Linear Algebra

198 Chapter 3. Maps Between Spaces
The point of Definition 1.2 is to generalize Example 1.1, that is, the point
of the definition is Theorem 1.4, that the matrix describes how to get from
the representation of a domain vector with respect to the domain’s basis to
the representation of its image in the codomain with respect to the codomain’s
basis. With Definition 1.5, we can restate this as: application of a linear map is
represented by the matrix-vector product of the map’s representative and the
vector’s representative.
1.6 Example With the matrix from Example 1.3 we can calculate where that
map sends this vector.
⎛
�v = ⎝ 4
⎞
1⎠
0
This vector is represented, with respect to the domain basis B, by
⎛ ⎞
0
RepB(�v) = ⎝1/2⎠
2
and so this is the representation of the value h(�v) with respect to the codomain
basis D.
⎛ ⎞
� � 0
−1/2 1 2
RepD(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
D
D
To find h(�v) itself, not its representation, take (9/2)(1 + x) − (9/2)(−1+x) =9.
1.7 Example Let π : R 3 → R 2 be projection onto the xy-plane. To give a
matrix representing this map, we first fix bases.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 −1
� � � �
B = 〈 ⎝0⎠
, ⎝1⎠
, ⎝ 0 ⎠〉
2 1
D = 〈 , 〉
1 1
0 0 1
For each vector in the domain’s basis, we find its image under the map.
⎛
⎝ 1
⎞
0⎠
0
π
� �
1
↦−→
0
⎛
⎝ 1
⎞
1⎠
0
π
� �
1
↦−→
1
⎛
⎝ −1
⎞
0 ⎠
1
π
� �
−1
↦−→
0
Then we find the representation of each image with respect to the codomain’s
basis
� � � � � � � � � � � �
1 1
1 0
−1 −1
RepD( )= Rep
0 −1
D( )= Rep
1 1
D( )=
0 1
B