[Math] Rank of projection matrix $P$

Question

Let $A$ be an $m \times n$ matrix with $n \le m$ and $\newcommand{\rank}{\operatorname{rank}} \rank(A) = n$. How can we show that the projection matrix $P = A \, (A^\top \! A)^{-1} \! A^\top$ has rank equal to $\rank(A)$?

I see that $\rank(P) \leq \rank(A)$ but I am having some trouble seeing how these two can be equal.

With the risk of repeating, essentially how would I prove:
if $X$ is an $n\times r$ matrix with $\rank(X) = r$, then $\rank(P) = r$.

Answer 1

HINT

Indicating with $W$ the subsbace spanned by the "$n$" linearly independent column vectors $w_i$ of $A$ (i.e. $rank(A)=n$), we can consider the $"m-n"$ basis vector $v_i$ for the orthogonal complement $W_\perp$ and show that for any $v_i$

$$Pv_i=0$$

then the null space of $P$ has dimension $m-n$.