cayley_hamilton

# cayley_hamilton - Cayley Hamilton theorem LinearAlg...

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Unformatted text preview: Cayley Hamilton theorem : LinearAlg Jonathan L.F. King University of Florida, Gainesville FL 32611-2082, USA [email protected] Webpage http://www.math.ufl.edu/ ∼ squash/ 17 November, 2011 (at 21:17 ) Ques. Q1. Suppose two F-matrices 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 : Cayley-Hamilton Theorem. Over field F we have an N × N-matrix M . Then ℘ ( M ) = N × N . So M is a “root” of its own char-poly. ♦ Proof when M is upper-triangular. In matrix M , let α 1 ,α 2 ,...,α 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 upper-triangular , the difference vector r j- 1 := M e j- α 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 characteristic-poly as ℘ = L j · R j , where the left&right are L j ( x ) := [ x- α N ] · [ x- α N- 1 ] · ... · [ x- α j +1 ] ; R j ( x ) := [ x- α j ] · [ x- α j- 1 ] · ... · [ x- α 2 ] · [ x- α 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- α j I ] · R j- 1 ( M ). Finally, to kill off e j note that R j ( M ) · e j = R j- 1 ( M ) · [ M- α 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 upper-triangular, the previous proof finishes the argument in the general case. Elementary proof using a cyclic subspace. The preceding argument used two non-trivial the- orems: JCFThm, as well as the result that a field has an algebraic closure. Here is an elementary proof of C-H thm , never leaving field F . Consider a trn T on a finite-dim’al F-vectorspace 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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