College Algebra Exam Review 384

# College Algebra Exam Review 384 - 394 8 MODULES Deﬁnition...

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Unformatted text preview: 394 8. MODULES Deﬁnition 8.6.1. The companion matrix of a monic polynomial a.x/ D x d C ˛d 1 x d 1 C C ˛1 x C ˛0 is the matrix 2 3 00 0 0 ˛0 61 0 0 0 ˛1 7 6 7 0 ˛2 7 60 1 0 6: : : :7 6: : :: ::: :7 :7 6: : 6 7 :: 40 0 0 :0 ˛d 2 5 00 0 1 ˛d 1 We denote the companion matrix of a.x/ by Ca . Lemma 8.6.2. Let T be a linear transformation on a ﬁnite dimensional vector space V over K and let a.x/ D x d C ˛d 1x d1 C C ˛1 x C ˛0 2 KŒx: The following conditions are equivalent: (a) V is a cyclic KŒx–module with annihilator ideal generated by a.x/. (b) V has a vector v0 such that V D span.fT j v0 W j 0g/ and a.x/ is the monic polynomial of least degree such that a.T /v0 D 0. (c) V Š KŒx=.a.x// as KŒx modules. (d) V has a basis with respect to which the matrix of T is the companion matrix of a.x/. Proof. We already know the equivalence of (a)-(c), at least implicitly, but let us nevertheless prove the equivalence of all four conditions. V is a cyclic module with generator v0 , if, and only if, V D KŒxv0 D ff .T /v0 W f .x/ 2 KŒxg D spanfT j v0 W j 0g. Moreover, ann.V / D ann.v0 / is the principal ideal generated by its monic element of least degree, so ann.V / D .a.x// if, and only if, a.x/ is the polynomial of least degree such that a.T /v0 D 0. Thus conditions (a) and (b) are equivalent. If (b) holds, then f .x/ 7! f .x/v0 is a surjective module homomorphism from KŒx to V , and a.x/ is an element of least degree in the kernel of this map, so generates the kernel. Hence V Š KŒx=.a.x// by the homomorphism theorem for modules. In proving that (c) implies (d), we may assume that V is the KŒx– module KŒx=.a.x//, and that T is the linear transformation f .x/ C .a.x// 7! xf .x/ C .a.x//: ...
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