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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 isomorphism 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...
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 Fall '08
 EVERAGE
 Algebra, Multiplication, Scalar

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