Show that $X^TX(X^TX)^g$ is a projection onto the column space of $X^T$

Question

Show that $$X^TX(X^TX)^g $$ is a projection onto the column space of $X^T$

I'm having a bit of trouble getting this, I am aware of the three conditions to prove that it is a projection but I don't know how to prove it.

Answer 1

Let $G$ be the generalized inverse of $A$. Then $AGA=A$. Therefore, $$(AG)^2=(AGA)G=AG$$ Hence $AG$ is idempotent.

For any two matrices $A$ and $B$, $\mathcal{C}(AB)\subseteq\mathcal{C}(A)$. Hence $\mathcal{C}(AG)\subseteq\mathcal{C}(A)$. However $$\mathrm{rank}(AG)\ge\mathrm{rank}(AGA)=\mathrm{rank}(A)\ge\mathrm{rank}(AG)$$ Hence we have $\mathrm{rank}(A)=\mathrm{rank}(AG)$ and hence $\mathcal{C}(AG)=\mathcal{C}(A)$.

This proves that $AG$ is the projector matrix for $\mathcal{C}(A)$.

Now we need to show that $\mathcal{C}(X^TX)=\mathcal{C}(X^T)$. Of course we know $\mathcal{C}(X^TX)\subseteq\mathcal{C}(X^T)$.

Now $$X^TXx=0\implies x^TX^TXx=0\implies(Xx)^T(Xx)=0\implies Xx=0$$ $$Xx=0\implies X^TXx=0$$ Hence $\mathrm{nullity}(X^TX)=\mathrm{nullity}(X)$. So $$\mathrm{rank}(X^TX)=\mathrm{rank}(X)=\mathrm{rank}(X^T)$$ Thus we have $\mathcal{C}(X^TX)=\mathcal{C}(X^T)$.