Linear Algebra Exercises-n-Answers.pdf

96 Linear Algebra, by Hefferon
and, as
Rep B (1 − 3x + 2x 2 ) =
the matrix-vector multiplication calculation gives this.
⎛ ⎞
1 1 0
Rep D (h(1 − 3x + 2x 2 )) = ⎜1 2 1
⎟
⎝0 0 0 ⎠
0 0 −1
⎛
⎞
⎝ 1 −3⎠
2
B,D
⎛
B
⎞
⎝ 1 −3⎠
2
B
⎛ ⎞
−2
= ⎜−3
⎟
⎝ 0 ⎠
−2
Thus, h(1 − 3x + 2x 2 ) = −2 · 1 − 3 · x + 0 · x 2 − 2 · x 3 = −2 − 3x − 2x 3 , as above.
Three.III.1.15 Again, as recalled in the subsection, with respect to E i , a column vector represents
itself.
(a) To represent h with respect to E 2 , E 3 we take the images of the basis vectors from the domain,
and represent them with respect to the basis for the codomain.
⎛ ⎞ ⎛ ⎞
⎛ ⎞ ⎛ ⎞
2
Rep E3
( h(⃗e 1 ) ) = Rep E3
( ⎝2⎠) =
0
These are adjoined to make the matrix.
(b) For any ⃗v in the domain R 2 ,
and so
is the desired representation.
⎝ 2 2
0
⎠ Rep E3
( h(⃗e 2 ) ) = Rep E3
(
Rep E2,E 3
(h) =
⎛
⎝ 2 0
⎞
2 1 ⎠
0 −1
( ) ( )
v1 v1
Rep E2
(⃗v) = Rep E2
( ) =
v 2 v 2
Rep E3
( h(⃗v) ) =
⎛
⎝ 2 0
⎞
( )
2 1 ⎠ v1
v
0 −1 2
⎛
= ⎝ 2v ⎞
1
2v 1 + v 2
⎠
−v 2
D
0
⎝ 1 ⎠) =
−1
⎝ 0 1
−1
Three.III.1.16 (a) We must first find the image of each vector from the domain’s basis, and then
represent that image with respect to the codomain’s basis.
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
0
1
0
0
Rep B ( d 1
dx ) = ⎜0
⎟
⎝0⎠
Rep B( d x
dx ) = ⎜0
⎟
⎝0⎠
Rep B( d x2
dx ) = ⎜2
⎟
⎝0⎠
Rep B( d x3
dx ) = ⎜0
⎟
⎝3⎠
0
0
0
0
Those representations are then adjoined to make the matrix representing the map.
⎛
⎞
0 1 0 0
Rep B,B ( d
dx ) = ⎜0 0 2 0
⎟
⎝0 0 0 3⎠
0 0 0 0
(b) Proceeding as in the prior item, we represent the images of the domain’s basis vectors
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
0
1
0
0
Rep B ( d 1
dx ) = ⎜0
⎟
⎝0⎠
Rep B( d x
dx ) = ⎜0
⎟
⎝0⎠
Rep B( d x2
dx ) = ⎜1
⎟
⎝0⎠
Rep B( d x3
dx ) = ⎜0
⎟
⎝1⎠
0
0
0
0
and adjoin to make the matrix.
⎛ ⎞
0 1 0 0
Rep B,D ( d
dx ) = ⎜0 0 1 0
⎟
⎝0 0 0 1⎠
0 0 0 0
Three.III.1.17 For each, we must find the image of each of the domain’s basis vectors, represent
each image with respect to the codomain’s basis, and then adjoin those representations to get the
matrix.
⎠