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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.


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.

$$ J \quad\text{is Hurwitz if} \quad \exists K \quad \text{s.t.} \quad J+J^T \prec 0 $$

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up vote 3 down vote accepted

I assume that by a Hurwitz matrix, you mean a stable matrix, i.e. a matrix such that all of its eigenvalues have negative real parts. Please correct me if I am wrong.

I did some numerical experiments too, but found that your conjecture is wrong. Here is a counterexample: $$ J=\begin{pmatrix} -1.4268 & -0.6777 & 0.7134 & 0.4497\\ -0.6777 & -0.3444 & 0.2280 & 0.1722\\ -3.7490 & 0 & 0 & 0\\ 0 & -8.6780 & 0 & 0 \end{pmatrix}. $$ You may verify that the eigenvalues of $A$ are $-1.7529$ and $-0.0183$, the eigenvalues of $B$ are $0.8620$ and $0.0236$, and the eigenvalues of $J$ are $-0.9401\pm1.8348i$ and $\color{red}{0.0545\pm0.3906i}$.

As to the stability of $$M = \begin{bmatrix}A & B \\ -B^T &0\end{bmatrix}$$ for negative definite $A$, I assume that the $A$ and $B$ here are real matrices. The proof is easy. Suppose $v^T=(x^T,y^T)$ is a unit eigenvector of $M$ corresponding to the eigenvalue $\lambda$. Then \begin{align} \lambda = v^\ast Mv &=(x^\ast,y^\ast)\begin{bmatrix}A & B \\ -B^T &0\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}\\ &=x^\ast Ax + x^\ast By - y^\ast B^Tx\\ &=x^\ast Ax + x^\ast By - (y^\ast B^Tx)^T\\ &=x^\ast Ax + x^\ast By - x^T B\bar{y}\\ &=x^\ast Ax + x^\ast By - \overline{x^\ast By}\\ &=x^\ast Ax + 2\,\mathrm{Imag}(x^\ast By). \end{align} Therefore the real part of $\lambda$ is given by $x^\ast Ax$, which is negative because $A$ is negative definite.

Edit: In the modified question, you ask whether $J=\begin{bmatrix}A & B \\ -cI &0\end{bmatrix}$ is always stable for sufficiently small $c>0$, given that $A=-(B+B^T)$ is negative definite. After some rough calculations, the answer seems to be negative. (OK, I was wrong. For the 2-by-2 case, actually $J$ is stable when $c$ is small.) Consider $A=\mathrm{diag}(-2a,-2b)$ and $B=\begin{pmatrix}a&-w\\w&b\end{pmatrix}$ with $a,b>0$ and $w\in\mathbb{R}$. It can be shown that $J$ is nonsingular and its characteristic equation is given by $\det(x^2 I - xA + cB)=0$, or equivalently, $$f(x) := x^4 + 2(a+b)x^3 + (ca+cb+4ab)x^2 + 4cabx + c^2(ab+w^2) = 0.$$ One can further show that $f$ has no purely imaginary root. So, we may employ Routh–Hurwitz theorem (see also here) to test the stability of $f$. I have done some calculations, but they are too long to fit here (uh, ... I've started to sound like Fermat). The result seems to show that when $w$ is large and $c>0$ is small, the related Cauchy index is always zero (thus $J$ is not stable).

I have also done some computer experiments. The results apparently point to the same conclusion. For instance, when $B=\begin{bmatrix}1 & -320\\ 320&1\end{bmatrix}$, the resulting $J$ always seems to have an eigenvalue with nonnegative real part when $0<c\le\frac12$.

Another edit: I may have messed up something, but it is too tedious to double-check the calculations.

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you are right, I have edited my post with an extra condition. – noether Nov 29 '12 at 22:29
Look at my first edit. For your counter example $BK + K^TB^T$ has eigenvalues -9.07036 and 0.7325. If you put as an extra condition $BK + K^TB^T < 0$, so far, I have not found any counter example. – noether Nov 29 '12 at 22:44
Well, it seems that choosing $K = -3.749I$, it is a counter example too, and it satisfies $BK + K^TB^T < 0$. – noether Nov 29 '12 at 22:56
It seems, that for a $K = -cI$ sufficiently small, then the matrix J is Hurwitz, now what I have to found is the condition on $c$. – noether Nov 29 '12 at 23:01
@noether: See my edit. Sorry for not having any water-tight conclusions, but I have to take care of other things and won't be able to spend further time on this quesiton. So, good luck! – user1551 Nov 30 '12 at 19:06

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