If $A$ is a symmetric matrix, then rank($A^{n}$) = rank($A$). Is, this statement true?

Question

If $A$ is a symmetric matrix over $\mathbb{R}$, then rank($A^{n}$) = rank($A$). Is, this statement true?

For $n=2$ and $n=3$ the statement seems to be true but I have no idea how to prove or disprove for the general case.

Thanks in advance!

Answer 1

Let $A$ be a symmetric $m \times m$ - Matrix.

If $x \in \ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x \in \ker(A)$.

This gives $\ker(A)=\ker(A^2)$, hence $\ker(A)=\ker(A^k)$ for all $k \ge 1$.

It follows (why ?): $\mbox{im}(A)=\mbox{im}(A^k)$ for all $k \ge 1$.

Hence $\mbox{rank}(A)=\mbox{rank}(A^k)$ for all $k \ge 1$.