Linear Algebra, 2020a

Section II. Similarity 403
represented with respect to the ending basis (also B)
⎛
⎜
0
⎞
⎛
⎟
⎜
0
⎞
⎛
⎟
⎜
0
⎞
⎟
Rep B (2x) = ⎝2⎠ Rep B (1) = ⎝0⎠ Rep B (0) = ⎝0⎠
0
1
0
gives the representation of the map.
⎛
⎜
0 0 0
⎞
⎟
T = Rep B,B (d/dx) = ⎝2 0 0⎠
0 1 0
Next, computing the matrix for the right-hand side involves finding the effect
of the identity map on the elements of B. Of course, the identity map does
not transform them at all so to find the matrix we represent B’s elements with
respect to D.
⎛ ⎞
⎛ ⎞
⎛ ⎞
Rep D (x 2 )=
⎜
⎝
−1
0
1
⎟
⎠ Rep D (x) =
⎜
⎝
−1
1
0
⎟
⎠ Rep D (1) =
⎜
1 ⎟
⎝0⎠
0
So the matrix for going down the right side is the concatenation of those.
⎛
⎞
−1 −1 1
⎜
⎟
P = Rep B,D (id) = ⎝ 0 1 0⎠
1 0 0
With that, we have two options to compute the matrix for going up on left
side. The direct computation represents elements of D with respect to B
⎛
⎜
0
⎞
⎛
⎟
⎜
0
⎞
⎛
⎟
Rep B (1) = ⎝0⎠ Rep B (1 + x) = ⎝1⎠ Rep B (1 + x 2 ⎜
1
⎞
⎟
)= ⎝0⎠
1
1
1
and concatenates to make the matrix.
⎛ ⎞
0 0 1
⎜ ⎟
⎝0 1 0⎠
1 1 1
The other option to compute the matrix for going up on the left is to take the
inverse of the matrix P for going down on the right.
⎛
⎞
⎛
−1 −1 1 1 0 0
⎜
⎟
⎜
1 0 0 0 0 1
⎞
⎟
⎝ 0 1 0 0 1 0⎠ −→ · · · −→ ⎝0 1 0 0 1 0⎠
1 0 0 0 0 1
0 0 1 1 1 1