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Unformatted text preview: Cayley Hamilton theorem : LinearAlg Jonathan L.F. King University of Florida, Gainesville FL 326112082, USA squash@ufl.edu Webpage http://www.math.ufl.edu/ squash/ 17 November, 2011 (at 21:17 ) Ques. Q1. Suppose two Fmatrices are conjugate over the algebraic closure of F . Are they conjugate over F ? Notation. Use ( x ) := Det( x I M ) for the characteristic poly of M . Let denote the zero matrix or trn. Use for the zero vector. 1 : CayleyHamilton Theorem. Over field F we have an N Nmatrix M . Then ( M ) = N N . So M is a root of its own charpoly. Proof when M is uppertriangular. In matrix M , let 1 , 2 ,..., N F be the diagonal entries; these are the eigenvalues of M . Using the std basis, let E j := Spn { e 1 ,..., e j } ; so E = { } . Since M is uppertriangular , the difference vector r j 1 := M e j j e j is in E j 1 , 2 : for each j [ 1 ..N ] . We want to show that each such e j is annihilated by ( M ). For j [ ..N ] , factor the characteristicpoly as = L j R j , where the left&right are L j ( x ) := [ x N ] [ x N 1 ] ... [ x j +1 ] ; R j ( x ) := [ x j ] [ x j 1 ] ... [ x 2 ] [ x 1 ] . [ So L j () = () and R j () = 1. ] All powers of M mutu ally commute, so we may write ( M ) = L j ( M ) R j ( M ) . Hence ISTS that R j ( M ) annihilates E j . Q [ j ] : Since all transformations annihilate E , we need to prove Q [ j 1] Q [ j ], for each j = 1 , 2 ,...,N . Induction. Fix a j [ 1 ..N ] . Note R j ( M ) an nihilates e 1 ,..., e j 1 , since R j 1 ( M ) does, and R j ( M ) = [ M j I ] R j 1 ( M ). Finally, to kill off e j note that R j ( M ) e j = R j 1 ( M ) [ M j I ] e j = R j 1 ( M ) r j 1 . This last product is , courtesy (2) and Q [ j 1]. Proof using JCF. We now handle a general M by means of JCF , the Jordan Canonical Form thm . Let G denote the algebraic closure of F . View ing M as acting on G N , our M is conjugate ( linearly isomorphic ) to its Jordan Canonical Form. Since the JCF is uppertriangular, the previous proof finishes the argument in the general case. Elementary proof using a cyclic subspace. The preceding argument used two nontrivial the orems: JCFThm, as well as the result that a field has an algebraic closure. Here is an elementary proof of CH thm , never leaving field F . Consider a trn T on a finitedimal Fvectorspace and let be its characteristic poly. Fixing a vector v 6 = , our goal is to show that ( T )( v ) equals . 3 : Exercise: Why does this suffice?...
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 Fall '07
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 Math, Calculus

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