Diagonalizing a Matrix

Date: 04/12/2001 at 02:14:11
From: john cholewa
Subject: Regarding a class of linear algebra problem.

Hi Dr. Math -

In Herstein's _Topics in Algebra_ there is a problem in section 6.10 
on page 349 stating that:

   If Q is a real symmetric matrix satisfying Q^k = I for some integer 
   k >= 1, prove that Q^2 = I.

Also in an online problem archive there is a problem stating:

   If A is a 2x2 matrix with integer entries, and A^k = I for some 
   k >= 1, then A^12 = I.

The obvious thing that jumps to mind is that the Det() of the matrix 
is +-1, and its invertible. With A^(k-1) = A^(-1).

However, to go any farther I need something on the other direction of 
the Caley-Hamilton to force something, like if one knows TQT^(-1) = I 
for some T, then it's the unique diagonalizing matrix for Q. (Actually 
looking for a generalized theorem that given a certain condition on A 
[normal?], and A^k = I for some k >= 1, then A^(n) = I for some n.)

Hope you can help me here. Thanks.

Date: 04/12/2001 at 13:16:38
From: Doctor Rob
Subject: Re: Regarding a class of linear algebra problem.

Thanks for writing to Ask Dr. Math, John.

Any real symmetric matrix can be diagonalized. If Q is that matrix,
there is a real orthogonal matrix P such that

   P'*Q*P = D

where D is diagonal and P' means the transpose of P, and
P*P' = P'*P = I.  Then

   Q = P*D*P'.

Now taking the kth power,

   I = Q^k = P*D^k*P',
   I = P'*I*P = D^k.

Since D is diagonal and real, that means all the diagonal entries
of D must satisfy d^k = 1, so each must be 1 or -1.  That implies
that d^2 = 1, and so D^2 = I, and so Q^2 = I.

Two-by-two integer matrices are a different story. Usually they cannot 
be diagonalized. On the other hand, they do satisfy quadratic 
characteristic equations, by the Cayley-Hamilton Theorem. Thus there 
are integers a and b such that

   a = trace(Q),
   b = det(Q),
   Q^2 - a*Q + b*I = 0.

On the other hand, you know that Q^k = I for some k, so

   det(Q^k) = det(Q)^k = 1,
   det(Q) = 1 or -1.

Thus b = 1 or -1.  The task at hand is to show that the only
possibilities for the minimal polynomial of Q are

   x - 1,
   x + 1,
   x^2 - 1,
   x^2 + 1,
   x^2 - x + 1,
   x^2 + x + 1,

and all of these are divisors of x^12 - 1.  That I leave to you.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/