If $A$ and $B$ are symmetric, $A^5=B^5$ then $A=B$

Question

Let $A$, $B$ be real symmetric $n\times n$ matrices such that $A^5=B^5$.

Prove that $A=B$.

Here are my attempts:

The following identity holds $(A-B)(A^4+A^3B+A^2B^2+AB^3+B^4)=0$ and yields at least one non-invertible matrix. What to do with that ?

Since $A$ and $B$ are symmetric, they can be orthogonally diagonalized as $D_1$ and $D_2$. Moreover $D_1^5$ and $D_2^5$ are orthogonally similar. I'm stuck here.

Thanks for your suggestions

Answer 1

If $A$ is symmetric, then it is orthogonally diagonalized is $A=U^*DU$, where $U^*U=I$ and $$ D=\mathrm{diag}(d_1,\ldots,d_n), $$ where the $d_i$'s are real. Then $A^5=U^*D^5U$.

If we know that $A$ is symmetric and we know $A^5$ we can UNIQUELY reproduce $A$ since $$ A^5=U^*\mathrm{diag}(d_1^5,\ldots,d_n^5)U, $$ and although the equation $x^5=d_j^5$ has five roos in general, only one of them if real.