Math 323
Exam 2 Sample Problems
Solution Guide
October 31, 2013
Note that the following provides a guide to the solutions on the sample problems, but in
some cases the complete solution would require more work or justification.
1. Determine if the following sets of vectors are or are not linearly independent. Explain
how you know a set is or is not linearly independent. If a set is linearly dependent,
then find a linear dependency among the vectors.
    
1 
 3
(a) 4 , 0


5
9
        
1
−1
1 
 3







4 , 0 , 2 , 1
(b)


5
9
0
1
        
1
2
1
−1 


        

3   2  −3  0 

(c)   ,   ,   ,  
−4
−4
2
1 





2
0
−4
0
Solution: (a) These two vectors are linearly independent . To see this we see that if
3
1
0
α 4 + β 0 = 0 , then 4α + 0β = 0, and so α = 0. Then it follows that β = 0.
5
9
0
(b) These vectors must be linearly dependent . To see this, we have 4 vectors in a 3dimensional space (where a linearly indpendent set can have no more than 3 elements).
For a linear dependency we solve
 
 
 
   
3
1
−1
1
0









a 4 + b 0 + c 2 + d 1 = 0
5
9
0
1
0
for a, b, c, d. This gives us three linear equations
3a + b − c + d = 0
4a + 2c + d = 0
5a + 9b + d = 0.
Solving these equations we find a = −5, b = 1, c = 2, and d = 16, is one of many
possible solutions.
(c) These vectors are linearly dependent . To see this, take the 4 × 4 determinant of
the matrix whose columns are the four vectors given. We find that the determinant is
0, and so the vectors must be linearly dependent. Setting up equations as we did in
part (b), we find that one of many dependencies is
 
 
 
   
1
2
1
−1
0
3
2
−3
 0  0

 
 
   
4
−4 − 3 −4 + 2  2  + 0  1  = 0 .
2
0
−4
0
0
2. For (a)–(c) in problem 1, find bases for the spans of the given sets of vectors.
n o
3
1
4 , 0
Solution: (a) Since the vectors are linearly independent, we see that
5
9
is a basis for the span.
(b) Let V be the span of the four given vectors. Since the first two vectors are repeated
from part (a) and are linearly independent, we see that 2 ≤ dim V ≤ 3. We calculate


3 1 −1
det 4 0 2  = −80 6= 0,
5 9 0
so the first three vectors in our set are linearly independent. So
n −1 o
3
1
4 , 9 ,
2
5
0
0
forms a basis for V .
(c) We wish to find a bais for the column space of the matrix


1
2
1 −1
3
2 −3 0 
.
A=
−4 −4 2
1
2
0 −4 0
Applying Gaussian elimination, we find that the

1 0 −2
0 1 3
2
B=
0 0 0
0 0 0
reduced row echelon form of A is

0
0
.
1
0
Since the number of pivots (leading 1’s) of B is 3, the dimension of the column space
of A must be 3. The columns of A corresponding to the three pivots of B will be a
1 2 −1 3
2
basis for the column space of A: namely
, −4
, 01
.
−4
2
0
0
3. Given
1
v1 =
,
2
v2 =
−2
,
−3
S=
4 −5
,
−5 −6
find vectors w1 and w2 so that S is the transition matrix from {w1 , w2 } to {v1 , v2 }.
Solution: In the notation we have used in class S = [I]E
F , where F = {w1 , w2 } and
2
E = {v1 , v2 }. That is, for all v ∈ R we should have
[v]E = S[v]F .
The problem does not say to find w1 and w2 with respect to the
standarrd basis
1 −2
necessarily, but that is what we will do. Let U = [I]E = 2 −3 be the transition
matrix from E to the standard basis. Let V = [I]F = (w1 , w2 ) be the transition
matrix from F to the standard basis. Then
−1
−1
V.
S = [I]E
F = ([I]E ) [I]F = U
Therefore U S = V . Thus
V =
1 −2
4 −5
14 7
=
.
2 −3
−5 −6
23 8
7
Thus we can take w1 = ( 14
23 ) and w2 = ( 8 ) .
n o
n o
1
0
0
1
1
0
0 , 1 , 0
1 , 0 , 1
4. Let S =
be the standard basis of R3 . Let A =
0
0
1
0
1
1
n 0 o
1
1
1
1 , 0 ,
and B =
.
−1
0
1
(a) Show that A is a basis for R3 . Show that B is not a basis for R3 .
(b) Find a linear equation (or equations) whose set of solutions is Span(B).
3
(c) Find the coordinates of 4 with respect to the basis A. (You may want to do
5
the next part first.)
(d) Find the change of basis matrix which converts coordinates in the S-basis into
coordinates in terms of the A-basis.
Solution: (a) We calculate


1 1 0
det 1 0 1 = −2 6= 0,
0 1 1
and so A is a linearly independent set of vectors. Since A consists of 3 vectors, which
is also the dimension of R3 , we see that A must be a basis of R3 . On the other hand,
we calculate


1 1 0
det 1 0 1  = 0,
0 1 −1
and so B is a linearly dependent set. Therefore, B cannot be a basis.
1 1 0 a
1
(b) Let B = 10 01 −1
, and so Span(B) is the column space of B. A vector cb is in
the column space of B if and only if the system of equations with augmented matrix


1 1 0 a
 1 0 1 b 
0 1 −1 c
is consistent. We apply row operations to put the matrix in row echelon form:


1 1 0
a
 0 1 1
a − b .
0 0 0 a−b−c
This is consistent if and only if a − b − c = 0, which gives the defining equation for the
column space of B.
(d) Using the hint, we do part (d) first. We know that the change of basis matrix that
converts coordinates in terms of the A-basis into the S-basis is


1 1 0
[I]SA = 1 0 1 .
0 1 1
To convert from the standard basis to coordinates with respect to A, we need

 1
1
− 12
2
2
S −1
[I]A
=  12 − 12 12  ,
S = ([I]A )
1
− 21 12
2
(c) Let v =
3
4
5
. Then
 
 
 
 1
1
3
1
3
− 12
2
2
A 
1
1
1  


−2 2
4
= 2 .
[v]A = [I]S 4
=
2
1
1
1
−2 2
5 S
3 A
5 S
2
In other words,
   
 
 
3
1
1
0
4 = 1 + 2 0 + 3 1 ,
5
0
1
1
which can be checked easily.
5. Let


5 −4 1 8
A = 1 −2 1 2 .
3 0 −1 4
(a) Find bases for the row space and the column space of A.
(b) What is the rank of A?
(c) What is the nullity of A?
c
(d) Let b = 86 . For what value or values of c will Ax = b be consistent?
Solution: (a) We put A in reduced row echelon form and find


1 0 − 31 43
U = 0 1 − 32 − 13  .
0 0 0
0
From this we see that {(1, 0, − 31 , 43 ), (0, 1, − 32 , − 13 )} is a basis for the row space of A,
n −4 o
5
1 , −2
and
is a basis for the column space of A.
3
0
(b) The rank is the dimension of the row space of A, which is 2 .
(c) Since rank(A) + nullity(A) = 4 = the number of columns of A, we see that the
nullity of A is also 2 .
(d) The system of equations Ax = b is consistent if and only if b is in the column
space of A. After setting up the equations, we see that this happens only if c = 22 :
 
   
5
−4
22





2 1 − 3 −2 = 8  .
3
0
6
6. Let A and B be n × n matrices, and let O be the n × n zero matrix.
(a) Prove that AB = O if and only if the column space of B is a subspace of the null
space of A.
(b) Show that if AB = O, then rank(A) + rank(B) ≤ n.
Solution: (a) (⇒): Assume that AB = O. Let v be in the column space of B. Since
AB = O, if we multiply A by any column of B we get 0. Therefore, A times any linear
combination of columns of B will be 0, and so Av = 0. Thus, v ∈ N (A).
(⇐): Suppose that the column space of B is contained in the null space of A. Then in
particular A times any column of B is 0. Therefoer AB = O.
(b) Suppose that AB = O. By part (a) we see that the column space of B is contained
in the null space of A. We know that
rank(A) + dim N (A) = n.
The column space of B is contained in the null space of A, and the rank of B is the
dimension of its column space. Thus rank(B) ≤ dim N (A). From the equation above,
we see then that
rank(A) + rank(B) ≤ n.
7. Consider the function L : R3 → R2 defined by
 
a
2a + 6b




b
L
=
.
3c − 9a
c
(a) Is L a linear operator? Explain.
(b) Find a matrix A so that L(x) = Ax for x ∈ R3 .
(c) What is the dimension of the kernel and image of L. Explain.
Solution: (a) Yes The reason is that
   
a1
a2
2(a
+
a
)
+
6(b
+
b
)
1
2
1
2
L   b1  +  b2   =
3(c1 + c2 ) − 9(a1 + a2 )
c1
c2
 
 
a1
a2
2a1 + 6b1
2a2 + 6b2






b1
b2  
=
+
=L
+L
3c1 − 9a1
3c2 − 9a2
c1
c2
(b) We know that we can take
2 6 0
.
A = L(e1 ), L(e2 ), L(e3 ) =
−9 0 3
(c) The dimension of the kernel of L is equal to the nullity of A. The dimension of the
image of L is equal to the dimension of the column space of A, which is the same as
the rank of A. It is easy to see that the second and third columns of A are linearly
independent, so the rank of A is 2. Therefore dim Im(L) = 2 . Since
rank(A) + nullity(A) = 3,
the nullity of A is 1, and so dim ker(L) = 1 .
8. Let T : R2 → R2 be defined by
T (x) =
2 1
x,
1 −3
where x ∈ R2 is written with respect to the standard basis. Let B = {( 11 ) , ( 12 )}. What
is the matrix representing T with respect to the basis B? (I.e., what is [T ]BB ?)
Solution: We know that the transition matrix that converts vectors in coordinates
with respect to B into the standard basis S is
1 1
S
.
[I]B =
1 2
Therefore
[T ]BB = ([I]SB )−1 · [T ]SS · [I]SB
−1 8 13
1 1
2 1
1 1
.
=
=
−5 −9
1 2
1 −3
1 2
9. Suppose that A and B are similar n × n matrices. Let c ∈ R, and let I denote the
n × n identity matrix. Show that A − cI and B − cI are similar matrices.
Solution: Since A and B are similar matrices, there exists an invertible n × n matrix
S so that
B = S −1 AS.
Note then that
S −1 (B − cI)S = S −1 BS − S −1 (cI)S = A − cI.
Therefore, A − cI and B − cI are similar.
10. For each of the following matrices, find the eigenvalues and determine the corresponding
eigenspaces.
1 3
(a)
4 2


1 1 −1
(b) 2 0 1 
1 1 0
Solution: (a) Let A be the matrix given. Then
1−λ
3
det(A − λI) = det
= (1 − λ)(2 − λ) − 12 = λ2 − 3λ − 10.
4
2−λ
We solve
λ2 − 3λ − 10 = (λ − 5)(λ + 2) = 0,
and so the eigenvalues of A are λ = 5, −2 . We look for eigenvectors for each of these
eigenvalues. For λ = 5 we solve
1 3
a
a
=5
,
4 2
b
b
which amounts to the system a + 3b = 5a and 4a + 2b = 5b. This becomes the
homogeneous system
−4a + 3b = 0
4a − 3b = 0.
The second equation is redundant, and so ( ab ) is an eigenvector with eigenvalue 5 if and
only if a = 34 b. Thus v = ( 34 ) is one, and the space is 1-dimensional, so the eigenspace
with eigenvalue 5 is R · ( 34 ) . Similarly, for λ = −2 we solve
1 3
a
a
= −2
,
4 2
b
b
which leads to the system of equations
3a + 3b = 0
4a + 4b = 0.
1
) is an eigenvector with eigenvalue −2, and
Thus we need a = −b, and so w = ( −1
1
the corresponding eigenspace is R · ( −1
).
(b) Let B be the matrix given in the problem. Then


1 − λ 1 −1
−λ 1  = −x3 + x2 + 2x − 2.
det(B − λI) =  2
1
1 −λ
We solve
x3 − x2 − 2x + 2 = (x − 1)(x2 − 2) = 0,
√
and find that the eigenvalues of B are λ = 1, ± 2 . In a similar manner as in part (a),
0
we find that the eigenspace of eigenvalue λ = 1 is spanned by 1 . The eigenspace of
1
1√
√
√
eigenvalue λ = 2 is spanned by 1+2√ 2 , and the eigenspace of eigenvalue λ = − 2
2+ 2
is spanned by
1√
1−2√ 2
2− 2
.