Linear Algebra, 2020a

192 Chapter Three. Maps Between Spaces
1.3 Example The domain and codomain can be other than spaces of column
vectors. Both of these are homomorphisms; the verifications are straightforward.
(1) f 1 : P 2 → P 3 given by
a 0 + a 1 x + a 2 x 2 ↦→ a 0 x +(a 1 /2)x 2 +(a 2 /3)x 3
(2) f 2 : M 2×2 → R given by
(
a
c
)
b
↦→ a + d
d
1.4 Example Between any two spaces there is a zero homomorphism, mapping
every vector in the domain to the zero vector in the codomain.
We shall use the two terms ‘homomorphism’ and ‘linear map’ interchangably.
1.5 Example These two suggest why we say ‘linear map’.
(1) The map g: R 3 → R given by
⎛ ⎞
x
⎜ ⎟
⎝y⎠
↦−→ g
3x + 2y − 4.5z
z
is linear, that is, is a homomorphism. The check is easy. In contrast, the
map ĝ: R 3 → R given by
⎛ ⎞
x
⎜ ⎟
⎝y⎠
↦−→ ĝ
3x + 2y − 4.5z + 1
z
is not linear. To show this we need only produce a single linear combination
that the map does not preserve. Here is one.
⎛
⎜
0
⎞ ⎛ ⎞
⎛
1
⎟ ⎜ ⎟
⎜
0
⎞ ⎛
⎟ ⎜
1
⎞
⎟
ĝ( ⎝0⎠ + ⎝0⎠) =4 ĝ( ⎝0⎠)+ĝ(
⎝0⎠) =5
0 0
0 0
(2) The first of these two maps t 1 ,t 2 : R 3 → R 2 is linear while the second is
not. ⎛ ⎞
⎛ ⎞
x ( ) x ( )
⎜ ⎟
⎝y⎠ t 1 5x − 2y ⎜ ⎟
↦−→
⎝y⎠ t 2 5x − 2y
↦−→
x + y
xy
z
z
Finding a linear combination that the second map does not preserve is
easy.