Let
$ A= \begin{pmatrix} g^{11} & g^{12} & \cdots & 0& 0 \\ g^{21} & g^{22} & \cdots & 0&0 \\ \vdots & \vdots & \ddots & \vdots &\vdots \\ 0 & 0 & \cdots &1 & 0\\ 0 & 0 & \cdots&0& 1 \end{pmatrix}$\
where $g^{ij}$ are the components of the inverse of a Riemannian metric $g_{ij}$
Given $\Phi=(\phi_{1},\phi_{2}....\phi_{n},\phi)$ where $\phi_{m}=\frac{\partial\phi}{\partial x^{m}}$ is the "energy" defined as:
$$E=\int\Phi^{T}A\Phi$$
Is $E$ a coercive quadratic form (i.e.,$\int\Phi^{T}A\Phi\ge\int\mu\Phi^{T}\Phi$ for $\mu\in\mathbb{R}^{+}$ )?
If not, which are the geometrical conditions on $g_{ij}$ such that this is satisfied?
This question arises from the fact that this is a key condition in symmetric hyperbolic systems in order to prove energy inequalities. Also, I am aware that second order linear wave equations can be rewritten as first order systems.
However, it is not clear to me that this "energy" is coercive.
A Riemannian metric (or its "inverse" metric) is by definition a positive definite quadratic form. In finite dimensions this immediately guarantees that the eigenvalues have a lower bound and that the form is coercive. There is nothing to prove here.