Non-linear waves and solitons: verifying solution for Nonlinear Schrödinger Equation 
If the initial solution is the unstable background $(2),$ the corresponding fundamental solution of the Lax pair is \begin{align*}\Psi_0(\lambda)&=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\,e^{it\sigma_3}\left[\begin{matrix}e^{\Theta(\lambda)} &-(\mu+\lambda)e^{-\Theta(\lambda)}\\(\mu+\lambda)e^{\Theta(\lambda)}&e^{-\Theta(\lambda)}\end{matrix}\right],\qquad (36)\\  \Theta(\lambda)&\equiv i\mu(x+2\lambda t), \end{align*} where $\sigma_3=\operatorname{diag}(1, -1)$ is the Pauli matrix and $\mu, \lambda$ are complex parameters satisfying the constraint $$\mu^2=1+\lambda^2.\qquad (37)$$ EsRW 05. Verify that $(36)$ is the fundamental solution of $(4)$ corresponding to $(2),$ and satisfying $\det(\Psi_0)=1.$

I'm verifying the Nonlinear Schrödinger equation using the Zakharov-Shabat (ZS) Lax pair condition. I've almost finished but I can't solve this last thing.
The last equation has three matrices with the same determinant, so I thought to diagonalize them in order to find the solution, which is $\mu^2=1+\lambda^2.$
Could you help me to find that please?
Ok, I'm going to post the entire exercise, I hope I am clear this time! Thank for your help!


 A: You have a lot of moving pieces, here.


*

*Construct $U$ and $V$ with the given expressions, and $u_0(x,t)=e^{2it}.$ Note that $u_x=0$ and $\overline{u}_x=0.$
\begin{align*}
U&=\left[\begin{matrix}0 &u\\ \overline{u} &0\end{matrix}\right]=\left[\begin{matrix}0 &e^{2it}\\ e^{-2it} &0\end{matrix}\right] \\
V&=\left[\begin{matrix}|u|^2 &iu_x \\ -i\,\overline{u}_x &-|u|^2\end{matrix}\right]
=\left[\begin{matrix}1 &0 \\ 0 &-1\end{matrix}\right].
\end{align*}

*Next we build the AKNS operators $X$ and $T$ as follows:
\begin{align*}
X&=-i\lambda\sigma_3+iU=-i\lambda\left[\begin{matrix}1 &0\\0&-1\end{matrix}\right]+i\left[\begin{matrix}0 &e^{2it}\\ e^{-2it} &0\end{matrix}\right]=\left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\end{matrix}\right] \\
T&=2\lambda X+iV =2\lambda \left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\end{matrix}\right]+i\left[\begin{matrix}1 &0 \\ 0 &-1\end{matrix}\right]
=\left[\begin{matrix}i-2\lambda^2i &2\lambda i e^{2it} \\ 2\lambda i e^{-2it} &-i+2\lambda^2 i\end{matrix}\right].
\end{align*}

*Moving on, we calculate $\Psi_0;$ but it has an annoying term $e^{it\sigma_3}$ in it. That is, we have to exponentiate a matrix. Fortunately, we do not have to go to the trouble of finding the eigenvalues and eigenvectors and exponentiating the result. Mathematica has the MatrixExp function that will do this for us. We get
$$e^{it\sigma_3}=\left[\begin{matrix}e^{it} &0 \\ 0 &e^{-it}\end{matrix}\right]. $$
Hence, 
\begin{align*}
\Psi_0&=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}e^{it} &0 \\ 0 &e^{-it}\end{matrix}\right]\left[\begin{matrix}e^{i\mu(x+2\lambda t)} &-(\mu+\lambda)e^{-i\mu(x+2\lambda t)}\\(\mu+\lambda)e^{i\mu(x+2\lambda t)}&e^{-i\mu(x+2\lambda t)}\end{matrix}\right] \\
&=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}&e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right].
\end{align*}

*The only missing pieces to $(4)$ are $\Psi_x$ and $\Psi_t,$ which for our case, of course, are really $\dfrac{\partial}{\partial x}\Psi_0$ and $\dfrac{\partial}{\partial t}\Psi_0,$ respectively. I get
\begin{align*}
\dfrac{\partial}{\partial x}\Psi_0&=
\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[
\begin{matrix}
i\mu e^{i[\mu(x+2\lambda t)+t]} 
&-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]}
&-i\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}\right] \quad\text{and} \\
\dfrac{\partial}{\partial t}\Psi_0&=
\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[
\begin{matrix}
 (i(2\mu\lambda+1))e^{i[\mu(x+2\lambda t)+t]} 
&(\mu+\lambda)(i(2\mu\lambda-1))e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)(i(2\mu\lambda-1))e^{i[\mu(x+2\lambda t)-t]}
& -i(2\mu\lambda+1)e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right].
\end{align*}
To see that $\Psi_0$ satisfies 
\begin{align*}
\Psi_x&=X\Psi \\
\Psi_t&=T\Psi
\end{align*}
should now be a matter of algebra.

*Speaking of algebra, we can note that if $\Psi\mapsto\alpha\Psi,$ whether $\Psi$ is a solution or not remains unchanged, because $\partial_x(\alpha\Psi)=\alpha\,\partial_x\Psi,$ and $X(\alpha\Psi)=\alpha X\Psi$, and similarly for the $t$ equation. Because M.SE has a rather narrow window for typing up equations, this has a practical value in that we can ignore constants out in front of $\Psi$. That is, we need to check that 
\begin{align*}
&\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}
i\mu e^{i[\mu(x+2\lambda t)+t]} 
&-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]}
&-i\mu e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right] \\
=&\left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\end{matrix}\right]\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}&e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right],
\end{align*}
or
\begin{align*}
&\left[
\begin{matrix}
i\mu e^{i[\mu(x+2\lambda t)+t]} 
&-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]}
&-i\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
=&\left[
\begin{matrix}
-i\lambda &ie^{2it}\\ 
ie^{-2it} &i\lambda
\end{matrix}
\right]
\left[
\begin{matrix}
e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]} &e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right].
\end{align*}
Performing the matrix multiplication on the RHS yields
\begin{align*}&X\Psi_0= \\ 
&i\left[
\begin{matrix}
-\lambda e^{i[\mu(x+2\lambda t)+t]}+(\mu+\lambda)e^{2it}e^{i[\mu(x+2\lambda t)-t]}
&\lambda(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}+e^{2it}e^{-i[\mu(x+2\lambda t)+t]} \\
e^{-2it}e^{i[\mu(x+2\lambda t)+t]}+\lambda(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}
&-(\mu+\lambda)e^{-2it}e^{-i[\mu(x+2\lambda t)-t]}+\lambda e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right]
\end{align*}
The $e^{2it}$ and $e^{-2it}$ have the effect of enabling the exponentials to be like terms, with the following spectacular simplification:
\begin{align*}&X\Psi_0 \\ 
&=i\left[
\begin{matrix}
-\lambda e^{i[\mu(x+2\lambda t)+t]}+(\mu+\lambda)e^{i[\mu(x+2\lambda t)+t]}
&\lambda(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}+e^{-i[\mu(x+2\lambda t)-t]} \\
e^{i[\mu(x+2\lambda t)-t]}+\lambda(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}
&-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)+t]}+\lambda e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
&=i\left[
\begin{matrix}
\mu e^{i[\mu(x+2\lambda t)+t]}
&[\lambda(\mu+\lambda)+1]e^{-i[\mu(x+2\lambda t)-t]} \\
[\lambda(\mu+\lambda)+1]e^{i[\mu(x+2\lambda t)-t]}
&-\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
&=i\left[
\begin{matrix}
\mu e^{i[\mu(x+2\lambda t)+t]}
&[\lambda\mu+\mu^2]e^{-i[\mu(x+2\lambda t)-t]} \\
[\lambda\mu+\mu^2]e^{i[\mu(x+2\lambda t)-t]}
&-\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
&=i\mu\left[
\begin{matrix}
e^{i[\mu(x+2\lambda t)+t]}
&(\lambda+\mu)e^{-i[\mu(x+2\lambda t)-t]} \\
(\lambda+\mu)e^{i[\mu(x+2\lambda t)-t]}
&-e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
\end{align*}
At this point, a bit of simplification of $\partial_x\Psi_0$ is in order. At last count, we had 
\begin{align*}
&\phantom{=}\partial_x\Psi_0 \\
&=\left[
\begin{matrix}
i\mu e^{i[\mu(x+2\lambda t)+t]} 
&-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]}
&-i\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
&=\left[
\begin{matrix}
i\mu e^{i[\mu(x+2\lambda t)+t]} 
&i\mu(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\
i\mu(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}
&-i\mu e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right] \\
&=i\mu\left[
\begin{matrix}
 e^{i[\mu(x+2\lambda t)+t]} 
&(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\
(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}
&-e^{-i[\mu(x+2\lambda t)+t]}
\end{matrix}
\right].
\end{align*}
Now you can see that these are, indeed, equal. The $\partial_t \Psi_0=T\Psi_0$ case should work out similarly.

