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### Diagonalizing a Matrix

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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.
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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/
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Associated Topics:
College Linear Algebra

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