Let $M = \left[ \begin{matrix} 0 & I \\ -A & -B\end{matrix}\right]$, where both $A$ and $B$ are positive definite matrices of desired order. Trying different random matrices for $A>0$ and $B>0$, I get the impression that $M<0$ always holds. Since one of the block diagonals in $M$ is zero, I struggle to use the Schur complement. So if $M<0$ really holds, can anyone please prove it or give me some hints? Thanks.
Negative definiteness of a 2$\times$2 block matrix with one zero block on diagonal
determinanteigenvalues-eigenvectorslinear algebrapositive definite
Best Answer
Since $vMv^T=v((M+M^T)/2)v^T,$it follows that$M$ is negative-definite iff $(M+M^T)/2$ is negative-definite iff $M+M^T$ is negative-definite. Suppose $A$ and $B$ are 2x2 matrices. Then $M+M^T$ has an upper left 2x2 block entirely 0. Let $v =[1, 1, 0, 0].$ Then $v(M+M^T)v^T=[0],$ so $M+M^T$ is not negative-definite and thus $M$ is not negative-definite.