For $B$ with negative eigenvalues and anti-symmetric $T$, such that $B^\top-TB^\top=B+BT$, show that $tr(TB)\leq0$ Let stable real matrix (i.e., its eigenvalues have negative real parts) $B$ and real anti-symmetric matrix $T$ satisfy
$$B^\top - T B^\top = B + B T.$$


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*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?)
I had previously posted this in mathoverflow here, which did not get any responses, but now I think it may be a more appropriate question for MSE.
For more details about the motivation of the question, see the original post.
 A: Suggestion The case of a $2\times 2$ is easy to solve: starting with a general $B$, the condition that $T$ is antisymetric and $B+BT$ is symetric gives a single possible matrix $T$.
In general, the symetry of $B+BT$ gives $\frac{n(n-1)}{2}$ equations while the antisymetry of $T$ gives $\frac{n(n-1)}{2}$ unknowns for the coefficients of $T$. It is likely that in many cases, this system of equations defines a unique matrix $T$ depending on $B$.
I suggest to solve completely the system in the case of a general $3\times 3$ matrix $B$. The system seems to have a nice structure that could be generalized to higher dimensions.
A: (Too long for a comment.) In the first part of alqp's answer on MO, rather than proving the existence and uniqueness of $T$ using a dimension argument, we can actually solve for $T$ directly. Note that we may rewrite the equation as a Sylvester equation  $BT+TB^\top=B^\top-B$. In vectorised form, it can be uniquely solved as $\operatorname{vec}(T)=(I\otimes B+B\otimes I)^{-1}\operatorname{vec}(B^\top-B)$. This also makes explicit the fact that $T$ is a continuous function in $B$. Furthermore, by taking transposes on both sides of $BT+TB^\top=B^\top-B$, we see that if $T$ is a solution, so must be $-T^\top$. Thus the skew-symmetry of $T$ also follows from its uniqueness.
The equation implies that $S=B(I+T)$ is symmetric. The most crucial observation in alqp's proof is that $S$ is negative definite. This is established using four facts: (1) the set of all stable matrices is connected (because the line segment joining every stable matrix to $-I$ is a subset of stable matrices), (2) the continuity of $S$ as a function of $B$ (because $S=B(I+T)$ and $T$ varies continuously with $B$), (3) that $S$ is always nonsingular (because $B$ is stable and $I+T$ is necessarily nonsingular), so that their eigenvalue paths never cross zero, and (4) $S=-I$ is negative definite when $B=-I$. This connectedness argument is nothing special; it is the observation that $S$ is negative definite is brilliant.
The final part, while involves no difficult details, is clever too. The idea is to split $BT$ into the sum of $S\big[(I+T)^{-1}\big]^\top T\big[(I+T)^{-1}\big]$ and $S\big[T(I+T)^{-1}\big]^\top \big[T(I+T)^{-1}\big]$. The first summand is the product of a symmetric matrix $S$ and a skew-symmetric matrix; hence its trace is zero. The second summand is the product of a negative definite matrix and a Gram matrix. Hence its trace is non-positive and it is zero if and only if the Gram matrix is zero (meaning that $T=0$ or $B$ is symmetric).
A: This question was answered in mathoverflow:
https://mathoverflow.net/a/288373/116256
First let us check that $T$ exists and is unique. Let $\mathrm{Sym}_n$ be the space of symmetric matrices (with real coefficients), $\mathrm{M}_n$ the space of all matrices and $\mathrm{Alt}_n$ the space of antisymmetric matrices.
Claim: the map $\mathrm{Alt}_n \rightarrow \mathrm{M}_n / \mathrm{Sym}_n, T \mapsto BT$ is injective. Let $T$ be in the kernel, i.e. $BT$ is symmetric. Up to conjugating by an orthogonal matrix, we can assume that $D = BT$ is diagonal, with exactly the first $r$ coefficients non-zero, for a well-defined $0 \leq r \leq n$. Let $C = B^{-1}$. The fact that $CD$ is antisymmetric implies that $C_{i,j} = 0$ for $i>r$ and $j \leq r$, i.e. $C$ is block upper diagonal. If $r >0$, the matrix $C' = (C_{i,j})_{1 \leq i,j \leq r}$ also has the properties that all of its eigenvalues have negative real part. But the fact that $CD$ is antisymmetric implies that $C'$ has vanishing diagonal, contradiction.
Comparing dimensions, the above linear map is an isomorphism, and the preimage of (the class of) $-B$ gives the unique $T \in \mathrm{Alt}_n$ such that $S := B(1+T)$ is symmetric.
The set of $B$ satisfying the assumption is open in $\mathrm{M}_n$ and path-connected ($t \mapsto t-1 + t B$ connects $-1$ to $B$), and the construction of $T$ is continuous, so the invertible symmetric matrix $S$ stays negative definite.
Now write
$$ BT = S(1+T)^{-1} T = S T(1-T)(1-T^2)^{-1} = ST(1-T^2)^{-1} - ST^2(1-T^2)^{-1}. $$
The matrix $T(1-T^2)^{-1}$ is antisymmetric (it is the commuting product of an antisymmetric and a symmetric matrix) so by symmetry of $S$ we have $\operatorname{tr}(ST(1-T^2)^{-1}) = 0$, and $\operatorname{tr}(BT) = \operatorname{tr}(-ST^2(1-T^2)^{-1})$. Now $-S$ is positive definite and $T^2(1-T^2)^{-1}$ is negative semi-definite, so this trace is $\leq 0$ (write $-S$ as the square of a positive definite matrix). We also get that we have equality iff $T = 0$, i.e. iff $B$ is symmetric.
