Orthogonal Projection onto Range of Matrix

Question

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

Answer 1

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 $x \in ker(A^T)$.

Answer 2

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

• $\forall x\in \mathbb R^n, Ax\in R(A)\subseteq\mathbb R^m$.
• 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.