4
$\begingroup$

Prove, for real $X$, that $\text{rank}(X^TX)=\text{rank}(X)$.

Could anyone please help me with this problem? I've tried to use full-rank factorization and rank-related theorems mentioned in my book but still failed to solve this. I am learning linear algebra by myself and my book has no solutions manual so I find it really hard to get used to solve linear algebra problems.

Many thanks in advance for your help!

$\endgroup$
1
$\begingroup$

If $X^TXu=O$ then $u^TX^TXu = u^TO=O$

Write $Xu=v$ & notice $u^TX^T=v^T$

So, $u^TX^TXu = v^Tv=O$ which implies $v=Xu=O$

Thus, $null(X^TX)\subset null(X)$

Proving the reverse inclusion is trivial.

So, $null(X^TX) = null(X)$

$\endgroup$
1
  • $\begingroup$ Thank you so much for your help! I can now finally move on to other problems :). $\endgroup$ – Sophil Jun 4 '18 at 17:00
3
$\begingroup$

Consider the map $T: Im(X)\to Im(X^TX)$ given by $T(w)=X^Tw$. We show that $T$ is an isomorphism.

$T$ is obviously onto, since given any $v\in Im(X^TX)$, we have $v=X^TXv'=X^T(Xv')=T(Xv')$ for some $v'$.

It remains to check that $T$ is one-to-one.

Suppose $T(w)=0$. We have $w=Xv$ for some $v$, so $X^TXv=0$. In particular, $(X^TXv, v)=0$. Thus $(Xv, Xv)=0$, so by positivity o the inner product, we conclude that $Xv=w=0$ as desired.

$\endgroup$
1
  • $\begingroup$ Thank you so much for your answer! Unfortunately, I haven't come to the section that mentioned in your solution (map, isomorphism...) yet. I would come back to your solution later when I learn those materials. $\endgroup$ – Sophil Jun 4 '18 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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