Name
ID
Math 311
Exam 2
Spring 2002
Section 503
P. Yasskin
Solutions
1.
10 points
1
/10
5
/10
2
/15
6
/15
3
/10
7
/25
4
/15
Which one of the following is NOT a vector space? Why?
a. Q = w, x, y, z  ∈ R 4 ∣ w + 2x + 3y + 4z = 0 
b. S = X ∈ M2, 2  ∣ AXA = X 
where M2, 2  is the set of 2 × 2 matrices and A =
c. T = p ∈ P 3 ∣ p1  = p0  + 1 
0
1
−1 0
.
where P 3 is the set of polynomials of degree less than 3.
T is NOT a vector space.
 p1  = p0  + 1 and q1  = q0  + 1
p, q ∈ T
 p + q 1  = p1  + q1  = p0  + 1 + q0  + 1 = p + q 0  + 2 ≠ p + q 0  + 1
 p+q ∉ T
2.
15 points
For ONE of the two vector spaces listed in #1 (say which), give a basis and the dimension.
Q is a vector space which is a 3-plane in R 4 thru the origin. The augmented matrix of the equation is
1 2 3 4 | 0 . So the parametric solution is
−2r − 3s − 4t
w
x
y
r
=
+s
0
t
−2 1 0 0
−3
1
=r
s
z
Basis is
−2
0
,
−3 0 1 0
0
1
−4
0
+t
0
0
1
dim Q = 3
−4 0 0 1
,
OR
S is a vector space.
If X =
a b
c d
So X ∈ S means
So X =
Basis is
a
−d c
b −a
b
=a
b −a
1
0
, then AXA =
0
0 −1
,
1
1 0
a b
0
−1 0
c d
−1 0
=
a b
0
+b
0 −1
0 1
1
c d
1
=
−d c
b −a
, or d = −a and c = b.
0 1
1 0
dim S = 2
1
3.
10 points
Which one of the following is NOT a linear function? Why?
a. I : C0, 2π   R
where C0, 2π  is the set of continuous real valued functions on the interval 0, 2π 
2π
and If  = ∫ x + fx  dx.
0
c. E : P 3  P 3
given by Epx  = x
2π
.
0 0
dpx 
− px .
dx
I is NOT linear because
Iaf  = ∫ x + afx  dx
1 1
given by ZX  = BXB where B =
b. Z : M2, 2   M2, 2 
2π
2π
0
0
aIf  = a ∫ x + fx  dx = ∫ ax + afx  dx
but
0
which are not equal.
4.
15 points For ONE of the two linear functions listed in #3 (say which), find the kernel and the image (as the
span of some vectors) and determine if the function is onto and/or one-to-one.
a b
Z is linear. If X =
, then ZX  = BXB =
c d
a b
KerZ  = X ∣ ZX  = 0  =
=
a
1
0
−1 0
ImZ  = ZX  =
+b
c d
0 1
0 0
a+c a+c
0
0
Z is NOT onto because ImZ  ≠ M2, 2 .
1 1
a b
1 1
0 0
c d
0 0
∣ a+c = 0
+d
=
0 0
0 1
a + c 
=
= Span
1 1
0 0
a
=
a+c a+c
0
0
.
b
−a d
1
0
−1 0
= Span
,
0 1
0 0
,
0 0
0 1
1 1
0 0
Z is NOT one-to-one because KerZ  ≠ 0 .
OR
E is linear. If px  = a + bx + cx 2 , then Epx  = xb + 2cx  − a + bx + cx 2  = −a + cx 2 .
KerE  = p ∣ Ep  = 0  = a + bx + cx 2 ∣ −a + cx 2 = 0  = a + bx + cx 2 ∣ a = c = 0 
= bx  = Spanx 
ImE  = Ep  = −a + cx 2  = Span1, x 
E is NOT onto because ImZ  = Span1, x  ≠ P 3 .
E is NOT one-to-one because KerE  ≠ 0 .
2
5.
10 points Let P 3 be the vector space of polynomials of degree less than 3. Which one of the following is NOT
HINT: Let p = a + bx + cx 2 and find which is not positive definite.
an inner product on P 3 ? Why?
1
a. ⟨p, q ⟩ 1 = ∫ x px qx  dx
0
b. ⟨p, q ⟩ 2 = p0 q0  + p1 q1 
c. ⟨p, q ⟩ 3 = p−1 q−1  + p0 q0  + p1 q1 
⟨p, q ⟩ 2 is NOT an inner product, because if p = a + bx + cx 2 then
⟨p, p ⟩ 2 = p0  2 + p1  2 = a  2 + a + b + c  2
So if p = x − x 2 then a = 0, b = 1, c = −1 and ⟨p, p ⟩ 2 = 0  2 + 0 + 1 − 1  2 = 0, but p ≠ 0
6.
15 points
For ONE of the two inner products listed in #5 (say which), find the angle between the polynomials
rx  = 4x
and
sx  = 6x 2
1
⟨p, q ⟩ 1 = ∫ x px qx  dx is an inner product.
0
1
1
1
0
1
0
1
1
0
1
0
⟨r, r ⟩ 1 = ∫ x rx  2 dx = ∫ 16x 3 dx = 4x 4  0 = 4
⟨s, s ⟩ 1 = ∫ x sx  2 dx = ∫ 36x 5 dx = 6x 6  0 = 6
1
⟨r, s ⟩ 1 = ∫ x rx sx  dx = ∫ 24x 4 dx =
0
0
⟨r, s ⟩ 1
24
cos θ =
=
= 2 6
5
|r||s|
5⋅2⋅ 6
5
24 x
5
1
0
|r| =
⟨r, r ⟩ 1 = 2
|s| =
⟨s, s ⟩ 1 =
= 24
5
θ = cos −1

6
2 6
≈ 0. 201 rad
5
OR
⟨p, q ⟩ 3 = p−1 q−1  + p0 q0  + p1 q1  is an inner product.
|r| = ⟨r, r ⟩ 3 = 32 = 4 2
⟨r, r ⟩ 3 = ⟨4x, 4x ⟩ 3 = 16 + 0 + 16 = 32
⟨s, s ⟩ 3 = ⟨6x 2 , 6x 2 ⟩ 3 = 36 + 0 + 36 = 72
⟨r, s ⟩ 3 = ⟨4x, 6x 2 ⟩ 3 = −24 + 0 + 24 = 0
⟨r, s ⟩ 3
0
cos θ =
=
=0

|r||s|
4 26 2
|s| =
⟨s, s ⟩ 3 =
72 = 6 2
θ = cos −1 0 ≈ π rad
2
3
7.
25 points Consider the vector space V = Spansin 2 θ, cos 2 θ, sin θ cos θ 
a. 3  Show e 1 = sin 2 θ, e 2 = cos 2 θ, e 3 = sin θ cos θ is a basis for V.
By definition of V, e 1 , e 2 and e 3 span V. Are they independent?
ae 1 + be 2 + ce 3 = 0
a sin 2 θ + b cos 2 θ + c sin θ cos θ = 0
θ = 0:

b=0
π
θ= :

a=0
2
a + b + c =0
θ = π:


c=0
2
2
2
4
So they are independent and they are a basis.
b. 7  Another basis is f 1 = 1, f 2 = sin 2θ, f 3 = cos 2θ. Find the change of basis matrices from the
e-basis to the f-basis and vice versa. Be sure to say which is which.
= sin 2 θ + cos 2 θ = e 1 + e 2
f 2 = sin 2θ = 2 sin θ cos θ
= 2e 3
f 3 = cos 2θ = cos 2 θ − sin 2 θ = −e 1 + e 2
f1 = 1
C
f
e
1 0 −1
1 0 0
1 0
1
0 1 0
0 2
0
0 0 1
1 0 −1
1
0
0 1
0
0
0
0 0
1
−1
1
2
=
2
1
2
0
1
2
0
−1
1
2
2
R2 − R1
0
1
2
0
1 0 −1
C

e f

R1 + R3
=
1 0
1
0 2
0
1 0 −1
1
0 0
2
−1 1 0
0 2
0
0
1R
2 3
1R
2 2
0 1
0 1 0
1
2
0
1
2
0
0 0 1
−1
1
2
1 0 0

0 0
2

0
1
2
0
0
1
2
0
4
(7 continued)
c. 4  Consider the derivative operator D : V  V given by Dg  =
Find the matrix of D relative to the e-basis.
De 1  = Dsin 2 θ 
= 2 sin θ cos θ
= 2e 3
De 2  = Dcos θ 
= −2 sin θ cos θ = −2e 3
De 3  = Dsin θ cos θ  = cos 2 θ − sin 2 θ = −e 1 + e 2
2
A

e e
=
dg
.
dθ
0
0
−1
0
0
1
2 −2
0
d. 6  Find the matrix of D relative to the f-basis in TWO ways.
i. by using the matrix of D relative to the e-basis:
A
f
f
= C
f
A
C
e ee ef
1
2
0
1
2
0
−1
1
2
=
2
0
=
0
0
1 0
1
0
0 2
0
1
1
2
0
1 0 −1
1 −1 0
0
0
0
0
−1
1 0 −1
0
0
1
1 0
1
2 −2
0
0 2
0
0 0
=
0
0 0 −2
0 2
0
ii. by differentiating basis vectors:
D f 1  = D 1 
=0
=0
Df 2  = Dsin 2θ  = 2 cos 2θ = 2f 3
Df 3  = Dcos 2θ  = −2 sin 2θ = −2f 2
0 0
A
f
f
=
0
0 0 −2
0 2
0
e. 5  Consider the function gθ  = 5 sin 2θ + 3 cos 2θ. Compute Dg  in TWO ways:
i. by differentiating:
Dg  = D5 sin 2θ + 3 cos 2θ  = 10 cos 2θ − 6 sin 2θ
ii. by using the matrix of D relative to the f-basis:
0
g  f =
5
3
0 0
Dg  f =
f
A g  f =
f
0
0
0 0 −2
5
0 2
3
0
0
=
−6
10
Dg  = −6f 2 + 10f 3 = −6 sin 2θ + 10 cos 2θ
5