= Solution
First suppose $W[\boldsymbol\xi]>0$ for every nonzero admissible <displacement>. The conserved <energy> from part (b) then bounds both $T$ and $W$ by $E$. In particular,
$$
\|\dot{\boldsymbol\xi}(t)\|_\rho^2\leq2E,\qquad 2W[\boldsymbol\xi(t)]\leq2E,
$$
where $\|\boldsymbol\xi\|_\rho^2=\langle\boldsymbol\xi,\boldsymbol\xi\rangle$. The positive quadratic form $\|\dot{\boldsymbol\xi}\|_\rho^2+2W[\boldsymbol\xi]$ defines the perturbation <energy> norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\|\boldsymbol\xi(t)\|_\rho\leq\|\boldsymbol\xi(0)\|_\rho+\sqrt{2E}\,t$, excluding exponential <displacement> growth. For any admissible <normal mode> $\boldsymbol\xi=\boldsymbol\zeta e^{-i\omega t}$, multiplication of $-\omega^2\boldsymbol\zeta=F\boldsymbol\zeta$ by $\boldsymbol\zeta^*$ gives directly
$$
\omega^2=\frac{2W[\boldsymbol\zeta]}{\|\boldsymbol\zeta\|_\rho^2}>0,
$$
so all such frequencies are real. No variational principle has been assumed.
To prove instability, suppose some admissible $\boldsymbol\xi_0$ has $W_0=W[\boldsymbol\xi_0]<0$, and put $N_0=\|\boldsymbol\xi_0\|_\rho^2>0$. Choose the initial <velocity>
$$
\dot{\boldsymbol\xi}(0)=\alpha\boldsymbol\xi_0,\qquad\alpha=\sqrt{-2W_0/N_0}>0.
$$
Its <kinetic energy> exactly cancels $W_0$, so $E=0$ for the entire subsequent motion. Write $N(t)=\|\boldsymbol\xi(t)\|_\rho^2$. Since $F$ is self-adjoint,
$$
N'=2\operatorname{Re}\langle\boldsymbol\xi,\dot{\boldsymbol\xi}\rangle,\qquad N''=2\|\dot{\boldsymbol\xi}\|_\rho^2+2\langle\boldsymbol\xi,F\boldsymbol\xi\rangle.
$$
Zero <energy> gives $\langle\boldsymbol\xi,F\boldsymbol\xi\rangle=\|\dot{\boldsymbol\xi}\|_\rho^2$, and hence $N''=4\|\dot{\boldsymbol\xi}\|_\rho^2\geq0$. Initially $N'(0)=2\alpha N_0>0$, so $N$ stays positive and its logarithm is defined. The <Cauchy-Schwarz inequality> now gives
$$
\begin{aligned}
(\ln N)''&=\frac{NN''-(N')^2}{N^2}\\
&=\frac{4\left[N\|\dot{\boldsymbol\xi}\|_\rho^2-(\operatorname{Re}\langle\boldsymbol\xi,\dot{\boldsymbol\xi}\rangle)^2\right]}{N^2}\geq0.
\end{aligned}
$$
Therefore $(\ln N)'(t)\geq(\ln N)'(0)=2\alpha$, and integration proves
$$
\boxed{N(t)\geq N_0e^{2\alpha t},\qquad\|\boldsymbol\xi(t)\|_\rho\geq\|\boldsymbol\xi_0\|_\rho e^{\alpha t}.}
$$
This is an exponentially growing perturbation. Scaling the initial <displacement> and <velocity> down leaves $\alpha$ unchanged, so arbitrarily small initial perturbations display the same linear growth mechanism. It proves that a <negative-energy direction gives exponential growth in a self-adjoint wave equation>, without assuming a discrete spectrum or quoting a trial-function variational theorem.
\b[Positive $W$ on nonzero displacements gives energetic stability; a negative direction gives instability.] For mathematical precision, a uniform bound on the mass-weighted <displacement> norm follows if $W\geq\tfrac12\omega_{\min}^2\|\boldsymbol\xi\|_\rho^2$ for some $\omega_{\min}>0$, giving $\|\boldsymbol\xi\|_\rho^2\leq2E/\omega_{\min}^2$. Strict positivity alone in an infinite-dimensional space need not supply such a uniform <spectral gap>; the standard <magnetohydrodynamic energy principle> asserts energetic stability and absence of growing modes. Cases with zero directions are marginal and require separate consideration.
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