# Eigenvalues of Eigenvectors of Projection and Reflection Matrices

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.

• The other Eigenspace is just the orthogonal of A, which answers 1 and 2. P is not invertible unless the null space is the 0 vector. Commented Dec 6, 2019 at 3:28
• The eigenspace corresponding to which eigenvalue(s)? Can you expand on your comments in an answer? Commented Dec 6, 2019 at 3:50
• As far as invertibility goes, do you know the relationship of the determinant of a matrix to its eigenvalues?
– amd
Commented Dec 6, 2019 at 7:59
• Ah indeed, I overlooked that one. $P$ is an orthogonal projection matrix, and hence diagonalizable. Then the determinant of $P$ is simply the determinant of its diagonal matrix, and hence the product of those diagonal entries (its eigenvalues). Since $0$ is an eigenvalue, it follows that $P$ is not invertible. That deals with 3) nicely. Commented Dec 6, 2019 at 8:12