Name
ID
MATH 311
Exam 2
Fall 2001
Section 200
Solutions
P. Yasskin
1.
1
/10
4
/10
2
/10
5
/60
3
/10
10 points Let P 1 be the vector space of polynomials of degree ≤ 1. Suppose L : P 1
linear map which satisfies
L 1 + 4t = −2.
L 2 + 3t = 1,
Compute L 5 − 2t .




 
 

 


5
= 1
R
is a

5 − 2t = a 2 + 3t + b 1 + 4t = 2a + b + 3a + 4b t
2a + b = 5
3a + 4b = −2
a
b

3 4
a
3 4
b
−1
2 1
=
2 1

5

−2
=


=
1
8−3
5
−2
−1
4
−3 2
−2
5
22
−19
5 − 2t = 22 2 + 3t − 19 1 + 4t
5
5
L 5 − 2t = 22 L 2 + 3t − 19 L 1 + 4t = 22 1 − 19 −2 = 22 + 38 = 12
5
5
5
5
5

2.




 
 
      = f0 
f ∈ C −1, 1  | f 0  = f0 
10 points Which of the following is not a subspace of C 1 −1, 1 ? Why?
f −1 + f 1
P = f ∈ C 1 −1, 1 | f −1 = f 1
Q = f ∈ C 1 −1, 1 |
2
R=
    
f ∈ C −1, 1  | ∫ ft  dt = 1
1
1
S=
0
1

′
1
1

R is not a subspace because if f, g ∈ R, then ∫ f t dt = 1 and ∫ g t dt = 1
1
  
0
0
but ∫ f + g t dt = 2. So f + g ∉ R.
0
1
3.
10 points Duke Skywater is flying the Millennium Eagle through the Asteroid Belt. At the current
time, his position is ⃗r = 4, −1, 2 and his velocity is ⃗v = 3, 2, −1 . He measures that the electric
field and its Jacobian are currently




12
⃗=
E
and
2
⃗=
JE
9
2 0 1
0 4 3
.
1 2 9
Use a linear (affine) approximation to estimate what the electric field will be 2 sec from now.

⃗ ⃗r t
E

⃗ ⃗r t 0
≈E
⃗
+ JE
12
≈
2
0
2 0 1
+
9
4.
⋅ ⃗rt − ⃗rt 
0 4 3
⋅
1 2 9
10 points Let L : R 5
obtains the matrix
R
4

⃗ ⃗r t 0
≈E
3
2
−1
2  =
⋅  
⃗ ⃗v t 0 t − t 0
+ JE
12
2

5
+2
9
5
−2
22
=
12
5
be a linear map whose matrix is A. If A is row reduced, one
1 3 0 0 2
0 0 1 0 −1
0 0 0 1 2
0 0 0 0 0
What is the dimension of the kernel of L?
Be sure to explain why.
What is the dimension of the image of L?
 
 This has 2 free parameters since there are 2 columns without leading 1’s.

dim ImL = # linearly independent columns = column rank = row rank
Ker L = X ∣ AX = 0
So dim Ker L = 2.
= # linearly independent rows = # leading 1’s = 3.
It is OK to do one of these and then use the Nullity-Rank Theorem:
dim Ker L + dim Im L = dim Dom L = 5.



2
5.
 
60 points Let M 2, 2 be the vector space of 2 × 2 matrices. Let P 2 be the vector space of
P 2 given by
polynomials of degree ≤ 2. Consider the linear map L : M 2, 2
 
 
LM =
1x M

a b
Hint: For some parts it may be useful to write M =
a.
1
x
and/or p x = α + βx + γx 2 .
c d
3  Identify the domain of L, a basis for the domain, and the dimension of the domain.

 
Dom L = M 2, 2


1 0
Basis_Dom L =
0 1
,
0 0
,
0 0
0 0
0 0
,
1 0
0 1
dim Dom L = 4
b.
3  Identify the codomain of L, a basis for the codomain, and the dimension of the codomain.
CodomL  = P
Basis_CodomL  = 1, x, x 
dim CodomL  = 3
6  Identify the kernel of L, a basis for the kernel, and the dimension of the kernel.
2
2
c.
L
a b
c d
 
a b
= 1x
  
a b
Ker L = M ∣ L M = 0 =
= Span
0
1
x
c d
c d
a + bx
c + dx
= 1x


∣ a + b + c x + dx 2 = 0

1
0
Basis_Ker L =
−1 0


= a + b + c x + dx 2
1
−1 0
=
0
b
−b 0

dim Ker L = 1
6  Identify the image of L, a basis for the image, and the dimension of the image.
ImL  = LM  = a + b + c x + dx  = Span1, x, x  = P
Basis_ImL  = 1, x, x 
dim ImL  = 3
e. 2  Is the function L one-to-one? Why?
L is not one-to-one because KerL  ≠ 0 .
f. 2  Is the function L onto? Why?
L is onto because ImL  = P = CodomL .
g. 2  Verify the dimensions in a, b, c and d agree with the Nullity-Rank Theorem.
1+3 = 4
dim KerL  + dim ImL  = dim DomL 
d.
2
2
2
2
2
3
h.
6  Find the matrix of L relative to the standard bases: (Call it A .)
E←e
1 0
e1 =
0 0
0 1
e2 =
,
,
0 0
and E 1 = 1,
a b
L
a b
= 1x
c d
 
L e  = x = E
L e  = x = E
Le4  = x = E
6  Another basis for P
1
x
c d
L e1 = 1 = E1
2
2
3
2
2
i.
0 0
e3 =
,
1 0
E 2 = x,
 
0 0
e4 =
for M 2, 2
0 1
E 3 = x 2 for P 2


a + bx
c + dx
= 1x
= a + b + c x + dx 2
1 0 0 0
A=
0 1 1 0
E←e
0 0 0 1
3
is F 1 = 1 + x, F 2 = 1 + x 2 , F 3 = x + x 2 . Find the change of basis
matrices between the E and F bases. (Call them C and C .) Be sure to identify which is which!
2
F←E
F1 = 1 + x = E 1 + E 2
1 1 0
C=
F2 = 1 + x 2 = E 1 + E 3
F3 = x + x = E 2 + E 3
1 0 0
1 0 1
0 1 0
0 1 1
0 0 1
−1 R
2 3

1 0 1
E←F
2
1 1 0
E←F
R2
R3
R1

0 1 1
1 0 1
0 1 0
0 1 1
0 0 1
1 1 0
1 0 0
1 0 1
0
1
0
0 1 1
0
−1
2
0
1
2
1
1
2
0 0 1

R 3 −R 1 −R 2

0
1
0
0 1
1
0
0
1
1
2
1
2
−1
2
0 1 0
0 0 1
1
1
−1
2
2
2
1
−1 1
C=
2
2
2
F E
−1 1
1
2
2
2
j. 6 Consider the polynomial q = 2 + 4x. Find q
E and F bases, respectively. Check q F .
1
0 0 −2
1 0 0
R 1 −R 3
R 2 −R 3
1 0
1
2
−1
2
1
2
1 −1 −1
−1
2
1
2
1
2
←

q = 2E 1 + 4E 2


q 
2
E
=
4
0

 
3F 1 − 1F 2 + 1F 3 = 3 1 + x − 1 1 + x 2
E

and q
F
1
2
1
q F =C q E =
2
F E
−1
2
2
+ 1 x + x = 2 + 4x = q

 
←


, the components of q relative to the
1
2
−1
2
1
2
−1
2
1
2
1
2
2
4
0
3
=
−1
1
4
k.
5  Find the matrix of L relative to the e basis for M2, 2  and the F basis for P . (Call it B .)
2
F←e
B
=
F←e
l.
=
C
A
F←E
E←e
1
2
1
2
−1
2
−1
1
2
−1
2
1
2
2
1
2
1
2
1
2
1
2
−1
2
1 0 0 0
0 1 1 0
=
0 0 0 1
1
2
−1
2
1
2
1
2
−1
2
1
2
−1









2
1
2
1
2
5  Find B by a second method.
F←e
 
L e  = x
L e  = x
Le4  = x
L e1 = 1
2
= dF 1 + eF 2 + fF 3
3
= dF 1 + eF 2 + fF 3
= gF 1 + hF 2 + iF 3
2
B=
F←e
m.
     
= d1 + x  + e1 + x  + fx + x 
= d1 + x  + e1 + x  + fx + x 
= g1 + x  + h1 + x  + ix + x 
= aF 1 + bF 2 + cF 3 = a 1 + x + b 1 + x 2 + c x + x 2
1
2
1
2
−1
2
1
2
−1
2
1
2
1
2
−1
2
1
2
2
2
2
2
2
2






= 1 1 + x + 1 1 + x2 − 1 x + x2





−1
2
1
2
1
2
6  Consider the matrix N = 13 24 . Find N  , the components of N relative to the e-basis
and LN  , the components of LN  relative to the F basis. Use LN  to find LN ?
e
F
F
1
N=
1 2
= 1e 1 + 2e 2 + 3e 3 + 4e 4
3 4
N 
e
=
2
3
4
1
1
1
−1
1
2
2
2
2
2
1
−1 −1 1
LN F = B N e =
2
2
2
2
3
F e
−1 1
1
1
4
2
2
2
2
L N = 1F 1 + 4F 3 = 1 1 + x + 4 x + x 2 = 1 + 5x + 4x 2
  
n.
←

1
=
0
4

   
2  Recompute LN  using the definition of L.

LN =
1x
1 2
3 4
1
x



2
2
2
1
1
2
=
1+x −
1 + x + 1 x + x2
2
2
2
1
1
2
=
1+x −
1 + x + 1 x + x2
2
2
2
1
1
2
= − 1+x +
1 + x + 1 x + x2
2
2
2
= 1 + 5x + 4x 2
5
