Linear Algebra, 2020a

194 Chapter Three. Maps Between Spaces
Proof For any input ⃗v ∈ V let its expression with respect to the basis be
⃗v = c 1
⃗β 1 +···+c n
⃗β n . Define the associated output by using the same coordinates
h(⃗v) =c 1 ⃗w 1 + ···+ c n ⃗w n . This is well defined because, with respect to the
basis, the representation of each domain vector ⃗v is unique.
This map is a homomorphism because it preserves linear combinations: where
⃗v 1 = c 1
⃗β 1 + ···+ c n
⃗β n and ⃗v 2 = d 1
⃗β 1 + ···+ d n
⃗β n , here is the calculation.
h(r 1 ⃗v 1 + r 2 ⃗v 2 )=h((r 1 c 1 + r 2 d 1 )⃗β 1 + ···+(r 1 c n + r 2 d n )⃗β n )
=(r 1 c 1 + r 2 d 1 )⃗w 1 + ···+(r 1 c n + r 2 d n )⃗w n
= r 1 h(⃗v 1 )+r 2 h(⃗v 2 )
This map is unique because if ĥ: V → W is another homomorphism satisfying
that ĥ(⃗β i )=⃗w i for each i then h and ĥ have the same effect on all of the
vectors in the domain.
ĥ(⃗v) =ĥ(c 1
⃗β 1 + ···+ c n
⃗β n )=c 1 ĥ(⃗β 1 )+···+ c n ĥ(⃗β n )
= c 1 ⃗w 1 + ···+ c n ⃗w n = h(⃗v)
They have the same action so they are the same function.
QED
1.10 Definition Let V and W be vector spaces and let B = 〈⃗β 1 ,...,⃗β n 〉 be a
basis for V. A function defined on that basis f: B → W is extended linearly
to a function ˆf: V → W if for all ⃗v ∈ V such that ⃗v = c 1
⃗β 1 + ···+ c n
⃗β n , the
action of the map is ˆf(⃗v) =c 1 · f(⃗β 1 )+···+ c n · f(⃗β n ).
1.11 Example If we specify a map h: R 2 → R 2 that acts on the standard basis
E 2 in this way
( ) ( ) ( ) ( )
1 −1 0 −4
h( )=
h( )=
0 1
1 4
then we have also specified the action of h on any other member of the domain.
For instance, the value of h on this argument
( ) ( ) ( ) ( ) ( ) ( )
3
1 0
1
0 5
h( )=h(3 · − 2 · )=3 · h( )−2 · h( )=
−2
0 1
0
1 −5
is a direct consequence of the value of h on the basis vectors.
Later in this chapter we shall develop a convenient scheme for computations
like this one, using matrices.