How to determine $A\in S_n (\mathbb{R})$ such that $A^5=I_n$?

Question

Let $A$ be a symmetric real matrix $A\in S_n (\mathbb{R})$ and the polynomial $A^5-I_n=0$. The problem is that this ring is not an integral domain so I cannot factorize. The only thing I know is that $A$ in a special basis has coefficients on the diagonal but they have to be real.

Thanks in advance !

Answer 1

We have $A=SDS^{-1}$ for a diagonal matrix $D=diag(d_1,\ldots ,d_n)$ and hence $$ A^5=(SDS^{-1})^5=SD^5S^{-1}. $$ hence $A^5=I$ says that $D^5=I$, i.e., $d_i^5=1$ for all $i$. Now you can solve this equation over the real numbers.