For a symmetric matrix,
$$M = \left(\begin{array}{cc}
A & C\\
C^{\top} & D
\end{array}\right)$$
it is well known that $M$ is positive definite if and only if $A$ and the Shur complement $M\backslash A = D-CA^{-1}C^T$ are positive definite.
Is there a generalization of this fact, for non-symmetric matrices? Can we claim that:
$$M = \left(\begin{array}{cc}
A & C\\
B & D
\end{array}\right)$$
has all eigenvalues with positive real part, if and only if $A$ and $M\backslash A = D – CA^{-1}B$ have eigenvalues with positive real parts too?
I am particularly interested in matrices $M$ which are not symmetric, but for which $B=-C^T$.
Best Answer
There is a counterexample where $A$ and $M\backslash A$ have both positive eigenvalues, but $M$ has an eigenvalue with negative real part, namely $$ \left( \begin{matrix} 1 & -2 \\ 2 & -2 \end{matrix} \right) $$ A has eigenvalue $1$ and $M\backslash A$ has eigenvalue 2, but $M$ has two eigenvalues with real part $-\frac{1}{2}$.