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.
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.