How do I show that $P_{range(A)}$ is an orthogonal projector onto A?

Question

I have the first part of the proof, which is that P is indeed an orthogonal projector ($P = P^2 = P^T$). Now, how do I show that $Px \in range(A) \forall x \in R^n$? The linear algebra class I took before numerical was... less than ideal, so I don't know much about the properties of the range.

Answer 1

$P_{\text{range}(A)} (x) = A \big( (A^T A)^{-1} A^T x \big) = Ay$, for $y = (A^T A)^{-1} A^Tx $.

$P_{\text{range}(A)} (x)$ is therefore in the range of $A$, because it is equal to $Ay$ for some $y$.

Side note: You are assuming that $(A^TA)$ is invertible, which means that the columns of $A$ are linearly independent. That is not necessarily the case in general.