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.

  • $\begingroup$ 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. $\endgroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.