Linear Algebra, 2020a

Section III. Computing Linear Maps 213
With these, for any member ⃗v of the domain we can compute h(⃗v).
( ) ( )
2 1
h(⃗v) =h(c 1 · + c 2 · )
0 4
( ) ( )
2
1
= c 1 · h( )+c 2 · h( )
0
4
⎛
⎜
1
⎞ ⎛ ⎞ ⎛
⎟
= c 1 · (0 ⎝0⎠− 1 0
⎜ ⎟ ⎜
1
⎞ ⎛ ⎞ ⎛ ⎞ ⎛
1 0
⎟ ⎜ ⎟ ⎜ ⎟ ⎜
1
⎞
⎟
⎝−2⎠+ 1 ⎝0⎠)+c 2 · (1 ⎝0⎠− 1 ⎝−2⎠+ 0 ⎝0⎠)
2
0 0 1
0 0 1
⎛
⎜
1
⎞
⎛ ⎞
⎛
⎟
=(0c 1 + 1c 2 ) · ⎝0⎠ +(− 1 0
2 c ⎜ ⎟
⎜
1
⎞
⎟
1 − 1c 2 ) · ⎝−2⎠ +(1c 1 + 0c 2 ) · ⎝0⎠
0
0
1
Thus,
if Rep B (⃗v) =
For instance,
( ) ( )
4 1
since Rep B ( )=
8 2
( )
c 1
then Rep
c D ( h(⃗v))=
2
B
⎛
⎞
0c 1 + 1c 2
⎜
⎟
⎝−(1/2)c 1 − 1c 2 ⎠.
1c 1 + 0c 2
⎛ ⎞
( ) 2
4 ⎜ ⎟
we have Rep D ( h( ))= ⎝−5/2⎠.
8
1
We express computations like the one above with a matrix notation.
⎛ ⎞
⎛
⎞
0 1 ( ) 0c 1 + 1c 2
⎜ ⎟ c 1 ⎜
⎟
⎝−1/2 −1⎠
= ⎝(−1/2)c 1 − 1c 2 ⎠
c 2
1 0
B 1c 1 + 0c 2
B,D
In the middle is the argument ⃗v to the map, represented with respect to the
domain’s basis B by the column vector with components c 1 and c 2 . On the
right is the value of the map on that argument h(⃗v), represented with respect to
the codomain’s basis D. The matrix on the left is the new thing. We will use it
to represent the map and we will think of the above equation as representing an
application of the map to the matrix.
That matrix consists of the coefficients from the vector on the right, 0 and
1 from the first row, −1/2 and −1 from the second row, and 1 and 0 from the
third row. That is, we make it by adjoining the vectors representing the h(⃗β i )’s.
⎛
⎜
⎝
.
.
Rep D ( h(⃗β 1 )) Rep D ( h(⃗β 2 ))
.
.
⎞
⎟
⎠
D