PROBLEM SET
Problems on Span, Independence, and Matrix Spaces
Math 3351, Fall 2010
Sept. 27, 2010
ANSWERS
i
Problem 1. Consider the matrix

−15 −80 −145

 −89 −19
51

A=
 66 −62 −190

77
81
85

22
−181
50

132 

.
−96 

150
78
−8
A Find a basis of the nullspace of A.
B Find a basis of the rowspace of A.
C Find a basis of the column space of A. Write the other columns of A as linear
combinations of these basis vectors.
D What is the rank of A?
Answer :
The RREF of A is

1
0
−1
0

 0

R=
 0

0
1
2
0
0
0
1
0
0
0
−1


3 

.
2 

0
To find a basis of the nullspace of R, which is the same as the nullspace of
A, solve the system Rx = 0. The variables x3 and x5 are free parameters, say
x3 = α and x5 = β. The rows of R give us the following equations
x4 + 2x5 = 0 =⇒ x4 = −2x5 = −2β
x2 + 2x3 + 3x5 = 0 =⇒ x2 = −2x3 − 3x5 = −2α − 3β
x1 − x3 − x5 = 0 =⇒ x1 = x3 + x5 = α + β.
Hence, the solutions of the system are given by the two parameter family

 





x1
α+β
1
1

 





 x2   −2 α − 3 β 
 −2 
 −3 

 






 





 x3  = 





α

 
 = α 1  + β  0 .

 





 x4  

 0 
 −2 
−2 β

 





x5
β
0
1
Hence, as basis for the nullspace of A (which is the same as the nullspace of R)
1
is given by the two vectors

1


1



 −3 




 0 




 −2 


1


 −2 




 1 ,




 0 


0
A basis of the rowspace of A is given by the nonzero rows of the RREF R,
hence a basis is
1 0 −1 0 −1 ,
0 1 2 0 3 ,
0 0 0 1 2 .
Thus, the rowspace has dimension 3.
In the following, let ai denote the ith column of A and let ri denote the ith
column of R.
The leading entries in R are in columns 1, 2 and 4. Hence the corresponding
columns a1 , a2 and a4 of A are a basis of the columnspace of A. Thus, a basis
of the columnspace of A is given by






−15
−80
22






 −19 
 50 
 −89 






, 
, 
.

 −62 
 78 
 66 






77
81
−8
Thus, the columnspace has dimension 3.
The columns of R satisfy the same linear relations as the columns of A.
Looking at the third column of R, we see that r3 = −r1 + 2r2 . Thus, we must
have
a3 = −a1 + 2a2 .
Looking at the the fifth column of R, we see r5 = −r1 + 3r2 + 2r4 , so we also
have
a5 = −a1 + 3a2 + 2a4
The rank of A is the common dimension of the rowspace and the columnspace
of A, which is 3 in this case.
2
Problem 2. Consider the matrix

8 9

 7 1


 5 5

A=
 2 7


 4 4

4
1

19
4
22 −25
−4
6
10
2
19
4
8
6
−1
6

10 


13 −12 


12 −33 


6 −14 

0
1
3
A Find a basis of the nullspace of A.
B Find a basis of the rowspace of A.
C Find a basis of the columnspace of A. Write the other columns of A as linear
combinations of these basis vectors.
D What is the rank of A?
Answer :
The dimension of the nullspace is 3 and the rank is 3.
r
Problem

5

 6

v1 = 
 2

3. Consider the vectors





7
11






 8 
 12 





 , v2 = 
 , v3 = 
,

 1 
 −1 





0
4
12

12



 14 


v4 = 
,
 3 


4

3



 3 


v5 = 
.
 8 


1
Let S = span(v1 , v2 , v3 , v4 , v5 ), which is a subspace of R4 .
A. Find a basis of S. What is the dimension of S?
B. Express the vectors in the list v1 , . . . , v5 that are not part of the basis as
linear combinations of the basis vectors.
C. Consider the vectors

7




 9 


w1 = 
,
 0 


3
12



 13 


w2 = 
.
 27 


−1
Determine if these vectors are in S. If the vector is in S, express it as a
linear combination of the basis vectors found above.
3
Answer :
Put v1 , . . . , v5 in a matrix and add w1 and w2 at the right. This gives the matrix


5 7 11 12 3 7 12


 6 8 12 14 3 9 13 


A=
.
 2 1 −1 3 8 0 27 


0 4 12 4 1 3 −1
The RREF of A is


1
0
−2
1
0
0

 0

R=
 0

0
1
3
1
0
0
0
0
1
0
0
0
0

0 −1 

.
0 3 

1 0
2
Looking at the first 5 columns, corresponding to the vi ’s, we have leading entries
in columns 1, 2 and 5. Hence a basis of S is given by v1 , v2 and v5 .
Use r1 , . . . , r7 for the columns of R. Looking at R, we see r3 = −2r1 + 3r2 .
Hence, we must have
v3 = −2v1 + 3v2
for the expression of v3 in terms of the basis vectors. Similarly we have r4 =
r1 + r2 , so we have
v4 = v1 + v2 .
For w1 and w2 , we see that r6 (corresponding to w1 ) is not in the span of
previous columns, so we conclude that w1 ∈
/ S.
Looking at r7 (corresponding to w2 ), we see that r7 = 2r1 − r2 + 3r5 , thus
w2 = 2v1 − v2 + 3v5 ,
so w2 ∈ S and this is its expression in terms of the basis vectors.
Problem

5

 6

v1 = 
 2

4. Consider the vectors





7
11






 8 
 12 





 , v2 = 
 , v3 = 
,

 1 
 −1 





0
4
12

12


 14 


v4 = 
,
 3 


4
Let S = span(v1 , v2 , v3 , v4 , v5 ), which is a subspace of R5 .
A. Find a basis of S. What is the dimension of S?
4


3



 3 


v5 = 
.
 8 


1
B. Express the vectors in the list v1 , . . . , v5 that are not part of the basis as
linear combinations of the basis vectors.
C. Consider the vectors

7


12



 13 


w2 = 

 27 


−1


 9 


w1 = 
,
 0 


3
Determine if these vectors are in S. If the vector is in S, express it as a
linear combination of the basis vectors found above.
Answer :
The basis of S is v1 , v2 , v3 , v5 . You should find w1 ∈ S and w2 ∈
/ S.
Problem 5. Determine if the following vectors are independent or dependent.
If they are independent, complete the list to a basis of R5 by
adding on a standard basis vector. If the vectors are dependent, find a linear
relation between them.








1
−1
4
13








 −2 
 1 
 −1 
 −10 























v1 =  24  , v2 =  1  , v3 =  0  , v4 =  8 









 0 
 2 
 −9 
 −2 








−4
0
0
−11
Answer :
Put the vectors in a matrix and augment it

13
4 −1 1

 −10 −1 1 −2


1
0
8
V =
 24

 −9 −2 2
0

The RREF of V is

1
−11
0
0
0
0
−4
0

 0 1 0 0


R=
 0 0 1 0

 0 0 0 1

0 0 0 0
by the identity matrix. This gives

1 0 0 0 0

0 1 0 0 0 


0 0 1 0 0 
.

0 0 0 1 0 

−4 0 0 0 0 1
0
12
0 −23
8
−10
0
31
−2
4
−9/2
11
0 −33
11/2
0
1
−1/2
5
0
23


−44 


119 
2
.


− 127
2

−1/2
The leading entries of R are in columns 1, 2, 3, 4 and 6. Thus, the corresponding
columns of V are a basis for the column space of V . It follows that the vectors
v1 , v2 , v3 , v4 and e2 are independent. Since there are 5 of them, they form a
basis of R5 .
Thus, our original vectors v1 ,v2 ,v3 and v4 are independent, and adding the
standard basis vector e2 to the list gives a basis of R5 .
Problem 6. Determine if the following vectors are independent or dependent.
If they are independent, complete the list to a basis of R5 by
adding on a standard basis vector. If the vectors are dependent, find a linear
relation between them.








−89
12
−63
−38








 −170 
 45 
 −26 
 91 
















 , v2 =  30  , v3 =  −14  , v4 =  64 
−1
v1 = 
















 24 
 60 
 10 
 63 








42
−35
22
−23
Answer :
The vectors are dependent. A linear relation is −2v1 + 3v2 + 2v3 − v4 = 0 (i.e.,
v4 = −2v1 + 3v2 + 2v3 ).
Problem 7. Consider the vectors



 

4
4
4



 

 −1 
 5 
 2 



 

v1 = 
,
 , v2 =   , v3 = 
 −3 
 3 
 0 



 

1
1
1




4
0




 5 
 6 




v5 = 
 , v6 = 

 1 
 0 




3
6

28



 26 


v4 = 
,
 12 


7
Do these vectors span R4 ? If so, select a basis of R4 out of this list of vectors.
Answer :
The span of these vectors is 3 dimensional, so the vectors do not span all of R4 .
6
Problem 8. Consider the vectors
 


9
3
 


 4 
 6 
 


v1 = 
 , v2 =   ,
 2 
 5 
 



3



 −8 


v3 = 
,
 −8 


−12
4
8





3
9






 4 
 10 





v5 = 
 , v7 = 
 , v6 = 

 8 
 0 





0
10

3



 0 


v4 = 
,
 7 


2

42

22 


47 

8
Do these vectors span R4 ? If so, select a basis of R4 out of this list of vectors.
Answer :
These vectors span R4 . The vectors v1 , v2 , v4 , v6 form a basis of R4 .
7