I have the next block matrix
$ J = \begin{bmatrix}A & B \\ K &0\end{bmatrix} $ all matrices are square, where $A < 0$ (definite negative), $B$ has all its eigenvalues with positive real part being $A = - (B + B^T)$, and $K$ is a diagonal matrix.
What I see from numerical simulation, is that if $K < 0 \iff J \text{ is Hurwitz}$.
Any clue about how can I prove this?
Another question, I have seen many times that
$ M = \begin{bmatrix}A & B \\ -B^T &0\end{bmatrix} $
is Hurwitz if $A < 0$, but I can not find the proof for it. I guess it is really related to the former question.
Many thanks in advance.
Edit
More ideas related to the second question. $M$ is Hurwitz and its characteristic polynomial is det$(\lambda^2I - A\lambda + BB^T)=0$. The cp of J is det$(\lambda^2I + (B + B^T)\lambda - BK) = 0$. Looking at the two cps, $-A > 0$ and $B+B^T > 0$, and $BB^T > 0$ and $-BK$ has its Eigenvalues with positive real part $\iff BK + K^TB^T < 0$.
Is this fact related to being Hurwitz ? I mean, the block matrix is Hurwitz if its cp det$(\lambda^2I + V\lambda + W)=0$ has $V > 0$ and $W > 0$ ?
Edit 2
Thank you very much for your response in Answer1. You are right, and also I have found several counterexamples to this conjecture.
However changing the condition (which has been proved wrong with a counter example) I have not found a counter example (yet). Let $K = -cI$, with $c>0$ being a scalar. In other words, $BK + K^TB^T < 0 \iff J \text{ is Hurwitz}$.
Finally, I have found a counter example (third comment to the answer 1). So this conjecture is wrong too.
However, it seems that for $c$ sufficiently small, $J$ is Hurwitz, now I have to found the condition in $c$ for that. Any clues or suggestions?
Edit 3
Finally, it has been found with counterexamples, that the last conjecture is false too.
Then, the only way (as far as I know) that I have for assessing the stability of $J$ is to check the next linear matrix inequality.
$ J \quad\text{is Hurwitz if} \quad \exists K \quad \text{s.t.} \quad J+J^T \prec 0 $