4
$\begingroup$

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

  • 1
    You should have mentioned that to begin with, you know.2011-12-07

1 Answers 1

1

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 $\lambda=0$.

  • 0
    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.2012-06-27