Is this true that $\operatorname{rank}A=\operatorname{rank}(AA^*)$?

Question

Let $A \in M_{n,m}$. Is this true that $\operatorname{rank}A=\operatorname{rank}(AA^*)$?

Just to be sure, is $A^*$ the complex conjugate of $A$ for you? — Daniel, May 13 '15 at 02:30
Rework the proof mentioned by @Yes and see if it can be generalized to $A^*$ instead of $A^T.$ — matt biesecker, May 13 '15 at 03:00
How to show that $rank(A)=rank(AA^*)$ by using the SVD ? @matt biesecker — user652838, Dec 03 '20 at 16:31

score 0 · Answer 1 · answered May 13 '15 at 03:57

It suffices to show that $A^*$ as a linear transformation is injective on $\text{Im}(A)$, i.e. $\text{ker}(A^*)\cap\text{Im}(A)=0$. Let $v\in\text{ker}(A^*)$. Then for all $w$, \begin{align*} \langle A^*v, w\rangle&=0\\ \langle v, Aw\rangle&=0 \end{align*} So $v\in\text{Im}(A)^\perp$. If $v$ is also in $\text{Im}(A)$, then it must be 0.

Is this true that $\operatorname{rank}A=\operatorname{rank}(AA^*)$?

1 Answers1