I am confused about why the following spectrum problem is self-adjoint:
\begin{matrix}y'(x) = \mathbf{M}(k,x)y(x)&(y:\mathbf{R}\rightarrow\mathbf{C}^2,x\in\mathbf{R},k\in\mathbf{C})\end{matrix}
and the operator
$\begin{matrix}\mathbf{M}(k,x) = \begin{bmatrix} -ik & q(x) \\ q(x)^* & ik \end{bmatrix} &(q:\mathbf{R}\rightarrow\mathbf{C})\end{matrix}$
is Hermitian, so that the discrete spectra lie on the real axis. I am a novice on functional analysis. The only explanation I have figured out is ugly: computing the eigensystem of $\mathbf{M}$ gives $\pm \sqrt{-\operatorname{Re}(k)^2-2i\operatorname{Re}(k)\operatorname{Im}(k)+\operatorname{Im}(k)^2+|q(x)|^2}$ as its eigenvalues. To make $\mathbf{M}(k,x)$ singular, it is necessary to set $\operatorname{Im}(k)$ zero. But this is irrelevant to the Hermitianess of $\mathbf{M}(k,x)$, which seems to me that
$\mathbf{M}(k,x)^* = \begin{bmatrix} ik^* & q(x)^* \\ q(x) & -ik^* \end{bmatrix} \ne \!\ \mathbf{M}(k,x)^T$
Is there a generalization of this "self-adjoint" problem in dimensions more than 2?
3x in advance!