Minimal polynomial and zero matrix

Let $A \in \mathbb{R}^{n \times n}$ be a symmetric matrix. Suppose that there exists an integer $m>0$ such that $A^m = \mathbb{O}_{n \times n}$. Prove that $A$ is the zero matrix.

Solution

Again since $A$ is symmetric it is diagonizable. The polynomial $P(x)=x^m$ has $A$ as a root and $P$ divides the minimal polynomial. Since $A$ is symmetric the minimal polynomial is $m_A(x)=x$. Since $A$ is also a root of the minimal the result follows.