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. 😀
Best Answer
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 $\lambda=0$.