Showing that $u_{tt} - \Delta u = 0$ is a hyperbolic PDE for $u: \mathbb{R}^{n} \times (0,\infty) \to \mathbb{R}$ A second order PDE can be expressed in the form
$$ F(D^{2}u , Du ,u , x) = 0.$$
if $F$ is linear in $D^{2}u$ then we can express
$$ F(D^{2}u , Du , u ,x) = L [u] + G(Du, u ,x),\tag{1}$$
with
$$ L[u] := -\text{tr}(A(x)D^{2}u),$$
for which $A(x)$ is a symmetric $n\times n $ matrix.
Problem statement
Let $u:\mathbb{R}^{n} \to \mathbb{R}$. Show that
$$ u_{tt}-\Delta u =0$$
is an hyperbolic PDE,
using the fact that a PDE is hyperbolic if for every $x, A(x) = A_{ij}(x)$ has nonzero eigenvalues ,and all but one have the same sign.
$\textbf{Hint}$ : Consider writing this operator as $-\text{tr}(A(x,t)D^{2}_{x,t} u) =0.$
Attempt at a solution
In the given problem statement $u=u(\vec{x},t)$ with $\vec{x} = (x_{1} , \dots , x_{n})$, therefore,
\begin{align} -\Delta u &= -(u_{x_{1}^{2}} + u_{x_{2}^{2}} + \dots + u_{x_{n}^{2}} + u_{tt}) \\
&= -\text{tr}(D^{2}_{x,t}u)
\end{align}
Therefore, following $(1)$,
\begin{align}
&u_{tt} -\text{tr}(D^{2}_{x,t}u) = -\text{tr}(A(x,t)D^{2}_{x,t}u) 
\end{align}
Since $A(x,t)$ is symmetric, I assume $A(x,t) = (a)_{ij} =0 $ for $i \neq j$ such that
\begin{gather}
u_{tt} = u_{tt} + \sum_{i=1}^{n} u_{ii} - a_{tt}u_{tt} - \sum_{i=1}^{n} a_{ii}u_{ii} \\
\implies a_{tt}u_{tt} = \sum_{i=1}^{n} u_{ii} (1-a_{ii})
\end{gather}
Now since the determinant of a symmetric matrix is the product of its eigenvalues and $a_{tt}\neq 0$ since we can't divide by $0$ , this implies $A(x)$ has nonzero eigenvalues.
Here I'm stuck and can't prove that all but one eigenvalues of $A(x)$ have the same sign. Moreover, I'm not sure of my solution.
Any indications or directions would be much appreciated.
 A: EDIT: I understand your confusion here. You're interpreting $\Delta u$ as the Laplacian in both space and time, i.e., $\Delta u = u_{x_1x_1} + \dots + u_{x_nx_n} + u_{tt}$. But in the wave equation (and often elsewhere), the Laplacian is only in the spatial variables, so $\Delta u = u_{x_1x_1} + \dots + u_{x_nx_n}$, without any $t$ derivatives.
Note that $u(x,t) = u(x_1,x_2,\dots,x_n,t)$ is not a function from $\mathbb{R}^n\to\mathbb{R}$ as you have written, but really a function from $\mathbb{R}^{n+1}\to\mathbb{R}$, where the $+1$ superscript indicates the fact that there's a time variable. Then the operator $u_{tt} -\Delta u$ can be written as
$$-\text{tr}(AD_{x,t}^2u)$$
where
$$A = \begin{bmatrix}
1 & & &\\
& \ddots & & &\\
& &1& \\
& & & -1\end{bmatrix},$$
and
$$D^2_{x,t}u=\begin{bmatrix}
u_{x_1x_1} & u_{x_1x_2} & &\\
u_{x_2x_1} & \ddots & &\\
& & u_{x_nx_n} &\\
& & & u_{tt}\end{bmatrix}.$$
The matrix $A$ clearly satisfies the criteria of one eigenvalue having different sign from the rest.
