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

Question

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.

Answer 1

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 $A^2=I$.

Answer 2

The minimal polynomial of $A$ divides $x^{10}-1$.

$x^{10}-1=(x^2-1)q(x)$, where $q(x)$ has no real roots.

The eigenvalues of a real symmetric matrix are all real and so its minimal polynomial is a product of linear real factors.

Therefore, the minimal polynomial of $A$ divides $x^2-1$ and so $A^2=I$.

Answer 3

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$.