minimal - Rational Normal Form May 11, 2007 Here is a brief...

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May 11, 2007 Here is a brief overview of the main ideas of the statement and proof of the theorem on rational normal form. Let V be a finite dimensional vector space and let T be an operator on V . A T -invariant subspace W of V is said to be cyclic if there exists a vector w W such that W is spanned by ( w, Tw, T 2 w, ··· ). If this is the case, and if m is the smallest integer such that T m w belongs to the span of ( w, Tw, T 2 w, ··· , T m - 1 w ), then in fact β w := ( w, Tw, T 2 w, ··· , T m - 1 w ) is a basis for W . The matrix for T W with respect to this basis is the companion matrix of the characteristic polynomial f T W of T W ; furthermore in this case f T W is also the minimal polynomial of T W . Theorem: If T is an operator on a finite dimensional vector space V , then V can be written as a direct sum of cyclic subspaces W i . Moreover, these subspaces can be chosen so that the characteristic polynomial of each T W i is a power of some irreducible factor of the minimal polynomial p T of T . Here is one way to attack the proof. Let p T := p e 1 1 ··· p e r r be the factorization of p T into irreducible factors. Lemma: Let V i := Ker p e i i ( T ). Then V = V 1 V 2 ··· ⊕ V r , and each V i is nonzero. Proof: Let f = p e 1 1 and let g = Q i> 1 p e i i . Since f and g have no common factors, there are polynomials p and q such that pf + qg = 1. Now if v V , let v g := p ( T ) f ( T )( v ) and let v f := q ( T ) g ( T )( v ). Then v = v f + v g . Moreover, f ( T )( v f ) = f ( T ) g ( T ) q ( T )( v ) = 0, so v f Ker ( f ( T )). Similarly v g Ker ( g ( T )). Thus V = Ker( f ( T )) + Ker( g ( T )). Moreover, if v Ker( f ( T )) Ker( g ( T )), then v = p ( T ) f ( T )( v ) + q ( T ) g ( T )( v ) = 0. The same equation shows that q ( T ) is inverse to g ( T ) on Ker( f ( T )). Note that if
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minimal - Rational Normal Form May 11, 2007 Here is a brief...

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