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For $x^{\prime }=Lx$ and $E=\Vert x{\Vert }_2^2$, the Hermitian matrix $H=(L+L^{\ast })/2$ gives $\dot{E}=2x^{\ast }Hx$. Thus some initial condition grows immediately iff the largest eigenvalue of $H$ is positive. All trajectories have nonincreasing energy iff $H$ is a negative semidefinite matrix. This differs from requiring negative real parts for the eigenvalues of $L$: a non-normal matrix can satisfy modal decay while allowing transient growth.

Differentiating $f(\mu ,\sigma )=\mathbb{E}[U(\mu +\sigma Z)]$ twice gives the Hessian matrixFor every vector $(a,b)$,Thus the Hessian is a negative semidefinite matrix, andEquivalently, with $W=-U^{\prime \prime }(X)\ge 0$, its determinant isby the Cauchy-Schwarz inequality, while both diagonal entries are nonpositive.

Differentiate the quadratic energy:For $x_1\ne 0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when ${\lambda }_2{r}^2+r+{\lambda }_1>0$. With both eigenvalues negative, such directions exist iff $1-4{\lambda }_1{\lambda }_2>0$. Writing $\delta =\sqrt{1-4{\lambda }_1{\lambda }_2}$, the growing initial conditions are the two opposite open wedgesIf $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly whenIndeed, this makes the symmetric part of a matrix $H$ a negative semidefinite matrix, so $\dot{E}\le 0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the instantaneous energy-growth criterion for a linear system.

Since $M^T=-M$, $(M^2{)}^T=(M^T{)}^2=M^2$, so $M^2$ is a real symmetric matrix. Moreover,proving that it is a negative semidefinite matrix.

For odd $n$, determinant invariance under transpose givesso $\det M=0$. Hence $M$ and $M^2$ have a zero eigenvalue. Finally, if $M^{2}w=0$, then $0=w^T{M}^{2}w=-\Vert Mw{\Vert }^2$, so $Mw=0$. Thuswhere equality of these kernels includes the reverse implication, which is immediate.