0 cos 2 x sin x dx =
π
∫0
2
3
|w 2 | =
2
3
⟨v 3 , w 2 ⟩ = ⟨cos 2 x, cos x⟩ =
π
∫ 0 cos 3 x sin x dx = 0
⟨v 3 , w 1 ⟩
⟨v , w ⟩
w 1 − 3 2 w 2 = cos 2 x − 2 1 − 0 = cos 2 x − 1
3⋅2
3
⟨w 1 , w 1 ⟩
⟨w 2 , w 2 ⟩
⟨w 3 , w 3 ⟩ = cos 2 x − 1 , cos 2 x − 1
3
3
=
2
cos 2 x − 1
3
2
sin x dx =
=
π
∫0
= 2 − 2 ⋅ 2 + 1 ⋅2 = 8
5
3 3
9
45
u1 = w1 =
|w 1 |
1
2
u2 = w2 =
|w 2 |
π
∫0
cos 2 x − 1
3
2
sin x dx
cos 4 x − 2 cos 2 x + 1
3
9
|w 3 | =
3 cos x
2
sin x dx
8
45
u3 = w3 =
|w 3 |
45
8
cos 2 x − 1
3
5