Check out some similar questions!

cool_dude · Jan 27, 2008, 09:24 PM

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

galactus · Jan 28, 2008, 05:55 PM

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 $k\geq{0}$ .

cool_dude · Jan 28, 2008, 09:01 PM

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.