Let Phi epsilon End(V) for a finite dim\'s vector space V, Prove that phi is manic phi is an isomorphism.
Solution
Given that is an endomorphism on a finite dimensional vector space V. Assume first that is monic it means that ker()={0} i.e. ker() is 0 dimensional, then by rank nullity theorem which says that
dim(V)=dim(ker())+dim(range())
we have dim(V)=dim(range()). Since range() is a subspace of V and its dimension is equal to whole of V, it is equal to V i.e. range()=V implies is epic.
Next assume that is epic then range()=V. Again by rank nullity theorem we have
dim(ker())=0
implies that ker()=0 and hence is monic.
Therefore, is is monic then it is epic and if it is epic then it is monic and hence any one of the conditions i.e. monic or epic will imply that is an isomorphism which completes the proof.
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Let Phi epsilon End(V) for a finite dim-'s vector space V- Prove that.docx
1. Let Phi epsilon End(V) for a finite dim's vector space V, Prove that phi is manic phi is an
isomorphism.
Solution
Given that is an endomorphism on a finite dimensional vector space V. Assume first that is
monic it means that ker()={0} i.e. ker() is 0 dimensional, then by rank nullity theorem which
says that
dim(V)=dim(ker())+dim(range())
we have dim(V)=dim(range()). Since range() is a subspace of V and its dimension is equal to
whole of V, it is equal to V i.e. range()=V implies is epic.
Next assume that is epic then range()=V. Again by rank nullity theorem we have
dim(ker())=0
implies that ker()=0 and hence is monic.
Therefore, is is monic then it is epic and if it is epic then it is monic and hence any one of the
conditions i.e. monic or epic will imply that is an isomorphism which completes the proof.
