Consider the following ODE:
$$P_n\frac{d f_n(x)}{d x} = \sum_{n' = 1}^NQ_{n,n'}f_{n'}(x),$$
for $n$ and $n' = 1, 2, ..., N$, and in which $P_n$ and $Q_{n,n'}$ are real constant parameters. If we expand this equation for every possible value of $n$ and $n'$, we would get a $N \times N$ system of ODE's that could be rewritten like:
$$\frac{d}{dx}\begin{bmatrix}f_1(x)\\\vdots\\f_N(x)\end{bmatrix}=\begin{bmatrix}Q_{1,1}/P_1&...&Q_{1,N}/P_1\\\vdots&\ddots&\vdots\\Q_{N,1}/P_N&...&Q_{N,N}/P_N\end{bmatrix} \begin{bmatrix}f_1(x)\\\vdots\\f_N(x)\end{bmatrix}.$$
So, if we call $\textbf{f}\equiv[f_1(x),...,f_N(x)]^T$ and the $N\times N$ matrix above (which we will consider henceforth to be non-singular) as $\textbf{M}$, then we could rewrite it again as:
$$\frac{d\textbf{f}}{dx}=\textbf{M}\textbf{f}.$$
Thus, we can suppose a solution to $\textbf{f}$ of the type:
$$\textbf{f} = \textbf{ξ}e^{\lambda x},$$
in which $\textbf{ξ}$ is the constant eigenvector associated with a particular eigenvalue $\lambda$ of matrix $\textbf{M}$. So, if matrix $\textbf{M}$ has $L$ eigenvalues $\lambda_l$ (so that $l = 1,2,...L$ and $L \le N)$ then the general solution for $f_n(x)$ would be:
$$f_n(x)=\sum_{l=1}^Lc_l ξ_n^{l}e^{\lambda_lx},$$
in which $ξ_n^{l}$ is the $n$-th coordinate of the eigenvector associated with the $l$-th eigenvalue of matrix $\textbf{M}$, and $c_l$ are constants determined by the boundary conditions of the problem, which are of no interest now.
My actual doubt:
Consider now the following PDE:
$$A_n\frac{\partial f_n(x,y,z)}{\partial x} + B_n\frac{\partial f_n(x,y,z)}{\partial y} + C_n\frac{\partial f_n(x,y,z)}{\partial z} = \sum_{n' = 1}^ND_{n,n'}f_{n'}(x,y,z),$$
for $n$ and $n' = 1, 2, ..., N$, and in which $A_n, B_n, C_n$ and $D_{n,n'}$ are real constant parameters. If we expand this equation for every possible value of $n$ and $n'$, we would get a $N \times N$ system of PDE's.
Is it possible to solve this system of PDE's through that same matricial approach that I presented for the ODE's, or through an equivalent one?
Consider solutions of the form $f_n(x,y,z) = u_n \exp(a x + b y + c z)$.
You get a system of linear equations $(A_n a + B_n b + C_n c) u_n - \sum_{n'} D_{n,n'} u_{n'} = 0$. These have nontrivial solutions if and only if $(a,b,c)$ makes the determinant $P(a,b,c)$ of the coefficient matrix zero. A difference between this and the ODE case is that instead of having a finite number of eigenvalues, $P(a,b,c) = 0$ is in general a two-dimensional variety in $\mathbb C^3$.