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I have the following generalized eigenvalue problem

$\det[P_i^TLP_j+zP_i^TP_j] = 0$

$L$ is a positive-semidefinite matrix with 1 eigenvalue at 0. More precisely, it is the combinatorial Laplacian matrix for a connected graph. $P_i$ ($P_j$) is the identity matrix with the i-th (j-th) column removed.

What, if anything, can be said about the generalized eigenvalues $z$ of this problem?

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What do you mean by $i$th column removed? Say it starts as an $n\times n$ matrix, does it becomes an $n\times n$ matrix with a 0 at its $(n,n)$ entry, or does it become an $n\times (n-1)$ matrix? Also, what is meant by "1 eigenvalue at 0"? Do you mean that it is positive-semidefinite with the eigenvalue 0 of multiplicity only 1? So that it is in fact rank (n-1)? –  Willie Wong Mar 3 '11 at 10:42
    
yes...$P_i$ is $n \times n-1$ matrix. $L$ has a real spectrum $0 = \lambda_1 < \lambda_2 \leq \lambda_3 \leq \ldots \leq \lambda_n$; so yes, it has rank $n-1$, and the 0 eigenvalue has multiplicity one. –  1yen Mar 3 '11 at 12:30
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