EXAM
Exam # 1
Math 3351, Spring 2003
Feb. 19, 2003
ANSWERS
i
70 pts.
Problem 1. Consider the matrix

2 3
 1 0

A=
 2 2
 1 2
2 1
The RREF of A is the matrix

1
 0

R=
 0
 0
0
−1
−2
−2
0
−3
3
3
4
1
5

1 −17
0 −2 

1 −12 

1 −10 
1 −7

0 −2
3 0 −2
1
1 −1 0 −5 

0
0
0 1
2 
.
0
0
0 0
0 
0
0
0 0
0
A. Find a basis for the nullspace of A.
Answer :
Call the variables x1 , . . . , x6 . From R we see that x1 , x2 and x5 are leading
variables and x3 , x4 and x6 are free variables, say
x3 = α
x4 = β
x6 = γ.
The third row of R gives us the equation x5 + 2x6 = 0, so x5 = −2x6 =
−2γ. The second row of R gives the equation x2 + x3 − x4 − 5x6 = 0, so
x2 = −x3 + x4 + 5x6 = −α + β + 5γ. Finally, the first row of R gives the
equation x1 − 2x3 + 3x4 − 2x6 = 0, so x1 = 2x3 − 3x4 + 2x6 = 2α − 3β + 2γ.
Thus, we have
 
 
 
  

2
−3
x1
2
2α − 3β + 2γ
5
1
−1
x2   −α + β + 5γ 
 
 
 
  

 
 
 
x3  

α
 =
 = α 1  + β  0  + γ  0 .


0

x4  


β
 
1
0
  

−2




x5  

0
0
−2γ
1
0
0
x6
γ
So, the three vectors

2
−1
 
1
 ,
0
 
0
0

 
−3
1
 
0
 ,
1
 
0
0
form a basis for the nullspace of A.
1

2
5
 
0
 ,
0
 
−2
1

B. Find a basis for the rowspace of A.
Answer :
The nonzero rows in R form a basis for the rowspace of R, which is the same
as the rowspace of A. Thus, a basis of the rowspace of A is given by
1 0 −2 3 0 −2 ,
0 1 1 −1 0 −5 ,
0 0 0 0 1 2 .
C. Find a basis for the columnspace of A
Answer :
The leading entries in R are in columns 1, 2 and 5. For a basis of the column
space of A, we take the corresponding columns of A. Thus, the vectors
 
 
 
2
3
1
1
0
0
 
 
 
2 ,
2 ,
1 ,
 
 
 
1
2
1
2
1
1
form a basis of the column space of A.
D. What is the rank of A?
Answer :
The rank of A is the common value of the dimension of the row space and
the dimension of the column space. In this case, these spaces have three
basis vectors, and so have dimension three. Thus, the rank of A is three.
50 pts.
Problem 2. Let S be the subspace of R4 spanned by the vectors
 
 
 
 
2
4
0
4
1
0
−2
0




v1 = 
v2 = 
v3 = 
v4 = 
1 ,
1 ,
−1 ,
1 .
1
1
−1
2
A. Pare down the list of vectors above to a basis for S. What is the dimension
of S?
2
B. For each of the following vectors, determine if the vector is in S and, if
so, express it as a linear combination of the basis vectors you found in the
previous part of the problem.
 
 
12
10
2
0


w1 = 
w2 = 
 4 ,
2
7
5
Answer :
We can solve the problem in one step by putting all the vectors in a matrix, say


2 4
0 4 12 10
 1 0 −2 0 2
0 
,
A = v1 v2 v3 v4 w1 w2 = 
 1 1 −1 1 4
2 
5
1 1 −1 2 7
The RREF of A is

R=
r1
r2
r3
r4
r5
1 0 −2 0
2
 0 1
−1
1
0
r6 = 
 0 0
0 1
3
0 0
0 0
0

0
0 
.
0 
1
From this, we see that r1 , r2 and r4 are linearly independent and r3 is a linear
combination of r1 , r2 and r4 . The same must be true of the columns of A, so
we conclude that v1 , v2 and v4 are a basis of S. Since there are three vectors
in a basis for S, the dimension of S is 3.
Looking at R, we see that r5 = 2r1 − r2 + 3r4 . The same must be true in
A, so we conclude that w1 = 3v1 − v2 + 3v4 and so w1 ∈ S.
Looking at R, we see that r6 is not a linear combination of r1 , . . . , r4 (because
of the last row). The same must be true in A, so w2 is not a linear combination
of v1 , . . . , v4 . Thus, w2 ∈
/ S.
50 pts.
Problem 3. Let A be a 9 × 7 matrix and let B be a 6 × 9 matrix.
A. What is the largest possible value of the rank of A?
Answer :
The rank is the common value of the dimension of the row space and the
dimension of the column space. In this case, there are seven columns, so
the dimension of the column space must be less than or equal to 7, thus the
maximum possible rank is 7.
3
B. If the nullspace of A has dimension 3, what is the rank of A?
Answer :
The rank theorem says
rank(A) + dim(N (A)) = number of columns in A.
The number of columns is 7 and the dimension of the nullspace is 3, so the
rank of A is 4.
C. If the columnspace of B has dimension 4, what is the dimension of the
nullspace of B?
Answer :
Since the columnspace has dimension 4, the rank is 4. Since there are 9
columns in B, the rank theorem says
rank(B) + dim(N (B)) = 9,
so we conclude that dim(N (B)) = 5.
40 pts.
Problem 4. Let
A=
16 −21
.
10 −13
Find the characteristic polynomial of A and the eigenvalues of A.
Answer :
The characteristic polynomial is
p(λ) = det(A − λI)
16 − λ
= det
10
−21
−13 − λ
= (16 − λ)(−13 − λ) − (−21)(10)
= −208 − 16λ + 13λ + 210
= λ2 − 3λ + 2.
The eigenvalues of A are the roots of p(λ). We have p(λ) = (λ − 2)(λ − 2), so
the eigenvalues of A are 1 and 2.
60 pts.
Problem 5. In each part you are given an matrix A and its eigenvalues. Find
a basis for each of the eigenspaces of A. Determine if A is diagonalizable, and if it is, find a matrix P and a diagonal matrix D so that
P −1 AP = D.
4
A.


1
0 0
A = −2 −1 3 ,
−2 −2 4
Eigenvalues = 1, 2.
Answer :
First consider the eigenvalue λ = 1. In this case we have


0
0 0
A − λI = A − I −2 −2 3 .
−2 −2 3
Using a calculator, the RREF of

1
0
0
this matrix is

1 −3/2
0
0 .
0
0
Call the variables x, y and z. We see from the RREF that y and z are free
variables, say y = α and z = β. The top row of the last matrix gives us the
equation x + y − (3/2)z = 0, so x = −y + (3/2)z = −α + (3/2)β. Thus, the
nullspace of A − I is parametrised by

 
 
  
−1
3/2
x
−α + 32 β
 = α 1  + β  0 .
y  = 
α
z
0
1
β
Thus, a basis for the λ = 1 eigenspace of A is
 
 
−1
3/2
 1 ,
 0 .
0
1
To avoid fractions, we can multiply the second vector by 2 to get the alternative basis
 
 
−1
3
 1 ,
0
0
2
for the λ = 1 eigenspace of A.
Next, consider the eigenvalue λ = 2. In this case we have


−1 0 0
A − λI = A − 2I = −2 −3 3 .
−2 −2 2
The calculator gives the RREF of

1
0
0
A − 2I to be

0 0
1 −1 .
0 0
5
Thus, z is a free variable, say z = α. The second row of the RREF gives us
the equation y − z = 0, so y = z = α. The top row of the RREF gives the
equation x = 0. Thus, the nullspace of A − 2I is parametrized by
   
 
x
0
0
y  = α = α 1 .
z
α
1
Thus, the λ = 2 eigenspace of A is one dimensional with basis vector
 
0
1 .
1
We’ve now found three linearly independent eigenvectors of the 3 × 3 matrix
A, so we conclude that A is diagonalizable. To get the diagonalizing
matrix P , put the eigenvectors we’ve found in as columns of P , so we can
take P to be


−1 3 0
P =  1 0 1 .
0 2 1
We will then have P −1 AP = D, where


1 0 0
D = 0 1 0 ,
0 0 2
(each column contains the eigenvalue for the corresponding column of P .)
B.

−12
A = −14
−13
3
4
3

7
8 ,
8
Eigenvalues = −2, 1.
Answer :
For the λ = 1 eigenspace, we want to find the

−13
A − λI = A − I = −14
−13
By calculator, the RREF of A − I

1
0
0
is

0 −1
1 −2 .
0 0
6
nullspace of

3 7
3 8 .
3 7
We see that z is a free variable, say z = α. The second row gives the
equation y − 2z = 0, so y = 2α. The top row gives the equation x − z = 0,
so x = z = α. Thus, the nullspace of A − I is parametrized by
   
 
x
α
1
y  = 2α = α 2 .
z
α
1
Thus, the λ = 1 eigenspace of A is one dimensional with basis vector
 
1
2 .
1
Next, consider the λ = 2 eigenspace. In this case,

−10 3
A − λI = A + 2I = −14 6
−13 3
we have

7
8 .
10
The calculator gives


1 0 −1
0 1 −1
0 0 0
for the RREF of A + 2I. We see that z is free, say z = α. The second row
gives the equation y − z = 0, so y = z = α. The top row gives the equation
x − z = 0, so x = z = α. Thus, the nullspace of A + 2I is given by
   
 
x
α
1
y  = α = α 1 .
z
α
1
Thus, the λ = −2 eigenspace is one dimensional with basis vector
 
1
1 .
1
Since we’ve found bases for all the eigenspaces but have not produced 3
linearly independent eigenvectors, we conclude that A is not diagaonalizable.
60 pts.
Problem 6.
Let U = u1
u2 be the ordered basis of R2 where
2
3
u1 =
,
u2 =
.
1
2
7
A. Find the change of basis matrices SEU and SU E .
Answer :
The change of basis matrix SEU is defined by
this means mat(U) = ISEU , so
2
SEU = mat(U) =
1
We then have
SU E = (SEU )−1 =
U = ESEU . In matrix terms
3
.
2
2 −3
,
−1 2
using a calculator to find the inverse.
B. Let L : R2 → R2 be the linear transformation such that
L(u1 ) = 2u1 − 4u2
L(u2 ) = 3u1 .
Find [L]U U , the matrix of L with respect to the basis U.
Answer :
The defining equation for [L]U U is L(U) = U[L]U U . We have
L(U) = L(u1 ) L(u2 ) = 2u1 − 4u2
so we conclude that
[L]U U =
2
−4
3u1
= u1
u2
2
−4
3
,
0
3
.
0
C. Find the matrix of L with respect to the standard basis E of R2
Answer :
The change of basis formula for the matrix of a linear transformation says,
in this case,
[L]EE = SEU [L]U U SU E .
We’ve found all these matrices above, so we have
2 3
2 3
2 −3
−22
[L]EE =
=
1 2 −4 0 −1 2
−15
using a calculator for the computation.
8
36
,
24