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If $Ainmathbb{R}^{mtimes n}$ and $P_{Rleft(Aright)}$ is the orthogonal projection onto the range of $A$, i.e. $Rleft(Aright)$, then show that $A^{T}P_{Rleft(Aright)}=A^{T}$. So far I have that
$$A^{T}P_{Rleft(Aright)}=A^{T}P_{Rleft(Aright)}^{T}=left(P_{Rleft(Aright)}Aright)^{T}.$$
I am stuck at the last step. How does $P_{Rleft(Aright)}A=A$? My textbook says that if $operatorname{rank}left(Aright)=r$ and $B_{mtimes r}$ is a matrix whose columns form a basis for $Rleft(Aright)$, then $P_{Rleft(Aright)}=Bleft(B^{T}Bright)^{-1}B^{T}$. Additionally, it says that if $operatorname{rank}left(Aright)=n$, then $P_{Rleft(Aright)}=Aleft(A^{T}Aright)^{-1}A^{T}$. It's easy to see that if $operatorname{rank}left(Aright)=n$, then by substituting, we get $P_{Rleft(Aright)}A=A$, but the problem statement does not mention anything about the rank of $A$.

We have $ mathbb R^n=R(A) oplus R(A)^{perp}$ and $R(A)^{perp}=ker(A^T)$.

Let $P=P_{R(A)}$, then we have

$Px=x$ for all $x in R(A)$ and $Px=0$ for all $x in ker(A^T)$.

Hence:

$A^TPx=A^Tx$ for all $x in R(A)$ and $A^TPx=0=A^Tx$ for all $x in ker(A^T)$.

["] How does $P_{R(A)}A=A$? [."]

• 
  $forall xin mathbb R^n, Axin R(A)subseteqmathbb R^m$.


• The behavior of $P$ on its range is just identity, i.e. $P_{R(A)}=I_{R(A)}.$ This is can be proved by $P^2=P$, since $P$ is projection.