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Cayley-Hamilton Theorem

Date: 04/19/99 at 13:03:32
From: Sarah Mastura
Subject: The Cayley-Hamilton Theorem - a proof

I was wondering if you could possibly send me a proof of the Cayley-
Hamilton Theorem, or if not point me in the right sort of direction 
of one. The proof in my notes is quite incomplete and I can't, right 
now, get hold of a book that it might be in.

Thanks very much.

Date: 04/19/99 at 16:10:08
From: Doctor Rob
Subject: Re: The Cayley-Hamilton Theorem - a proof

Thanks for writing to Ask Dr. Math.

Cayley-Hamilton Theorem: If A is an n-by-n matrix and f(x) its
characteristic polynomial, then f(A) = 0.


Let B = x*I - A, where I is the identity matrix of the same size as A.  
Let C be the classical adjoint of B, so that BC = (det B)I. Recall 
f(x) = det(B), by definition. We wish to show that f(A) = 0. It 
suffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the 
k-th unit vector, so fix a value of k in that range.

Now x = A implies that for all j,

   x*e(j) = A*e(j),

          = SUM A(i,j)*e(i),
      n                     n
   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),
     i=1                   i=1

   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,

   SUM B(i,j)*e(i) = 0,

   C(j,k)*SUM B(i,j)*e(i) = 0,

   SUM B(i,j)*C(j,k)*e(i) = 0.

Then, summing over all j, reversing the order of summation, and using
the definition of the classical adjoint,

    n   n
   SUM SUM B(i,j)*C(j,k)*e(i) = 0,
   j=1 i=1

    n    n
   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,
   i=1  j=1

   SUM [det(B)*delta(i,k)]*e(i) = 0,

   SUM [f(x)*delta(i,k)]*e(i) = 0,

   SUM [f(A)*delta(i,k)]*e(i) = 0,

   f(A)*SUM [delta(i,k)]*e(i) = 0,

   f(A)*e(k) = 0.

This proves that f(A) annihilates all the unit vectors, so must be
the zero matrix, and f(A) = 0.

- Doctor Rob, The Math Forum   
Associated Topics:
College Linear Algebra

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