Suppose you wish to solve a system of $n$ PDEs, which can be expressed in vector form as $$ A \boldsymbol{u}_x + B \boldsymbol{u}_y = \boldsymbol{c} \hspace{10mm} (*_1) $$ where $$ A = \begin{pmatrix} a_{11} & \dots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \dots & a_{nn} \end{pmatrix} \hspace{20mm} B = \begin{pmatrix} b_{11} & \dots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \dots & b_{nn} \end{pmatrix} \\ \boldsymbol{c} = \begin{pmatrix} c_1 \\ \vdots \\ c_n \end{pmatrix} \hspace{20mm} \boldsymbol{u} = \begin{pmatrix} u_x^{(1)} \\ \vdots \\ u_x^{(n)} \end{pmatrix} $$ The method for solving such a system is to multiply through $(*_1)$ by $A^{-1}$ (to the left of each term) so that it is in the form $$ \boldsymbol{u}_x + D \boldsymbol{u}_y = \boldsymbol{d} \hspace{10mm} (*_2) $$ where $D = A^{-1} B$ and $\boldsymbol{d} = A^{-1} \boldsymbol{c}$.
We then diagonalize the matrix $D$, expressing it in the form $$ \Lambda := P^{-1} D P \hspace{10mm} (*_3) $$ where $P$ is the matrix who's columns are the Eigenvectors of the matrix $D$, and define the transformation $$ \boldsymbol{u} = P \boldsymbol{v} \hspace{10mm} (*_4) $$
Then, through substituting in $(*_3)$ and $(*_4)$, the equation $(*_2)$ can be expressed in the form $$ \boldsymbol{v}_x + \Lambda \boldsymbol{v}_y = \boldsymbol{q} $$ where $$ \boldsymbol{q} = P^{-1} \boldsymbol{d} - P^{-1} P_x \boldsymbol{v} - P^{-1} D P_y \boldsymbol{v} $$ (I believe that this is known at the canonical form of the system of equations)
This is where I no longer know how to continue with this method. From this point, where we have a new system of equations in canonical form, how do we obtain a solution the the system of PDEs?