College Algebra Exam Review 151

# College Algebra Exam Review 151 - ±.˛v D ˛±.v for v 2 V...

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

3.3. VECTOR SPACES 161 But this follows from the closure of W under scalar multiplication; namely, if v C W D v 0 C W and, then ˛v ± ˛v 0 D ˛.v ± v 0 / 2 ˛W ² W . Thus ˛v C W D ˛v 0 C W , and the scalar multiplication on V=W is well-deﬁned. Theorem 3.3.9. If W is subspace of a vector space V over K , then V=W has the structure of a vector space, and the quotient map ± W v 7! v C W is a surjective linear map from V to V=W with kernel equal to W . Proof. We know that V=W has the structure of an abelian group, and that, moreover, there is a well-deﬁned product K ³ V=W ±! V=W given by ˛.v C W / D ˛v C W . It is straighforward to check the remaining vector space axioms. Let us indclude one veriﬁcation for the sake of illustration. For ˛ 2 K and v 1 ;v 2 2 V , ˛..v 1 C W / C .v 2 C W // D ˛..v 1 C v 2 / C W // D ˛.v 1 C v 2 / C W D .˛v 1 C ˛v 2 / C W D .˛v 1 C W / C .˛v 2 C W / D ˛.v 1 C W / C ˛.v 2 C W / Finally, the quotient map ± is already known to be a group homomorophism. To check that it is linear, we only need to verify that
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ±.˛v/ D ˛±.v/ for v 2 V and ˛ 2 K . But this is immediate from the deﬁnition of the product, ˛v C W D ˛.v C W / . n V=W is called the quotient vector space and v 7! v C W the quotient map or quotient homomorphism . We have a homomorphism theorem for vector spaces that is analogous to, and in fact follows from, the homomor-phism theorem for groups. Theorem 3.3.10. (Homomorphism theorem for vector spaces). Let T W V ±! V be a surjective linear map of vector spaces with kernel N . Let ± W V ±! V=N be the quotient map. There is linear isomorphism Q T W V=N ±! V satisfying Q T ı ± D T . (See the following diagram.) V T q q q q q V ± q q q q q ± ± ± ± ± ± q q q q q q q ∼ = Q T V=N...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online