$A\in M_n(\mathbb{R})$ is symmetric s.t. $A^{10}=I.$ Prove $A^2=I$ 
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.

You have shown that $A$ is similar to $D=diag(\lambda_1,\ldots ,\lambda_n)$, so that $A^{10}=I$ is equivalent to $\lambda_i^{10}=1$ for all $i$. Since the $\lambda_i$ are real, this is equivalent to $\lambda_i^{2}=1$ for all $i$, which in turn is equivalent to $A^2=I$.
