TITLE: $A\in M_n(\mathbb{R})$ is symmetric s.t. $A^{10}=I.$ Prove $A^2=I$
QUESTION [5 upvotes]: Let $A\in M_n(\mathbb{R})$ be symmetric, such that $A^{10}=I.$ Prove $A^2=I$

My thoughts:
Since $A$ is symmetric, $A^2$ is symmetric, so there exists an orthogonal $P\in M_n(\mathbb{R})$ such that $D=P^{-1}A^2P$ is a diagonal matrix.
I tried to work with that in order to find the "right" diagonal matrix such that after power manipulations, I could prove that $A^{10}$ is similar to $A^2$ and conclude the result, but got stuck.
Any help is appreciated.

REPLY [2 votes]: Your method was good.   
Indeed   $A^2$ is diagonalizable as symmetric.
$D=P^{-1}A^2P$      
$ A^2= PDP^{-1} $      
$ A^{10}=PD^5P^{-1}=I$     
$D^5=P^{-1}IP =I$   
But if for diagonal $D$ we have $D^5=I$ then the only possibility $D=I$.    
And consequently $  A^2= PIP^{-1} =I$.