Home Work 5
Solution
1. Find a basis for the column space, row space and the null space of the matrices A and B below and
compute their ranks.

2
A = 0
2

1
2
3
−1
1
3


1
1
−1

2,B = 
0
3
2

3 2 0
3 0 1 

3 2 1
0 2 −1

1
2 . Perform row operations to get a row echelon form of A. After,
3 

1 0.5 −0.5 0.5
1
2 . After R2 = R2 /2 and R3 − R3 − 2R2 ,
R1 = R1 /2 and R3 = R3 − 2R1 we get A =  0 2
0 2
4
2
we get 

1 0.5 −0.5 0.5
A = 0 1
0.5
1 .
0 0
3
0
Finally,
dividing
the
third
row by 3, we get

1 0.5 −0.5 0.5
U = 0 1
0.5
1  in row echelon form.
0 0
1
0
  
 

0
0
1
 0.5   1   0 
So, the basis for the row space of A is {
 ,  }.
,
1
0.5
−0.5
0
1
0.5
2
Solution: Consider A =  0
2
1
2
3
−1
1
3
Choosing the three columns corresponding to the leading one’s in the row echelon form we obtained,
namely,
firstthree
of A, we get a basis for the column space. It is

 columns
 
 the
−1
1
2
{ 0  ,  2  ,  1 }.
2
2
4
To get a basis for the null space of A, we need to solve the equation AX = 0. From the row echelon
form of A obtained above, we see that there is only one column, the fourth one, without a leading one. So,


0
 −0.5 
setting x4 = 1, we see, x3 = 0, x2 = −1, x1 = −0.5 − 0.5x2 = 0. So a basis for the null space is {
}.
0
1
Rank of A = dim Col space of A = 3
1
 −1
Consider the matrix B 
0
2
By doing the row operations

3
3
3
0
R2

2 0
0 1 

2 1
2 −1
= R2 + R1 and R4 = R4 − 2R1 , we get
1

1 3
2
0
2
1 
0 6


0 3
2
1
0 −6 −2 −1
By doing R4 = R4 − R2 and then R2 = R2 /6 and R3 = R3 − 3R2 , we get


1 3
2
0
 0 1 1/3 1/6 
U =

0 0
1
0.5
0 0
0
0
Now, the first three rows of this matrix U form a basis for the row space of B. The first three columns of U
contain leading 1’s. So the first three columns of B form a basis for the column space of B. Finally, the rank of
B =3. The null space of B, if obtained by solving BX = 0. We see x4 = 1, x3 = −0.5, x2 = −1/6−1/3x3 = 0
, and x1 = −2x3 − 3x2 = 1 − 0 = 1. Thus,
  
 

1
0
0
3  1   0 
Basis for row space of B = {  , 
,
}.
2
1/3
1
0
1/6
0.5


   
2
3
1
 −1   3   2 
A basis for the column space of B = {
}.
, ,
2
3
0
−2
0
2


1
 0 
A basis for the null space of B = {

−0.5
1

2. (10 points) f (x) = 2x3 − 5x2 + 4.
a. Prove that B = {f, f 0 , f 00 , f 000 } is a basis for P4 .
B = {2x3 − 5x2 + 4, 6x2 − 10x, 12x − 10, 12}. B contains four polynomials of distinct degrees 3,2,1 and
0 respectively. So these polynomials are linearly independent. Since the dimension of P4 is 4 and we have
four linearly independent vectors in B, B spans P4 and so form a basis for P4 .
b. Find the change of basis matrix from this basis B to the basis C = {x3 −1, x3 +x2 +1, x2 −x+1, x+2}
Let us choose an easy basis {x3 , x2 , x, 1} for P4 and write our vectors in this basis. We form the
augumented matrix


1 1 0 0 | 2
0
0
0
6
0
0 
 0 1 1 0 | −5


0 0 −1 1 | 0 −10 12
0
−1 1 1 2 | 4
0
−10 12
By performing the row operations R4 = R4 + R1 ; R4 = R4 − 2R2 ; R3 = −R3 we come to


1 1 0
0 | 2
0
0
0
0 | −5
6
0
0 
0 1 1


0 0 1 −1 | 0
10 −12 0
0 0 −1 2 | 16 −12 −10 12
Now we do R4 = R4 + R3 to get


1 1 0 0 | 2
0
0
0
0
0 
 0 1 1 0 | −5 6


0 0 1 −1 | 0
10 −12 0
0 0 0 1 | 16 −2 −22 12
Finally, perform R3 = R3 + R4 , R2 = R2 − R3 and R1 = R1 − R2 to get
2
1
0

0
0
So,

0
1
0
0
the
0 0
0 0
1 0
0 1
change

| 23
2 −34 12
| −21 −2 34 −12 

| 16
8 −34 12
| 16 −2 −22 12
of basis matrix is
−34
34
−34
−22
23
2
 −21 −2

16
8
16 −2


12
−12 

12
12
c. Find the coordinate vectors of x3 + x2 + x + 2 in the basis B and the basis C.
By choosing an easy basis {x3 , x2 , x, 1}, we begin with


1 1 0 0 | 1
 0 1 1 0 | 1

 By performing the row operations R4 = R4 + R1 , R4 = R4 − 2R2 , R3 = −R3
0 0 −1 1 | 1
−1 1 1 2 | 2


1 1 0 0 | 1
0 1 1 0 | 1 
and R4 = R4 + R3 , we get 

0 0 1 −1 | −1
0 0 0 1 | 0
Solving this we find, x4 = 0, x3 = −1 + 0 = −1, x2 = 1 − x3 = 2, x1 = 1 − x2 = −1. So, the coordinate
vector of in the basis C of x3 + x2 + x + 2

−1
 2 
XC (x3 + x2 + x + 2) = 

−1
0


2
 −1 
d. If the coordinate vector of p(x) in the basis B is 
, what is the coordinate of p(x) in the basis
0
1

C?
Multiplying by the change of basis matrix from B to C, we get,
23
2
 −21 −2
XC (v) = 
16
8
16 −2


 

56
2
−34 12
34 −12   −1   −52 
=


36
−34 12
0
46
1
−22 12
3. Find the matrix that represents L in the basis B = {1 − x, 1 − 2x + x2 , 2 + x} of P3 .
L : P3 → P3 is given by L(f (x)) = f (1) + 2xf (−1) + x2 f (2).
1. Computing L(B). L(1 − x) = 0 + 4x − x2
L(1 − 2x + x2 ) = 0 + 8x + x2
L(2 + x) = 3 + 2x + 4x2
b. Choosing the basis {1, x, x2 } for P3 , we start with the augumented matrix

1
 −1
0
1
−2
1
2 | 0
1 | 4
0 | −1
3

0 3
8 2
1 4
Now we perform R2 = R2 + R1 , R2 = −R2 , R3 = R3 − R2 to get


1 1 2 | 0
0
3
 0 1 −3 | −4 −8 −5 
0 0 3 | 3
9
9
Dividing the last row by 3 and R2 = R2 + 3R3 and R1 = R1 − 2R3 , we get


1 1 0 | −2 −6 −3
 0 1 0 | −1 1
4 
0 0 1 | 1
3
3
Finally, R1 = R1 − R2 brings us to

1 0 0
0 1 0
0 0 1
| −1
| −1
| 1
−7
1
3

−7
4 
3
So, the matrix of the linear transformation L in the basis B is


−1 −7 −7
 −1 1
4 
1
3
3
3
4. Write the matrix of T : R → R
2×2
 
a
a+b


in your choice of bases: T ( b ) =
c+a
c
a−b
.
0
I choose the easy bases E = {e1 , e2 , e3 } for R3 and {E1 1, E12 , E21 , E22 } for R2×2 .
Then
1 1
1 −1
0 0
L(e1 ) =
, L(e2 ) =
, L(e3 ) =
1 0
0 0
1 0
So, the matrix of L in my choice of bases is
1
1

1
0

1
−1
0
0

0
0

1
0
5. L : V → W is a linear transformation. v1 , v2 , . . . , vn are vectors in V. wi = L(vi ). Show that if
w1 , w2 , . . . wn are linearly independent, then v1 , v2 , . . . , vn are also linearly independent.
Proof: Let c1 v1 + c2 v2 + · · · + cn vn = 0.
Applying L to both sides, we get, L(c1 v1 + c2 v2 + · · · + cn vn ) = L(0).
Since L is a linear transformation, this becomes,
c1 L(v1 ) + c2 L(v2 ) + · · · + cn L(vn ) = L(0) = 0.
But L(vi = wi . So, we get
(c1 w1 + c2 w2 + · · · + cn wn ) = (0).
But we are given that w1 , w2 , . . . wn are linearly independent. So, c1 = 0, c2 = 0, · · · cn = 0.
Hence v1 , v2 , . . . vn are linearly independent.
4