Rank of $A^2$ is $A$ is symmetric

Question

Let $A$ be a symmetric matrix. Is $rank(A^2)=rank(A)$

I'm pretty sure this is true, and if it is, could someone guide me towards the method of proving it?

Answer 1

Hint: show the column spaces of $A^2$ and $A$ are same.

Answer 2

It suffices to show that $A|_{\text{Im}(A)}$ is an isomorphism, i.e. $\text{Im}(A)\cap\text{ker}(A)=\{0\}$. But $\text{Im}(A)^\perp=\text{ker}(A)$ for symmetric matrix because $v\in\text{ker}(A)\Leftrightarrow Av=0\Leftrightarrow \langle Av, w\rangle=0\text{ for all }w\Leftrightarrow \langle v, Aw\rangle=0\text{ for all }w\Longrightarrow v\in\text{Im}(A)^\perp$. This establishes the claim.

Answer 3

You know for a symmetric matrix $rank$ $=$ no. of non zero eigenvalues.

And eigenvalues of $A^2$ are squares of eigenvalues of $A$. Thus both have same no. of non zero eigenvalues and hence same rank.

Answer 4

Assume that $v$ is in the kernel of $A$, then $A^2v=0$, so $v$ is in the kernel of $A^2$. On the other hand, if $v$ is in the kernel of $A^2$, then $$ 0=v^TA^2v=v^TA^TAv=\|Av\|^2, $$ so $v$ is in the kernel of $A$. Thus, the nullspace of $A$ coincides with the nullspace of $A^2$. I guess you can conclude the result you want from this?