Review 1:
91. 4. 11
1.
 Find a basis of spanv1 , v 2 , , v k .
 Find a basis for the row space, column space and null space of a
matrix. Also find the rank and nullity of a matrix.
Example 1:
 1  3 2   1 

  2  3  2  

2

     



S   0 , 8 , 7 ,  0    v1 , v2 , v3 , v4 
.
  3  1   2   4  
     



  4
 
 4
 
 3
 
 3



Find a subset of S that is a basis for
W  spanv1 , v2 , v3 , v4 .
[solution:]
Step 1:
Form the equation
 1
 2

c1v1  c 2 v 2  c3 v3  c 4 v 4   0

 3
 4
3
2
8
1
4
2  1
 c1 
3 2   
c2
7 0    0
.
  c3 
2 4  
c 4 
3  3
Step 2:
The augmented matrix corresponding to the above equation is
1
1
 2

0

3
 4
1
3 2  1 0


2 3 2 0
0
row echelon form

0
8 7 0 0 inreduced


1 2 4 0
0

4 7  3 0
0
0 0
1 0
0 1
0 0
0 0
11
24
 49
24
7
3
0
0
0

0

0
0

0
Step 3:
Since the columns corresponding to the leading 1’s are the 1’st, 2’nd, and
  1   3  2 
      
   2  2  3 



v1 , v 2 , v3    0 , 8, 7 
3’rd columns,
 3  1 2  is a basis of W.
      

 4 4 3 

Example 2:
1 2  2 1,  3 0  4 3, 2 1 1  1,
S  v1 , v2 , v3 , v4 .v5   
.
 3 3  9 6, 9 3 7  6

Find a subset of S that is a basis for
W  spanv1, v2 , v3 , v4 , v5  .
[solution:]
Step 1:
Form the equation
2
c1v1  c2 v2  c3v3  c4 v4  c5v5  c1 1 2  2 1  c2  3 0  4 3
 c3 2 1 1  1  c4  3 3  9 6  c5 9 3 7
 6  0
 c1 
 1  3 2  3 9  
 2
 c 2 
0
1
3
3
 c3   0
 
 2  4 1  9 7   

 c4
3  1 6  6  
1
c5 
Step 2:
The augmented matrix corresponding to the above equation is
 1 3 2 3 9
 2
0
1
3
3

 2  4 1  9 7

3 1 6  6
1
1
0

0 reduced row echelon 0
   
0
0

0
0
0
1
2

1
1
2
0
0
0
0
3
3
2
2
0
0
0
2

5
0
2

0
0
0
0
3
Step 3:
Since the columns corresponding to the leading 1’s are the 1’st and the
2’nd columns,
v1 , v2   1
2  2 1,  3 0  4 3 is a basis
of W.
Example 3:
Find the basis for the row space, column space, and null space of
 1
 2

A  0

 3

 4
3
2
8
1
4
2
3
7
2
3
Also, determine the rank and nullity of A.
[solution:]
3
 1
2 

0 .

4 
 3

Step 1:
We transform the matrix A into the matrix in the reduced row echelon
form:
 1
 2

 0

 3
 4
1
 1

2 
0
in reduced row echelon form
8 7 0      0


1 2 4
0

4 3  3
0
3 2
2 3

24 
1 0  49 
24
7
0 1
3 
0 0
0 

0 0
0 
0 0
11
Step 2:
 The nonzero rows

1 0 0

11  
, 0 1 0
24  
 49  
7 
,
0
0
1
,
24  
3  
form a basis of the row space of A.
 Since the columns corresponding to the leading 1’s are the 1’st, 2’nd,
and
  1   3  2 
      
   2  2  3 



col1 ( A), col2 ( A), col3 ( A)   0 , 8, 7 
3’rd columns,
 3  1 2  is a basis
      

 4 4 3 

of the column space of A.
 The reduced row echelon matrix in the homogeneous system is
4

1

0

0

0


0
 x1 
x2 
0
0
1
0
0
1
0
0
0
0
11
24
 49
24
7
3
0
0

0

0

0

0

0

 11
 11
x4 
s
24
24
49
49
x4 
s
24
24
 x1   11s 24  11 24
 x   49s   49 

24    24  s
, sR  x   2  
 x3 
7
 7s 

3 
3
  
 x 4   s   1 
7
7
x4 
s
3
3
x4  s
x3 
 11 
 24
 49 24 
Thus,   7  is a basis for the null space of A.
 3
 1 


 Since the bases of the column space or row space of A consist of 3
vectors and the basis of the null space of A consists of 1 vector, then
rank ( A)  3, nullity ( A)  1 .
Note:
In the above example,
rank ( A)  nullity ( A)  3  1  4  the number of column vec tors of A
5
2.
 Find the coordinate vector of some vector with respect to the ordered
basis.
 Find the transition matrix from one basis to another basis
Example 4:
Let S  v1 , v2   t , t  3 and T  w1 , w2   t  1, t  1
be two bases for the vector space consisting of all the polynomials with
degrees  1 . Compute the transition matrix PS T .
[solution:]
PS T  w1 S
w2 S   
a1
a 2
b1 
b2  ,
where
w1  a1v1  a 2 v 2
w2  b1v1  b2 v 2 .
Therefore,
w1  t  1  a1t  a 2 t  3  a1  a 2 t  3a 2  a1  a 2  1
 a1 
2
1
, a2 
3
3
 3a 2  1
w2  t  1  b1t  b2 t  3  b1  b2 t  3b2  b1  b2  1
 3b2  1
Thus,
PS T
2
 3
 13

6

3
1  .
3
4
 b1 
4
1
, b2 
3
3 .