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 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.

  • 2
    $\begingroup$ I haven't gone through all of the paper but I have the feeling that the authors are just trying to avoid saying 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. $\endgroup$
    – percusse
    Commented Dec 12, 2017 at 3:16
  • $\begingroup$ I am not familiar with $\mathcal H_2$ norm, but a quick google search does suggest that it is related to the mean square norm. Do you know of a concise reference for the definition of $\mathcal H_2$, what is known about it, and the derivation of how it is related to the amplification factor above? $\endgroup$
    – stochastic
    Commented Dec 12, 2017 at 16:24
  • $\begingroup$ Equation (S5) is indeed a Lyapunov equation. I agree with your comment on the symbol $\Xi$. $\endgroup$
    – stochastic
    Commented Dec 12, 2017 at 16:26
  • $\begingroup$ I am as puzzled as you are about how the symmetry constrains $\mbox{tr}(TB)$, but it actually does. Try it for the 2x2 case. $\endgroup$
    – stochastic
    Commented Dec 12, 2017 at 16:29
  • $\begingroup$ Is everyone sure that this is a true statement? If so, this is a beautiful application of a property of a matrix to be stable. $\endgroup$ Commented Dec 12, 2017 at 17:35

1 Answer 1


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.


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