$ \newcommand\tr{\text{tr}} \newcommand\mean[1]{\left\langle{#1}\right\rangle}$ Let $ B\in\mathbb R^{n\times n}$ with its eigenvalues restricted to the left half plane, and anti-symmetric matrix $ T\in\mathbb R^{n\times n}$ satisfy $ $ B^\top-TB^\top=B+BT.$ $
Prove that $ \tr(TB)\leq0$ .
What are necessary and sufficient conditions on $ B$ such that $ \tr(TB)=0$ . (e.g. it is sufficient for $ B$ to be symmetric. Is that also a necessary condition?)
Note that for each $ B$ , there is at most one $ T$ satisfying this condition.
Motivation: There is a claim in this paper that nonorthogonality of the eigenvectors of linear stability operator of a stochastic dynamical system amplifies the effect of noise, which is proven for the $ 2\times2$ case, and stated for the general case. This claim can be reduced to the above statement. Here is how it goes:
Consider a linear stochastic dynamics described by $ $ d x = Ax\, dt+\sigma dW,$ $ where $ \sigma>0$ , $ t\in\mathbb R^+$ , $ x(t)\in\mathbb R^n$ , $ A\in\mathbb R^{n\times n}$ with eigenvalues in left half plane, and $ W$ is the $ n$ -dimensional Wiener process. This is an $ n$ -dimensional Ornstein-Uhlenbeck process.
If $ A$ is symmetric, the distribution of $ x$ at long time approaches a multivariate normal distribution with its covariance given by $ A^{-1}$ , and $ $ \mean{||x||^2} = -\frac12\sigma^2\tr(A^{-1}).$ $
When $ A$ is not symmetric, the covariance can be written as the inverse of a symmetric matrix $ GA$ , where $ $ \frac12(G^{-1}+(G^{-1})^\top) = I_{n\times n}.$ $ This relationship along with the symmetry of $ GA$ uniquely defines $ G$ . In this case $ $ \mean{||x||^2} = -\frac12\sigma^2\tr(G^{-1}A^{-1}).$ $
The ratio of mean squared norm of $ x$ to its value for a symmetric matrix with the same eigenvalues is what is called the amplification $ $ \mathcal H=\frac{\tr(G^{-1}A^{-1})}{\tr(A^{-1})}.$ $ The claim is that $ \mathcal H\geq 1$ .
Let $ B = A^{-1}$ , and $ T$ be the anti-symmetric part of $ G^{-1}$ . Now, $ GA$ is symmetric iff $ B^\top-TB^\top=B+BT$ , and $ \mathcal H\geq 1$ iff $ \tr(TB)\leq 0$ .
