 Solution

If $Lf=\lambda f$ and $f_t+Af=0$, differentiation and the Lax equation give $(L-\lambda )f_t=-A(L-\lambda )f$, so ${\lambda }_t=0$ after projection onto the normalized eigenfunction for a nondegenerate eigenvalue.

For $\Psi =(f,f_x{)}^T$,The compatibility ${\Psi }_{xt}={\Psi }_{tx}$ is precisely ${\partial }_t{M}_L-{\partial }_x{M}_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes ${\dot{M}}_L=[M_A,M_L]$, and cyclicity of trace gives $d\mathrm{Tr}(M_L^k)/dt=0$.