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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.

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  • $\begingroup$ Just to clarify, $A$ has exactly one zero eigenvalue, or at least one zero eigenvalue? $\endgroup$ – πr8 Jan 5 '17 at 21:08
  • $\begingroup$ $A$ has indeed exactly one zero eigenvalue, and I'm interested in knowing at least the bounds on the number of positive eigenvalues, i.e. not necessarily the exact number. $\endgroup$ – Omicron_Persei_11 Jan 6 '17 at 8:16
  • $\begingroup$ Yep, just checking. One technique is that, given the symmetry condition, you can actually assume $A$ to be diagonal, and indeed, even to be a diagonal projection matrix (i.e. $A=diag(1,\cdots,1,0)$, by considering an appropriate change of basis. Alternatively, you could apply a similar trick to $B$, though note that you can't expect (in general) to do this to both matrices at once. (This trick may not lead to an answer, but I find it to occasionally be useful in getting a feel for these questions). $\endgroup$ – πr8 Jan 6 '17 at 9:36
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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.

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