Linear Algebra, 2020a

Section III. Computing Linear Maps 219
Looked at in this row-by-row way, this new operation generalizes dot product.
We can also view the operation column-by-column.
⎛
⎞ ⎛ ⎞ ⎛
⎞
h 1,1 h 1,2 ... h 1,n c 1 h 1,1 c 1 + h 1,2 c 2 + ···+ h 1,n c n
h 2,1 h 2,2 ... h 2,n
c 2 ⎜
⎟ ⎜ ⎟
⎝ .
⎠ ⎝ . ⎠ = h 2,1 c 1 + h 2,2 c 2 + ···+ h 2,n c n
⎜
⎟
⎝
.
⎠
h m,1 h m,2 ... h m,n c n h m,1 c 1 + h m,2 c 2 + ···+ h m,n c n
⎛ ⎞ ⎛ ⎞
h 1,1
h 1,n
h 2,1
= c 1 ⎜ ⎟
⎝ . ⎠ + ···+ c h 2,n
n ⎜ ⎟
⎝ . ⎠
h m,1 h m,n
The result is the columns of the matrix weighted by the entries of the vector.
1.12 Example
( ) ⎛ ⎞
2 ( ) ( ) ( ) ( )
1 0 −1 ⎜ ⎟ 1 0 −1 1
⎝ −1⎠ = 2 − 1 + 1 =
2 0 3
2 0 3 7
1
This way of looking at matrix-vector product brings us back to the objective
stated at the start of this section, to compute h(c 1
⃗β 1 +···+c n
⃗β n ) as c 1 h(⃗β 1 )+
···+ c n h(⃗β n ).
We began this section by noting that the equality of these two enables us
to compute the action of h on any argument knowing only h(⃗β 1 ), ..., h(⃗β n ).
We have developed this into a scheme to compute the action of the map by
taking the matrix-vector product of the matrix representing the map with the
vector representing the argument. In this way, with respect to any bases, for any
linear map there is a matrix representation. The next subsection will show the
converse, that if we fix bases then for any matrix there is an associated linear
map.
Exercises
̌ 1.13 Multiply the matrix
⎛ ⎞
1 3 1
⎝0 −1 2⎠
1 1 0
by each vector, or state “not defined.”
⎛ ⎞
⎛ ⎞
2 ( ) 0
(a) ⎝1⎠
−2
(b) (c) ⎝0⎠
−2
0
0
1.14 Multiply this matrix by each vector or state “not defined.”
( ) 3 1
2 4