product of skew-symmetric and symmetic matrix: diagonal elements

Question

This is a second attempt, related to my earlier question zero diagonal of product of skew-symmetric and symmetric matrix with strictly positive identical diagonal elements where I think I asked the wrong question. I try to give more background: For a stability analysis of an ordinary differential equation, I encountered an expression

\begin{equation} \Delta = \sum\limits_{j=1}^{n} [(\mathbf{A}\mathbf{B})_{jj}]^2 \end{equation}

where $\mathbf{A}$ is a skew-symmetric matrix, and $\mathbf{B}$ is symmetric but not diagonal (i.e. there are non-zero off-diagonal elements). Both matrices are of size $n \times n$. There are more conditions on $\mathbf{B}$, but I'm not sure whether they are required, so I leave them out.

What I want to show is that for any symmetric, non-diagonal $\mathbf{B}$ I can find a skew-symmetric $\mathbf{A}$ such that $\Delta > 0$.

This should be analogous to the statement that

\begin{equation} \forall \mathbf{B}: \exists \mathbf{A}: \exists j: (\mathbf{A}\mathbf{B})_{jj} \neq 0. \end{equation}

To give more background: In my problem, $\Delta > 0$ indicates that I'm at a minimum or a saddle point which is what I want to show. The matrix $\mathbf{B}$ indirectly characterizes the different critical points. $\mathbf{A}$ parametrizes a small step on a manifold away from the critical point. If, for all $\mathbf{B}$ (for all critical points), I can find some $\mathbf{A}$ (some small step away from the critical point) such that $\Delta > 0$, I can conclude that all critical points are a minimum or a saddle.

So two questions this time:

1. Do I ask the right question for my problem?

2. How can I show the desired property? I have no idea how to approach such an existence statement.

Answer 1

Regarding point 1., yes this is correct.

For 2. fix a $j$ such that there exists at least a $B_{k,j}\neq 0$ with $k\neq j$. This $j$ exist by the hypothesis that $B$ is not diagonal.

Then form the following matrix

$$ A = \sum_n \operatorname{sign}(B_{n,j}) |j\rangle \langle n| - \operatorname{sign}(B_{n,j}) |n\rangle \langle j| $$

Such a matrix is antisymmetric by construction. Its matrix elements are

$$ A_{j,k} = \operatorname{sign}(B_{j,k}) - \operatorname{sign}(B_{j,j}) \delta_{j,k} $$

Now

\begin{align} \left ( A B \right )_{j,j} &= \sum_k A_{j,k} B_{k,j} \\ &= \sum_k \left | B_{k,j} \right | - \left | B_{j,j} \right | \\ &= \sum_{k\neq j} \left | B_{k,j} \right | \end{align}

The latter sum is non-zero (and positive).