If $A$ is a $12 \times 12$ real matrix such that $A^{17}=I$ , is $A$ diagonalizable ? Are all eigenvalues of $A$ real ?

Question

If $A$ is a $12 \times 12$ real matrix such that $A^{17}=I$ , is $A$ diagonalizable ? Are all eigenvalues of $A$ real ?

Answer 1

The minimal polynomial of $A$ divides $$x^{17} - 1 = \left(x - 1\right)\left(\sum_{i=0}^{16} x^i \right).$$ Since $x^{17} - 1$ has only one real root $x = 1$, the factorization over $\mathbb{R}$ of $x^{17} - 1$ is of the form $(x - 1)p_1(x) \ldots p_8(x)$ where $p_i$ are quadratics without real roots.

Since a matrix $A \in M_n(\mathbb{R})$ is diagonalizable if and only if the minimal polynomial of $A$ splits into linear factors, we see that a matrix $A \in M_n(\mathbb{R})$ that satisfies $A^{17} = I$ will be diagonalizable if and only if the minimal polynomial of $A$ is $(x - 1)$ which means that $A = I_n$. If $A \neq I_n$, then $A$ will have at least two conjugate non-real complex eigenvalues.