Let stable matrix (i.e., its eigenvalues have negative real parts) $B \in \mathbb R^{n \times n}$ and anti-symmetric matrix $T \in \mathbb R^{n \times n}$ satisfy

$$B^\top - T B^\top = B + B T$$

Prove that $\mbox{tr}(TB) \leq 0$.

What are necessary and sufficient conditions on $B$ such that $\mbox{tr}(TB) = 0$. (e.g. it is sufficient for $B$ to be symmetric. Is that also a necessary condition?)

Note that for each $B$, there is a unique $T$ 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

$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(A^{-1}).$$

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

$$\left\langle ||x||^2 \right\rangle = -\frac12\sigma^2\mbox{tr}(G^{-1}A^{-1}).$$

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 $$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})}.$$ 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 $\mbox{tr}(TB)\leq 0$: $$ \begin{cases} T = \frac12 (G^{-1}-(G^{-1})^\top)\\ I_{n\times n} = \frac12 (G^{-1}+(G^{-1})^\top) \end{cases}\implies G^{-1} = I_{n\times n}+T\\ (GA)^\top=GA\Leftrightarrow (G^{-1})^\top(A^{-1})^\top=A^{-1}G^{-1}\Leftrightarrow B^\top-TB^\top=B+BT.$$ $$\mathcal H=\frac{\mbox{tr}(G^{-1}A^{-1})}{\mbox{tr}(A^{-1})} = \frac{\mbox{tr}(B+TB)}{\mbox{tr}(B)}= 1+\frac{\mbox{tr}(TB)}{\mbox{tr}(B)}\geq 1\Leftrightarrow \mbox{tr}(TB)\leq 0.$$

I am having difficulty understanding how the equation $B^\top - T B^\top = B + B T$, puts a restriction on the $\mbox{tr}(TB)$, since taking the trace of both side of this equation gives no new information.

we are computing the $\mathcal{H}_2$norm of the system. The equation you wrote is basically saying that $B(I+T)$ symmetric. To switch to $TB$ term we can use $tr(XY) = tr(YX)$ and end up with $tr(B(I+T)) = tr((1+T)B) = tr(B) + tr(TB)$. Then $tr(TB)=tr(BT)$. But then I don't know what the constraint is. However, I refuse to read anything that uses $\Xi$ as a variable. A final nitpick: the equation (S5) is a Lyapunov equation. $\endgroup$3more comments