Assume the symmetric matrix:
$$ M= \begin{bmatrix} \frac{\sigma\omega\pi^2}{4L^2} + g & 0 & -\frac{\sigma + c^2Qg}{2\sigma} \\ 0 & a\mu & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right) \\ -\frac{\sigma + c^2Qg}{2\sigma} & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right)& \frac{c^2Q}{\sigma} \end{bmatrix} \ $$ with $\omega \in (0,1)$, $\sigma > 0$, $\mu > 0$.
The objective is to find some values of $Q,a > 0, g \in \mathbb{R}$ that make $M$ positive definite.
Attempt:
Using Sylvester's criterion, if the determinants:
\begin{align} d_1 &= \left( \frac{\sigma\omega\pi^2}{4L^2} + g\right)a\mu \\ d_2 &= \left[ \frac{a\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} + g -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} + g \right) - a\mu \left(\frac{\sigma + c^2Qg}{2\sigma} \right)^2 \end{align}
are positive then $M$ is positive definite.
$d_1$ is easy to make it positive, for $g = 0$ and $a = 1$ for example. But then, I'm having trouble finding a value for $Q$ such that:
$$ d_2 = \left[ \frac{\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} -\mu \right)^2\right]\frac{\sigma\omega\pi^2}{4L^2} - \frac{\mu}{4} > 0 $$
Any thoughts? Is my initial choice of $g,a$ good?
You may take $g=0$ and $a=Q^2$. Then $d_1$ is obviously positive and \begin{align} d_2 &= \left[ \frac{a\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} + g -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} + g \right) - a\mu \left(\frac{\sigma + c^2Qg}{2\sigma} \right)^2\\ &= \left[ \frac{\mu c^2Q^3}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} \right) - Q^2\mu \left(\frac{\sigma}{2\sigma} \right)^2\\ &= \frac{\mu c^2Q^3}{\sigma}\left( \frac{\sigma \omega \pi^2}{4L^2} \right) - Q^2\times\text{constant}\\ \end{align} is positive when $Q>0$ is sufficiently large.