Negative Definiteness of a real symmetric block matrix Here is a question I have in my mind. Would really appreciate it if anyone can help offer a clue.
Say we have a real symmetric $2n\times 2n$ matrix $M$, with the form of
$$M = \left[ \begin{array}{*{20}{c}}
A & B\\
B^T& C
\end{array} \right],$$
where $A$ and $C$ are symmetric and negative definite, with rank $n.$ Matrix $B$ is known to have zero trace.
Q: Is there anything we can say about the negative definiteness of the matrix $M$?
PS: In the specific problem I am looking at, matrix $C$ is a permutation of $A$. Therefore, you can say $C=P^T A P$, where $P$ is some permutation operation.
Update:
The B matrix has the following format:
$$B = \left[ \begin{array}{*{20}{c}}
0 &... &... &... \\
... & 0 & -\frac{3\sin((m+n)\theta)}{|\sin^3(\frac{m-n}{2}\theta)|^3} &...\\
... & -\frac{3\sin((n+m)\theta)}{|\sin^3(\frac{n-m}{2}\theta)|^3} & 0 &...\\
... &... &... & 0\\
\end{array} 
\right],$$
Basically, the diagonal terms are zero and the off diagonal terms $B_{m,n}=-\frac{3\sin((m+n)\theta)}{|\sin^3(\frac{m-n}{2}\theta)|^3}$, and $\theta=2\pi/N$.
The A matrix has the following format:
$$A = \left[ \begin{array}{*{20}{c}}
-c &... &... &... \\
... & -c & \frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} &...\\
... & \frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} & -c &...\\
... &... &... & -c\\
\end{array} 
\right],$$
where c is a positive constant value and the off diagonal terms $A_{m,n}=\frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} $.
 A: There's the counterexample of
\begin{align*}
A = C &=
\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix}\\
B &= 
\begin{bmatrix}
-2 & 0\\
0 & 2
\end{bmatrix}
\end{align*}
which results in an $M$ with eigenvalues $-3$ and $1$ ($[0, 1, 0, 1]^T$ is one of the eigenvectors with positive eigenvalue). More loosely, the eigenvectors of all the matrices can line up, and trace of $B$ being zero still allows for large positive eigenvalues, so for $M$ to be negative definite there has to be some constraint that prevents that from happening.
Update:
Since $\operatorname{Tr}(B)= 0$, and $B$ is not the zero matrix, there always is some eigenvector $v$ of $B$ with positive eigenvalue $\lambda>0$. Then,
\begin{align}
\begin{bmatrix} v^T&v^T\end{bmatrix} M\begin{bmatrix} v\\v\end{bmatrix}
&= v^TAv + v^TBv+v^TB^Tv + v^TCv\\
&= v^TAv + 2\lambda v^Tv + v^TCv
\end{align}
could still be positive if, for example, $v$ is an eigenvector of $A$ and $C$, and the corresponding eigenvalues have absolute value less than $\lambda$.
Propositions 2.1 and 2.2 involving the Schur complement described here might be useful. The fact that $B$ is symmetric simplifies the condition slightly to $A-B C^{-1} B$ being negative definite, but it appears some information about $A$ is still needed to conclude anything useful.
