Let $A \in \mathbb{R}^{n\times n}$ be a symmetric positive semi-definite matrix with exactly one zero eigenvalue and $B \in \mathbb{R}^{n\times n}$ be a symmetric matrix having $k$ positive eigenvalues.
Is it possible to infer the number of positive eigenvalues of the GEP
$Av = \lambda B v$
given the above information? Or some bounds on the number of positive eigenvalues?
I assume that the generalized eigenvalues will be real in this case, but I'm not sure about the proof. Following the classic proof for the basic eigenvalue problem results in
$u^{*T}Bu(\lambda^* - \lambda) = 0$
with $u^{*T}Bu$ not necessarily being nonzero if $B$ is just a real symmetric matrix.
A similar question assumes a general matrix $A$, not a real PSD one.
Another related question points out, that the number of generalized eigenvalues equal to zero will be the same as the number of such eigenvalues of $A$, but I don't understand the argumentation.
You can show that they are real-valued. For every finite nonzero generalized eigenvalue $\lambda$, we have $$A v = \lambda B v \quad \Rightarrow \quad v^\dagger A v = \lambda v^\dagger B v \quad \Rightarrow \quad \lambda = \frac{v^\dagger A v}{v^\dagger B v}.$$ Now if we compute $\lambda^*$, we find $$\lambda^* = \lambda^\dagger = \frac{v^\dagger A^\dagger v}{v^\dagger B^\dagger v} = \frac{v^\dagger A v}{v^\dagger B v} = \lambda,$$ since $A^\dagger = A$ and $B^\dagger = B$. Therefore, $\lambda^* = \lambda \in \mathbb{R}$.
However, I think you cannot be sure to get $k$ positive generalized eigenvalues. The matrix $A$ has a zero eigenvalue so $(A,B)$ will have a zero generalized eigenvalue as well. Based on this one can find examples where there are less than $k$ positive generalized eigenvalues. It might be possible to prove that there are at least $k-1$ if $A$ has exactly one zero eigenvalue but of this I am not sure.