= Solution
<Lax's equation> in the convention used here is
$$
\boxed{\dot{\mathcal L}=[\mathcal L,\mathcal A]
=\mathcal L\mathcal A-\mathcal A\mathcal L.}
$$
Let $\mathcal L\psi=\lambda\psi$ with $(\psi,\psi)=1$. Since $\mathcal L$ is <self-adjoint> and $\mathcal A$ is <anti-self-adjoint>, differentiating $\lambda=(\psi,\mathcal L\psi)$ gives
$$
\dot\lambda
=(\dot\psi,\mathcal L\psi)
+(\psi,\dot{\mathcal L}\psi)
+(\psi,\mathcal L\dot\psi).
$$
The first and third terms equal $\lambda,d(\psi,\psi)/dt=0$. For the middle term,
$$
(\psi,[\mathcal L,\mathcal A]\psi)
=(\mathcal L\psi,\mathcal A\psi)
-(\psi,\mathcal A\mathcal L\psi)=0.
$$
Thus $\dot\lambda=0$, proving the <Isospectral Lax equation> property: \b[every eigenvalue is independent of $t$].
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