Linear Algebra, 2020a

216 Chapter Three. Maps Between Spaces
were not true then we would adjust the definition to make it so. Nonetheless,
we need the verification.
1.8 Example For the matrix from Example 1.4 we can calculate where that map
sends this vector.
⎛
⎜
4
⎞
⎟
⃗v = ⎝1⎠
0
With respect to the domain basis B the representation of this vector is
⎛ ⎞
0
⎜ ⎟
Rep B (⃗v) = ⎝1/2⎠
2
and so the matrix-vector product gives the representation of the value h(⃗v) with
respect to the codomain basis D.
⎛ ⎞
(
) 0
−1/2 1 2 ⎜ ⎟
Rep D (h(⃗v)) =
⎝1/2⎠
−1/2 −1 −2
B,D 2
B
(
) ( )
(−1/2) · 0 + 1 · (1/2)+2 · 2 9/2
=
=
(−1/2) · 0 − 1 · (1/2)−2 · 2 −9/2
To find h(⃗v) itself, not its representation, take (9/2)(1 + x)−(9/2)(−1 + x) =9.
1.9 Example Let π: R 3 → R 2 be projection onto the xy-plane. To give a matrix
representing this map, we first fix some bases.
⎛ ⎞ ⎛
1
⎜ ⎟ ⎜
1
⎞ ⎛
⎟ ⎜
−1
⎞
( ) ( )
⎟
2 1
B = 〈 ⎝0⎠ , ⎝1⎠ , ⎝ 0 ⎠〉 D = 〈 , 〉
1 1
0 0 1
For each vector in the domain’s basis, find its image under the map.
⎛
⎜
1
⎞
( ) ⎛ ⎞
1 ( ) ⎛ ⎞
−1 ( )
⎟
⎝0⎠
↦−→
π 1 ⎜ ⎟ ⎝ 1⎠
↦−→
π 1 ⎜ ⎟ ⎝ 0 ⎠
π −1
↦−→
0
1
0
0
0
1
Then find the representation of each image with respect to the codomain’s basis.
( ) ( ) ( ) ( ) ( ) ( )
1 1
1 0
−1 −1
Rep D ( )= Rep
0 −1
D ( )= Rep
1 1
D ( )=
0 1
Finally, adjoining these representations gives the matrix representing π with
respect to B, D.
(
)
1 0 −1
Rep B,D (π) =
−1 1 1
B
D
B,D
D