If a linear system of two PDE’s $${\bf{A}}{{\bf{x}}_{{u_2}}} + {\bf{B}}{{\bf{x}}_{{u_1}}} = {\bf{0}}$$ (where ${\mathbf{x}} = {({x_1},{x_2})^T} \in {\mathbb{R}^2}$, and ${\mathbf{A}}$ and ${\mathbf{B}}$ are real 2-by-2 matrices that depend only on ${\mathbf{u}} = ({u_1},{u_2})$, real) has characteristics $\frac{{d{u_1}}}{{d{u_2}}} = {\lambda _i}({\mathbf{u}})$ that are functions of ${\mathbf{u}} = ({u_1},{u_2})$ alone, why does that imply that there exist functions ${v_i}({\mathbf{u}})$ that are constant along these characteristics (Riemann invariants)?
Looking at the directional derivative of the ${v_i}$’s along the characteristics: $\left( {\frac{\partial }{{\partial {u_2}}} + {\lambda _i}({u_1},{u_2})\frac{\partial }{{\partial {u_1}}}} \right){v_i}({u_1},{u_2}) = \frac{{\partial {v_i}}}{{\partial {u_2}}} + {\lambda _i}({u_1},{u_2})\frac{{\partial {v_i}}}{{\partial {u_1}}}$ , I don’t see why ${\lambda _i}$ must be a function of ${\mathbf{u}} = ({u_1},{u_2})$ alone for this to be zero for some ${v_i}({\mathbf{u}})$.
I also don't see how this system can have characteristics unless we know that ${\mathbf{A}}$ is invertible so that the system can be written ${{\mathbf{x}}_{{u_2}}} + {{\mathbf{A}}^{ - 1}}{\mathbf{B}}{{\mathbf{x}}_{{u_1}}} = {\mathbf{0}}$ with the characteristic velocities being defined as the eigenvalues of ${{\mathbf{A}}^{ - 1}}{\mathbf{B}}$, from the usual definition of characteristics of a system of the form ${{\mathbf{u}}_t} + {\mathbf{C}}{{\mathbf{u}}_x} = {\mathbf{f}}$, where the characteristic velocities are given by the eigenvalues of ${\mathbf{C}}$.