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I'm stuck on finding the eigenvalues of $$ \bar{A} = \begin{bmatrix} 0 & S\\ S^\top & A \end{bmatrix} $$ Both $S$ and $A$ are square matrices of the same dimension and are invertible. $A$ is symmetric positive definite.

Any help is appreciated. :-D

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I'm afraid not much can be said in such generality. Do you know anything about those matrices? – Patrick Da Silva Dec 7 '11 at 6:29
This isn't what's usually referred to as a block triangular matrix -- that would mean that the off-diagonal block is zero, not the diagonal block. – joriki Dec 7 '11 at 8:31
A better description might be that your matrix is "block antitriangular". In any event, is $\mathbf A$ (the $2,2$ block) symmetric? – J. M. Dec 7 '11 at 11:22
hmm, in my problem $A$ is symmetric positive definite. – Muhammad Fuady Dec 7 '11 at 12:32
You should have mentioned that to begin with, you know. – J. M. Dec 7 '11 at 15:15

We have for block matrices, with $A$ invertible, that $$\pmatrix{A&B\\ C&D}=\pmatrix{A&0\\ C&I}\cdot\pmatrix{I&A^{-1}B\\0&D-CA^{-1}B}.$$ In our case, when $\lambda\neq 0$, we get $$\det(\overline A-\lambda I_{2n})=(-\lambda)^n\det(A-\lambda I_n-S(\lambda)^{-1}I_nS^T)=\det(\lambda^2I_n-\lambda A+SS^T).$$ This formula is also true for $\lambda=0$.

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Here, I think you need the version where $\mathbf D$ is the one inverted, not $\mathbf A$... – J. M. Jun 27 '12 at 18:56
I computed the determinant of $\overline A-\lambda I$, for $\lambda\neq 0$, hence we get that the matrix on the top left is invertible. – Davide Giraudo Jun 27 '12 at 19:40

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