Suppose I have some matrix $A = \begin{bmatrix}1 & 0 \\ -1 & 1 \\1 & 1 \\ 0 & -2 \end{bmatrix}$, and I'm interested in the matrix $P$, which orthogonally projects all vectors in $\mathbb{R}^4$ onto $\mathcal{R}(A)$ (i.e. the column space of A), as well as the matrix $R$ which reflects vectors in $\mathbb{R}^4$ through $\mathcal{R}(A)$.
From a simple geometric argument (or the fact that with orthogonal projection matrices $M$ we have $M^T = M$ and $M^2 = M$, if you'd like), we can immediately gauge that the eigenvalues of $P$ are 0 and 1 (each with geometric multiplicity of 2) and the eigenvalues of $R$ are -1 and 1 (again each with geometric multiplicity of 2).
Now I was wondering:
1) Can we compute the corresponding eigenvectors of $P$ and $R$ without computing each of those matrices explicitly?
2) Can we determine a basis for the null space and the column space of $P$ directly from $A$?
3) Can we conclude that $P$ is invertible? Naturally, $R$ is invertible through another simple geometric argument, but I can't see something like this immediately for $P$.
Apologies if these 3 questions should be separate questions, but I thought maybe their solutions would build on each other, or at least give some intuition.
