Multiplying an eigenvalue equation by a non-invertible matrix: what eigenvalue characteristics are retained?

Question

Suppose I have an eigenvalue equation $$M v=\lambda v$$ and I have characterized the eigenvalues. Maybe $M$ is Hermitian and $\lambda$ is real, for example.

Given a non-invertible matrix $P$ (I'm mostly interested in the case where $P$ is diagonal positive semi-definite), consider this modified eigenvalue equation: $$P M v=\lambda' P v$$ Assume $v$ does not lie in the nullspace of either $P$ or $PM$.

Question: can we relate the generalized eigenvalues $\lambda'$ to the original eigenvalues $\lambda$? In particular, do the $\lambda'$ retain the characteristics of the $\lambda$ (such as being purely real when $M$ is Hermitian)?

Answer 1

Your equation has solutions where $\lambda'$ can be any complex number.

Your equation says $P(M v - \lambda' v) = 0$, i.e. $M v - \lambda' v$ is in the nullspace of $P$. If $\lambda'$ is an eigenvalue of $M$, this has nonzero solutions where $M v - \lambda' v = 0$. Of course if the eigenspace for $\lambda'$ is contained in the nullspace of $P$, you're not allowing that because $P v = 0$, and if $\lambda' = 0$ you're not allowing it because $PMv = 0$.

On the other hand, if $\lambda'$ is not an eigenvalue of $M$, $M - \lambda' I$ is invertible, so for any $w$ in the nullspace of $P$ we can take $v = (M - \lambda' I)^{-1} w$ and have $M v - \lambda' v = w$. The only exception is that if the nullspace of $P$ is contained in the nullspace of $M$, we'll have $v = - (\lambda')^{-1} w$ so $M v = 0$, which you don't allow.