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?

score 0 · Accepted Answer · answered Jun 06 '15 at 08:20

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

answered Jun 06 '15 at 08:20

PSPACEhard

10,283

Could you expend on how I would do that for this problem? – tzamboiv Jun 06 '15 at 08:38
http://math.stackexchange.com/questions/1299061/using-svds-to-prove-cxx-prime-cx/1299159#1299159 This link proves column space of $(A^2)$ is the same as column space of $(A)$. – PSPACEhard Jun 06 '15 at 08:42

score 0 · Answer 2 · answered Jun 06 '15 at 09:02

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.

score 0 · Answer 3 · answered Jun 09 '15 at 04:43

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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.

answered Jun 09 '15 at 04:43

Sry

722

score 0 · Answer 4 · answered Jun 09 '15 at 04:48

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?

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

4 Answers4