Linear Algebra

Section II. Homomorphisms 173Proof. We will define the map by associating ⃗ β 1 with ⃗w 1 , etc., and then extendinglinearly to all of the domain. That is, where ⃗v = c 1⃗ β1 + · · · + c n⃗ βn ,the map h: V → W is given by h(⃗v) = c 1 ⃗w 1 + · · · + c n ⃗w n . This is well-definedbecause, with respect to the basis, the representation of each domain vector ⃗vis unique.This map is a homomorphism since it preserves linear combinations; where⃗v 1 = c 1⃗ β1 + · · · + c n⃗ βn and ⃗v 2 = d 1⃗ β1 + · · · + d n⃗ βn , we have this.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 )And, this map is unique since if ĥ: V → W is another homomorphism suchthat ĥ(⃗ β i ) = ⃗w i for each i then h and ĥ agree on all of the vectors in the domain.ĥ(⃗v) = ĥ(c 1 ⃗ β 1 + · · · + c n⃗ βn )= c 1 ĥ( ⃗ β 1 ) + · · · + c n ĥ( ⃗ β n )= c 1 ⃗w 1 + · · · + c n ⃗w n= h(⃗v)Thus, h and ĥ are the same map.QED1.10 Example This result says that we can construct a homomorphism byfixing a basis for the domain and specifying where the map sends those basisvectors. For instance, if we specify a map h: R 2 → R 2 that acts on the standardbasis E 2 in this way(1h( ) =0)( )−11(0and h( ) =1)( )−44then the action of h on any other member of the domain is also specified. Forinstance, the value of h on this argument( ( ( ( ( ( )3 1 0 1 0 5h( ) = h(3 · − 2 · ) = 3 · h( ) − 2 · h( ) =−2)0)1)0)1)−5is a direct consequence of the value of h on the basis vectors.Later in this chapter we shall develop a scheme, using matrices, that isconvienent for computations like this one.Just as the isomorphisms of a space with itself are useful and interesting, sotoo are the homomorphisms of a space with itself.1.11 Definition A linear map from a space into itself t: V → V is a lineartransformation.