Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.