# 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 :). Jun 4, 2018 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. Jun 4, 2018 at 16:57