College Algebra Exam Review 152

College Algebra Exam Review 152 - B is a vector subspace of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
162 3. PRODUCTS OF GROUPS Proof. The homomorphism theorem for groups (Theorem 2.7.6 ) gives us an isomorphism of abelian groups Q T satisfying Q T ı ± D T . We have only to verify that Q T also respects multiplication by scalars. But this follows at once from the definitions: Q T .˛.x C N// D Q T .˛x C N/ D T.˛x/ D ˛T.x/ D ˛ Q T .x C N/ . n The next three propositions are analogues for vector spaces and linear transformations of results that we have established for groups and group homomorphisms in Section 2.7 . Each is proved by adapting the proof from the group situation. Some of the details are left to you. Proposition 3.3.11. (Correspondence theorem for vector spaces) Let T W V ! V be a surjective linear map, with kernel N . Then M 7! T ± 1 . M/ is a bijection between subspaces of V and subspaces of V containing N . Proof. According to Proposition 2.7.12 , B 7! T ± 1 . B/ is a bijection be- tween the subgroups of V and the subgroups of V containing N . I leave it as an exercise to verify that
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: B is a vector subspace of V if, and only if, T 1 . B/ is a vector subspace of V ; see Exercise 3.3.6 . n Proposition 3.3.12. Let T W V ! V be a surjective linear transformation with kernel N . Let M be a subspace of V and let M D T 1 . M/ . Then x C M 7! T.x/ C M denes a linear isomorphism of V=M to V = M . Equivalently, .V=N/=.M=N/ V=M; as vector spaces. Proof. By Proposition 2.7.13 , the map x C M 7! T.x/ C M is a group iso-morphism from V=M to V = M . But the map also respects multiplication by elements of K , as .v C M/ D v C M 7! T.v/ C M D T.v/ C M D .T.v/ C M/ We can identify V with V=N , by the homomorphism theorem for vector spaces, and this identication carries the subspace M to the image of M in V=N , namely M=N . Therefore .V=N/=.M=N/ V = M V=M: n...
View Full Document

Ask a homework question - tutors are online