Show $A^TP_{R(A)}=A^T$ for all $A$ in $\mathbb{R}^{m \times n}$?

Question

Prove that right multiplication of the orthogonal projector that maps vectors to the range of $A$ by $A^T$ backs to $A^T$. Algebraically,

$$A^TP_{R(A)}=A^T$$

where $R(A)$ is the range of matrix $A$.

My try:

$$A^T(I-P_{R(A)})=0$$

Therefore, $I-P_{R(A)}$ is the projector onto $R(A)^{\perp}=N(A^T)$. I do not know how to argue the rest.

Answer 1

For any matrix $M$, $M=0$ if and only if $v^T M u=0$ for all column vectors $u$ and $v$ (of appropriate dimension).

For this question, let $u\in\mathbb{R}^m$ and $v\in\mathbb{R}^n$, then $$v^TA^T(I-P_{R(A)})u=(Av)^T(I-P_{R(A)})u=0$$ since $Av\in R(A)$ and $(I-P_{R(A)})u\in R(A)^\bot$.