Consider the linear system $\dot x = Ax$ with $x \in {\mathbb C}^n$ and $A \in \mathcal{M}_{n}({\mathbb C})$ skew-Hermitian. Suppose $x$ is partitioned as $$x = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]$$ with $x_1 \in {\mathbb C}^{n_1}$, $x_2 \in {\mathbb C}^{n_2}$, and $n_1 + n_2 = n$. We can then write $$\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_ \times }}\\ { - A_ \times ^\dagger }&{{A_{22}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]$$ or equivalently $$\begin{array}{*{20}{c}} {{{\dot x}_1} = {A_{11}}{x_1} + {A_ \times }{x_2}}\\ {{{\dot x}_2} = {A_{22}}{x_2} - A_ \times ^\dagger {x_1}} \end{array}$$ where $A_{11}$ and $A_{22}$ are skew-Hermitian and $A_{\times}$ has no special properties.
It is well known that the solution to this set of equations has the property that $\left\|x(t)\right\|$ is constant for all $t$. But suppose that we swap out $A_{22}$—which has the diagonal form ${A_{22}} = {U^\dagger }\left( {iD} \right)U$ where $U$ is unitary and $D$ is real diagonal—with $${A_{22}'} = {U^\dagger } D_2 U$$ where $D_2$ is diagonal and has all elements with strictly negative real part (making $A_{22}'$ normal and Hurwitz stable).
In this amended system, if we were to decouple the two equations by setting $A_{\times} = 0$, then clearly $\left\|x_1(t)\right\|$ remains constant and $\frac{d}{dt}\left\|x_2(t)\right\| \leq 0$, with equality only at the origin. That is, if we treat the norm as a measure of the energy in the system, the energy in the $x_1$ subsystem remains constant, and the energy in the $x_2$ subsystem bleeds away to zero regardless of $x_2$.
If we were to set $A_{11} = A_{22} = 0$, then the system would behave like a harmonic oscillator, losslessly cycling energy between the $x_1$ and $x_2$ subsystems.
In the general case, we can regard ${{\dot x}_1} = {{A_{11}}{x_1} + {A_ \times }{x_2}}$ as the $x_1$ subsystem losslessly circulating its own energy while transferring some to/from the $x_2$ subsystem, and ${{\dot x}_2} = {A_{22}'}{x_2} - A_ \times ^\dagger {x_1}$ as the $x_2$ subsystem lossily circulating its own energy while transferring energy to/from the $x_1$ subsystem.
Based on this intuition, it is my conjecture that $\frac{d}{dt}\left\|x(t)\right\| = 0\;\forall\,t$ if and only if both of the following conditions are met:
$x_1(t)$ lies in the null space of $A_{\times}^\dagger$ for all $t$
$x_2(t) = 0\;\forall\,t$
That is, because the $x_2$ subsystem is constantly losing energy, if coupling between the subsystems occurs (i.e. any energy is transferred from the $x_1$ subsystem to the $x_2$ subsystem) (1), or if any energy is present in the $x_2$ subsystem at any time (2), the overall system must strictly lose energy at some point in time. We note that (1) implies (2) if $x_2(0) = 0$.
However, I've thus far been unable to prove this intuition mathematically. I'm at a loss for what approach to take in tackling it.
Any advice, assistance, counterexamples, etc. would be greatly appreciated.
I ultimately figured this out after a bit more thought.
Consider that $\left\| x\right\|^2$ and $\left\| x\right\|$ are monotonically related and that $\left\| x\right\|^2 = \left\| x_1\right\|^2 + \left\| x_2\right\|^2$. Then $$\begin{array}{*{20}{l}} {{\textstyle{d \over {dt}}}{{\left\| x \right\|}^2}}& = &{{\textstyle{d \over {dt}}}{{\left\| {{x_1}} \right\|}^2} + {\textstyle{d \over {dt}}}{{\left\| {{x_2}} \right\|}^2}}\\ {}& = &{\dot x_1^\dagger {x_1} + x_1^\dagger {{\dot x}_1} + \dot x_2^\dagger {x_2} + x_2^\dagger {{\dot x}_2}}\\ {}& = &{{{\left( {{A_{11}}{x_1} + {A_ \times }{x_2}} \right)}^\dagger }{x_1} + x_1^\dagger \left( {{A_{11}}{x_1} + {A_ \times }{x_2}} \right) + {{\left( {{A_{22}'}{x_2} - A_ \times ^\dagger {x_1}} \right)}^\dagger }{x_2} + x_2^\dagger \left( {{A_{22}'}{x_2} - A_ \times ^\dagger {x_1}} \right)}\\ {}& = &{x_1^\dagger A_{11}^\dagger {x_1} + x_2^\dagger A_ \times ^\dagger {x_1} + x_1^\dagger {A_{11}}{x_1} + x_1^\dagger {A_ \times }{x_2} + x_2^\dagger A_{22}'^\dagger {x_2} - x_1^\dagger {A_ \times }{x_2} + x_2^\dagger {A_{22}'}{x_2} - x_2^\dagger A_ \times ^\dagger {x_1}}\\ {}& = &{x_1^\dagger \left( {A_{11}^\dagger + {A_{11}}} \right){x_1} + x_2^\dagger \left( {A_{22}^\dagger + {A_{22}'}} \right){x_2}}\\ {}& = &{x_2^\dagger \left( {A_{22}'^\dagger + {A_{22}'}} \right){x_2}}\\ {}& \le &0 \end{array}$$ with equality holding only if $x_2 = 0$.
The penultimate step relies on the anti-Hermiticity of $A_{11}$. The inequality follows from the fact that ${A_{22}'^\dagger + {A_{22}'}}$ is negative definite.
Thus ${{\textstyle{d \over {dt}}}{{\left\| x \right\|}}} = 0 \;\implies\; {{\textstyle{d \over {dt}}}{{\left\| x \right\|}^2}} = 0 \;\implies\; x_2 \equiv 0$ and the second equation reduces to $${{\dot x}_2} = - A_ \times ^\dagger {x_1}$$ But $x_2 \equiv 0 \;\implies\; \dot x_2 = 0$, hence $A_ \times ^\dagger {x_1} = 0$, and $x_1$ must lie in the null space of $A_\times^\dagger$ for all time.