Linear Algebra MATH 225: Assignment 5
Problem 1 (2 points): Let
  
 0
B =  1 ,

0
V = R3 . Consider two bases of V :
     
  
0
1 
0
1 
 1








0
0
1 , 1
0  .
and C =



1
0
0
1
1
Let


3
v =  1 .
5
Find the coordinate vectors [v]B , [v]C and the base change matrices P , P .
B→C C→B
Solution. (a) To find [v]B we must solve the system
 
 
 
 
3
0
0
1
 1  = x 1  + y 0  + z 0 .
5
0
1
0
The solution is x = 1, y = 5, z = 3, hence


1
[v]B =  5  .
3
(b) To find [v]C we must solve the system
 
 
 
 
3
1
0
1
 1  = x 1  + y 1  + z 0 .
1
1
5
0
The solution is x = −0.5, y = 1.5, z = 3.5, hence


−0.5
[v]C =  1.5  .
3.5
(c) To find P
C→B
we must write every vector from C as a linear combination of vectors from
B. One has
                 
1
0
1
0
0
0
1
0
1
 1  =  1  +  0 ,  1  =  1  +  0 ,  0  =  0  +  0 
0
0
1
0
1
1
1
0
0
Hence the base change matrix is

P
C→B

1 1 0
=  0 1 1 .
1 0 1
1
(d) To find P
B→C
we must write every vector from B as a linear combination of vectors from
C. Easy computations yield

P
B→C

0.5 −0.5
0.5
0.5 −0.5  .
=  0.5
−0.5
0.5
0.5
Problem 2 (2 points): Prove that the map f : P3 → P3 defined by f (p(x)) = xp0 (x) is a
linear transformation.
Solution. (a) Let us first verify that f (p(x) + g(x)) = f (p(x)) + f (g(x)). The left hand side
is
f (p(x) + g(x)) = x(p(x) + g(x))0
= xp(x)0 + xg(x)0
Obviously, it is equal to the right hand side of the above equality.
(b) We also have to verify that f (ap(x)) = af (p(x)) for all scalars a and all polynomials
p(x). The left hand side is
f (ap(x)) = x(ap(x))0 = axp(x)0 .
Obviously, it is equal to the right hand side of the above equality.
Problem 3 (2 points): Let f : R3 → R3 be a linear transformation. Assume that we know
that
  

   
   
1
2
0
1
2
4
f  1  =  1 , f  1  =  0 , f  1  =  0 .
0
−1
2
2
3
2


 
3
a
Find f  2  and f  b  .
c
−1
Solution. To find the images of


 
3
a
w1 =  2  , w2 =  b 
−1
c
we first have to write them as a linear combinations of the vectors
 
 
 
1
0
2
v1 =  1  , v 2 =  1  , v 3 =  1  .
0
2
3
The augmented matrix of

1
 1
0

1
 0
0
the system w1 = x1 v1 + x2 v2 + x3 v3 is of the form



0 2 |
3
1 0
2 |
3
1 1 |
2  →  0 1 −1 | −1  →
2 3 | −1
0 2
3 | −1



1 0 0 | 13
0
2 |
3
5
1 −1 | −1  →  0 1 0 | − 45  .
1
0
5 |
1
0 0 1 |
5
Thus we have
13
4
1
v1 − v2 + v3 .
5
5
5
Applying f we find
4
1
13
v1 − v2 + v3 =
f (w1 ) = f
5
5
5

 
 

2
1
4
13
4
1
13 
4
1
1 −  0 +  0 =
f (v1 ) − f (v2 ) + f (v3 ) =
5
5
5
5
5
5
−1
2
2


26
1
13  .
5
−19
w1 =
Analogously, the augmented matrix of the equation w2 = x1 v1 + x2 v2 + x3 v3 is of the form




1 0 2 | a
1 0
2 |
a
 1 1 1 | b  →  0 1 −1 | b − a  →
0 2 3 | c
0 2
3 |
c




1
(4b
+
a
−
2c)
1 0 0 |
1 0
2 |
a
5
1
.
 0 1 −1 |
b−a → 0 1 0 |
5 (3b − 3a + c)
1
0 0
5 | −2b + 2a + c
0 0 1 | 5 (−2b + 2a + c)
Thus we have
1
1
1
w2 = (4b + a − 2c)v1 + (3b − 3a + c)v2 + (−2b + 2a + c)v3 .
5
5
5
Applying f we find
1
1
1
f (w1 ) = f
(4b + a − 2c)v1 + (3b − 3a + c)v2 + (−2b + 2a + c)v3 =
5
5
5
1
1
1
(4b + a − 2c)f (v1 ) + (3b − 3a + c)f (v2 ) + (−2b + 2a + c)f (v3 ) =
5
5


  5
 
2
1
4
1
1
1
(4b + a − 2c)  1  + (3b − 3a + c)  0  + (−2b + 2a + c)  0  =
5
5
5
−1
2
2


3b + 7a + c
1
4b + a − 2c  .
5
−2b − 3a + 6c
Problem 4 (2 points): Let V = F 3 . Let B = {v1 , v2 , v3 } where
 
 
 
1
0
2





1 , v2 =
1 , v3 =
1 
v1 =
0
2
3
be a basis of V . Assume that for two linear transformations f, g : V → V one has
f (v1 ) = g(v1 ), f (v2 ) = g(v2 ), f (v3 ) = g(v3 ).
Prove that f = g, i.e. for every vector v ∈ V one has f (v) = g(v).
Solution. Since B is a basis in V every vector v can be written uniquely as a linear combination v = a1 v1 + a2 v2 + a3 v3 where a1 , a2 , a3 are scalars. Then
f (v) = f (a1 v1 + a2 v2 + a3 v3 ) = a1 f (v1 ) + a2 f (v2 ) + a3 f (v3 ).
Similarly, we have
g(v) = g(a1 v1 + a2 v2 + a3 v3 ) = a1 g(v1 ) + a2 g(v2 ) + a3 g(v3 ).
But we are given that
f (v1 ) = g(v1 ), f (v2 ) = g(v2 ), f (v3 ) = g(v3 ),
hence the assertion follows.
Problem 5 (2 points): Let f : P3 → P3 be the linear transformation defined by f (p(x)) =
xp0 (x). Find its kernel and image. What are the dimensions of the kernel and image?
Solution. Let us first describe the kernel. By definition, the kernel of f consists of all
polynomials p(x) of degree at most 3 such that f (p(x)) = xp0 (x) = 0. It follows that the
kernel consists of all polynomial whose derivative is zero. Therefore kernel consists of all
constant polynomials. It is the span of one vector p(x) = 1, hence its dimension is 1.
We now describe the image of f . Take an arbitrary polynomial p(x) = a + bx + cx2 + dx3
of degree at most 3. Here a, b, c, d are coefficients. We have
f (p(x)) = xp(x)0 = x(b + 2cx + 3dx2 ) = bx + 2cx2 + 3dx3 .
It follows that the image of f consists of all polynomials whose free coefficient is zero. It is the
span of three polynomials x, x2 , x3 which are linearly independent. Hence dim(Im(f )) = 3.