
> 
> 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 
> 
> 
> $$B^\top - T B^\top = B + B T$$
> 
> 
> 1. Prove that $\mbox{tr}(TB) \leq 0$.
> 2. 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](http://guava.physics.uiuc.edu/~nigel/REPRINTS/2017/Biancalani%20NG%20giant%20amplification%20of%20fluctuation-induced%20patterns%20PRL%202017.pdf) 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](https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_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.



For a stable matrix $B$ and anti-symmetric $T$, such that $B(I+T)$ is symmetric, show that $\mbox{tr}(TB)\leq0$
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.


