Prove that $\text{rank}(X^TX)=\text{rank}(X)$

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.

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

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

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.

• 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. – Sophil Jun 4 '18 at 16:57