Effect on eigenvalues of a projection matrix when removing its main diagonal?

Question

Do you know what happens to a projection matrix when you remove the main diagonal? Consider $P=Z(Z'Z)^{-1}Z'$ and form P-diag(P-{i,i}).

Now, when you eigendecompose $P$ we get zeroes and ones as eigenvalues and the columns of $Z$ as eigenvectors (almost trivially I suppose).

But when we eigendecompose $P-D$ into $Q \Lambda Q'$, what do we get then?

There is some excellent discussion here on this topic already and the third answer to this question may be of help.

Answer 1

You assume that $P=[p_{i,j}]$ is a real orthogonal projection, that implies that $P$ is a real symmetric matrix. Let $D=diag(p_{i,i})$ and $Q=P-D$ (a symmetric matrix). Note that $p_{i,i}=<e_i,Pe_i>=\cos(\theta)\in[0,1]$. Since $spectrum(P)\subset \{0,1\}$, $X^TQX=X^TPX-\sum_i p_{i,i}{x_i}^2$.

Then $X^TQX\leq ||X||^2$ and $X^TQX\geq -||X||^2$. Then $spectrum(Q)\subset [-1,1]$.