The question I have is:

Let A be an n x n symmetric matrix. Show that A^k is symmetric if k is any nonnegative integer.

I know this is true but I'm a little confused in how to show it. I can take any symmetric matrix and then take any exponent n and see that its symmetric. But is there any other way to show this other than by example?

Thanks

Cool problem. I think we can prove this with induction on k.

If A is symmetric, then so is A^{k} when k=1.

(Also, since A^{0}=I, the result is true for k=0.)

Let's assume that A^{k} is symmetric for k=1,. n for some n\geq{1}.

Then, A^{n+1}=A(A^{n}) is the product of two symmmetric matrices. So,

by the properties of the transpose :

(A^{n+1})^{t}=(A^{n})^{t}A^{t}=A^{n}A=A^{n+1}

And therefore, hence and heretofore A^{n+1} is also symmetric.

Thus, A^{k} is symmetric for all k\geq{0}.

Neat solution. I was thinking somewhere along the lines of induction but ended up prooving it by just using some of the properties for symetric and transpose matrices. I handed it in so when I get it back I'll see if this was the correct approach. Thank you.