The wave equation in space, is it a first-order hyperbolic linear system? A first order linear system is any PDE of the form
$$q_t(x, t) + A q_x(x, t) = 0$$
Where $q(x,t):\mathbb{R}\times\mathbb{R} \to \mathbb{R}^n$ is a vector function, and $A$ is an $n\times n$ matrix satisfying the hyperbolicity condition, i.e. diagonalizable with real eigenvalues. 
$q_t, q_x$ designate the derivatives with respect to time $t$ and space coordinate $x$, respectively.
We can easily see that the second order -classical- wave equation in one dimension
$$u_{tt}= c^2u_{xx}$$
can be written as a first order linear system of two equations, simply by taking $v_t=cu_x\ ,\ v_x= \frac{1}{c}u_t$ ;
\begin{align*}
&v_{tx}= cu_{xx} \\
&v_{tx}= \frac{1}{c}u_{tt}
\end{align*}
To obtain $u_{tt}= c^2u_{xx}$ , so here the first order system is
\begin{align*}
&v_t- cu_x= 0 \\
&u_t- cv_x= 0
\end{align*}
Meaning we take $q= (u,v):\ \mathbb{R}\times\mathbb{R} \to \mathbb{R}^2$ and $A= \Big(\begin{matrix} 0 & -c \\
-c & 0 
\end{matrix}\Big)$
Now for the $3$-dimensional wave equation :
$$u_{tt}= c^2\Delta u= c^2(u_{xx}+u_{yy}+u_{zz})$$
My question is : Can we turn this equation into a first order linear system or no ?
 A: Maybe let's try first the 2D case: $u(x,y,t)$ and $u_{tt} = c^2\Delta (u_{xx} + u_{yy})$.
Then define $v^{(x)}_t = cu_x, v^{(y)}_t = cu_y$. For the spatial derivatives of $\boldsymbol v = \begin{pmatrix} v^{(x)} \\ v^{(y)} \end{pmatrix}$ require that $v^{(x)}_x + v^{(y)}_y = \frac 1c u_t$. Then for continuously differentiable $v^{(x)}, v^{(y)}$:
\begin{align} \frac 1c u_{tt} &= \partial_t \Big(v^{(x)}_x + v^{(y)}_y \Big) \\
&= v^{(x)}_{xt} + v^{(y)}_{yt} \\
&= v^{(x)}_{tx} + v^{(y)}_{ty} \\
&= cu_{xx} + c u_{yy}\end{align}
so that the original PDE (Wave equation) remains preserved.
So the first order system reads
$$ \partial_t\begin{pmatrix} v^{(x)} \\v^{(y)} \\ u \end{pmatrix} + \nabla \cdot \begin{pmatrix} -cu & 0 \\ 0 & -cu \\ -cv^{(x)} & -cv^{(y)} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} $$
where the $\nabla \cdot$ acts row-wise.
This is the general form of a conservation law in multiple dimensions in divergence form:
$$ \partial_t \boldsymbol u + \nabla \cdot \boldsymbol f(\boldsymbol u) = \boldsymbol{0}.$$
I am not sure how you would write a linear system in divergence form, since for a row-wise acting divergence $\nabla \cdot$ you need that $A \boldsymbol u \in \mathbb R^{m\times d}$, with $m$ the number of variables $| \boldsymbol u |$ and $d$ the spatial dimension. While you can ensure that $A$ has $m$ rows, there is no way for $A$ having $d$ columns if you multiply it with the column vector $\boldsymbol u \in \mathbb R^{m \times 1}$.
The extension to 3D is then straightforward, here you have
$$ \partial_t\begin{pmatrix} v^{(x)} \\v^{(y)} \\ v^{(z)} \\ u \end{pmatrix} + \nabla \cdot \begin{pmatrix} -cu & 0 & 0\\ 0 & -cu  &0 \\ 0 & 0 & -cu \\ -cv^{(x)} & -cv^{(y)} & -cv^{(z)} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0\end{pmatrix} $$
