Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 101
(b) The images of the vectors in the domain’s basis are e x d/dx
↦−→ e x and e
with respect to the codomain’s basis and adjoining gives this matrix.
Rep B,B ( d ( )
1 0
dx ) = 0 2
B,B
2x
d/dx
↦−→ 2e 2x . Representing
(c) The images of the members of the domain’s basis are 1 d/dx
↦−→ 0, x d/dx
↦−→ 1, e x d/dx
↦−→ e x , and xe x d/dx
↦−→
e x + xe x . Representing these images with respect to B and adjoining gives this matrix.
⎛
⎞
0 1 0 0
Rep B,B ( d
dx ) = ⎜0 0 0 0
⎟
⎝0 0 1 1⎠
0 0 0 1
Three.III.1.26 (a) It is the set of vectors of the codomain represented with respect to the codomain’s
basis in this way. ( ( ) ( )
1 0 x ∣∣ x ∣∣
{
x, y ∈ R} = { x, y ∈ R}
0 0)
y
0
As the codomain’s basis is E 2 , and so each vector is represented by itself, the range of this transformation
is the x-axis.
(b) It is the set of vectors of the codomain represented in this way.
( ( ) ( )
0 0 x ∣∣ 0 ∣∣
{
x, y ∈ R} = {
x, y ∈ R}
3 2)
y
3x + 2y
With respect to E 2 vectors represent themselves so this range is the y axis.
(c) The set of vectors represented with respect to E 2 as
( ( ) ( ) ( )
a b x ∣∣ ax + by ∣∣ 1 ∣∣
{
x, y ∈ R} = {
x, y ∈ R} = {(ax + by) · x, y ∈ R}
2a 2b)
y
2ax + 2by
2
is the line y = 2x, provided either a or b is not zero, and is the set consisting of just the origin if
both are zero.
Three.III.1.27 Yes, for two reasons.
First, the two maps h and ĥ need not have the same domain and codomain. For instance,
( )
1 2
3 4
represents a map h: R 2 → R 2 with respect to the standard bases that sends
( ( ( ( 1 1 0 2
↦→ and ↦→
0)
3)
1)
4)
and also represents a ĥ: P 1 → R 2 with respect to 〈1, x〉 and E 2 that acts in this way.
( (
1 2
1 ↦→ and x ↦→
3)
4)
The second reason is that, even if the domain and codomain of h and ĥ coincide, different bases
produce different maps. An example is the 2×2 identity matrix
( )
1 0
I =
0 1
which represents the identity map on R 2 with respect to E 2 , E 2 . However, with respect to E 2 for the
domain but the basis D = 〈⃗e 2 , ⃗e 1 〉 for the codomain, the same matrix I represents the map that swaps
the first and second components ( (
x y
↦→
y)
x)
(that is, reflection about the line y = x).
Three.III.1.28 We mimic Example 1.1, just replacing the numbers with letters.
Write B as 〈 β ⃗ 1 , . . . , β ⃗ n 〉 and D as 〈 ⃗ δ 1 , . . . , ⃗ δ m 〉. By definition of representation of a map with respect
to bases, the assumption that
⎛
⎞
h 1,1 . . . h 1,n
⎜
Rep B,D (h) =
. . ⎟
⎝ . . ⎠
h m,1 . . . h m,n
B,B