Let $L:=L(t)$ be a Lax operator, i.e., there exists an operator $P:=P(t)$ such that the pair (L,P) satisfy the Lax equation $$ \frac{dL}{dt}=PL-LP\,.$$
This operator satisfies the isospectral property, namely, there exists a one-parameter family of unitary operator $U(t)$ such that $$ U(t)^{-1}L(t)U(t)$$ is independent of $t$. In general, this property ensure that the eigenvalues of the Lax operator $L$ are invariant in time.
From Wikipedia, the evolution of the eigenfunctions $f_n$ of the Lax operator is obtained by $$ \partial_t f_n(t)=P f_n(t)\,.$$ But I don't quite understand why this is true.
In the case where the eigenspaces of $L$ corresponding to the eigenvalues $\lambda_n$ are one-dimensional, I came up with the following proof :
Taking the derivative of $Lf_n=\lambda_n f_n$ with request to time, and using the Lax equation, we find $$ PLf_n-LPf_n+L\partial_tf_n=\lambda_n\partial_t f_n.$$ And since $Lf_n=\lambda_n f_n$ we infer $$ (L-\lambda_n)(Pf_n-\partial_t f_n)=0$$ That means that $Pf_n-\partial_tf_n$ is an eigenfunction of $L$ corresponding to the eigenvalue $\lambda_n$. If the eigenspace of $\lambda_n$ is one-dimensional, then $$ Pf_n-\partial_t f_n=\alpha f_n$$ And $P$ can “absorb” $\alpha \mathrm{Id} \,$, since $P-\alpha \mathrm{Id}$ still commute with $L\,$, so that the Lax equation does not change.
But, I am struggling to prove the general case where the eigenspaces are not supposed to be of dimension 1...
I think the best answer is to read the first few pages of the original paper by Peter Lax, Integrals of nonlinear equations of evolution and solitary waves. It is readily available online.