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In my homework, the teacher asked us to show that Ker(M*)=the complement of Ran(M). What I think is that Ran(M)=Ran(M*), and since Ker(M)+Ran(M)=n, then it follows that Ker(M*)=the complement of Ran(M). Is this right?

A second question is we need to show Ker(M*M)=Ker(M) and Ker(MM*)=Ker(M*). How to prove these two? I have no idea about them and even do not know where to start. Would please anyone give me any examples of them?

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First, for a matrix $M$ the usual notation for for the image is $\operatorname{Im}(M)$.
Second remark is that $\dim\ker(M)+\dim\operatorname{Im}(M)=n$, not $\ker(M)+\operatorname{Im}(M)=n$.
As to your question, since $M$ is self-adjoint, $M=M^*$ it indeed holds that $\operatorname{Im}(M)=\operatorname{Im}(M^*)$, but the only thing it proves is that $\dim\ker(M^*)+\dim\operatorname{Im}(M)=n$, but this, by itself does not prove that $\ker(M^*)+\operatorname{Im}(M)=V$ (where $V=F^n$). To prove that $\ker(M^*)$ is a complement of $\operatorname{Im}(M)$ you need to prove further that $\ker(M^*)\cap\operatorname{Im}(M)=\{0\}$. Then it will follow that: $$\begin{align*}\dim\left(\ker(M^*)+\operatorname{Im}(M)\right)&=\dim\ker(M^*)+\dim\operatorname{Im}(M)-\dim\left(\ker(M^*)\cap\operatorname{Im}(M)\right)\\ &=n-0=0\end{align*}$$ And hence $\ker(M^*)+\operatorname{Im}(M)=V$.
As for your second question, prove by double-inclusion. Clearly, $\ker(M)\subseteq\ker(M^*M)$ (do you see why?). Now take any $v\in\ker(M^*M)$. Then $M^*Mv=0$, so $Mv\in\ker(M^*)$. Then $Mv\in\ker(M^*)\cap\operatorname{Im}(M)$. It follows that $Mv=0$ (why?). Hence $v\in\ker(M)$. Do the same for the second part as well.

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Would you please explain why it is ker(M) is a subset of ker(M*M)? I did not get that step. – Scorpio19891119 Nov 29 '12 at 22:44
If $v\in\ker(M)$ then $Mv=0$, so $M^*Mv=M^*0=0$. – Dennis Gulko Nov 30 '12 at 7:36

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