Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 97
(a) The basis vectors from the domain have these images
1 ↦→ 0 x ↦→ 1 x 2 ↦→ 2x . . .
and these images
⎛
are
⎞
represented with
⎛
respect
⎞
to the codomain’s
⎛ ⎞
basis in this way.
⎛ ⎞
0
1
0
0
0
0
2
0
0
0
0
Rep B (0) =
.
Rep ⎜.
B (1) =
.
Rep ⎟
⎜.
B (2x) =
.
. . . Rep ⎟
⎜.
B (nx n−1 0
) =
.
⎟
⎜ .
⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝n⎠
0
The matrix
⎞
0 1 0 . . . 0
Rep B,B ( d
0 0 2 . . . 0
⎜
⎟
⎛
dx ) = ⎜
⎝
.
0 0 0 . . . n
0 0 0 . . . 0
has n + 1 rows and columns.
(b) Once the images under this map of the domain’s basis vectors are determined
1 ↦→ x x ↦→ x 2 /2 x 2 ↦→ x 3 /3 . . .
then they can be represented with respect to the codomain’s basis
⎛ ⎞
0
1
Rep Bn+1
(x) =
0
⎜
⎝
⎟
. ⎠
⎛
⎞
0
0
Rep Bn+1
(x 2 /2) =
1/2
⎜
⎝
⎟
. ⎠
and put together to make the matrix.
∫
Rep Bn,B n+1
( ) =
⎟
⎠
⎛
. . . Rep Bn+1
(x n+1 /(n + 1)) =
⎜
⎝
⎛
⎞
0 0 . . . 0 0
1 0 . . . 0 0
0 1/2 . . . 0 0
⎜
⎝
⎟
.
⎠
0 0 . . . 0 1/(n + 1)
(c) The images of the basis vectors of the domain are
1 ↦→ 1 x ↦→ 1/2 x 2 ↦→ 1/3 . . .
and they are represented with respect to the codomain’s basis as
Rep E1
(1) = 1 Rep E1
(1/2) = 1/2 . . .
so the matrix is
∫
Rep B,E1 ( ) = ( 1 1/2 · · · 1/n 1/(n + 1) )
(this is an 1×(n + 1) matrix).
(d) Here, the images of the domain’s basis vectors are
1 ↦→ 1 x ↦→ 3 x 2 ↦→ 9 . . .
and they are represented in the codomain as
Rep E1
(1) = 1 Rep E1
(3) = 3 Rep E1
(9) = 9 . . .
and so the matrix is this.
∫ 1
Rep B,E1 ( ) = ( 1 3 9 · · · 3 n)
0
0
0
0
.
1/(n + 1)
(e) The images of the basis vectors from the domain are
1 ↦→ 1 x ↦→ x + 1 = 1 + x x 2 ↦→ (x + 1) 2 = 1 + 2x + x 2 x 3 ↦→ (x + 1) 3 = 1 + 3x + 3x 2 + x 3 . . .
which are represented as
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
0
1
2
0
0
Rep B (1) =
⎜0
Rep
⎟ B (1 + x) =
⎜0
Rep
⎟ B (1 + 2x + x 2 1
) =
⎜0
. . .
⎟
⎜.
⎟
⎝.
⎠
0
⎜.
⎟
⎝.
⎠
0
⎜.
⎟
⎝.
⎠
0
⎞
⎟
⎠