Eigenvalue relation with conjugated variable

Question

I am currently facing a problem where I would like to find, for a given complex square matrix $H$, the vectors $x$ and values $\lambda$ satisfying the "eigen-relationship" like

$$Hx=\lambda \overline{x}$$

Does this equation have a name ? Is there a theory on this ? Can it be linked to a more traditionnal problem in some way ?

For the background, the matrix $H$ models the mode coupling for electromagnetic wave scattering in a given system. A solution to the above equation is a mode combination such that the propagated wave is the same as the incoming one but in the reverse direction (up to a multiplicative factor).

Thank you for any reference or help you may provide.

Answer 1

This is a first step to answering your question. Separate real and imaginary parts. Write $H=A+iB$, $\lambda=\mu+i\nu$, and a solution $z=x+iy$. Then we get (if I haven't made a sloppy error) \begin{align*} Ax-By &= \mu x - \nu y \\ Bx+Ay &= \nu x - \mu y. \end{align*} So this turns into an interesting question of finding values $\mu$ and $\nu$ so that the matrix $$\left[\begin{array}c A-\mu I & -B+\nu I \\ B-\nu I & A+\mu I \end{array}\right]$$ is singular. (But computing the determinant of this matrix will be difficult, unless $A$ and $B$ happen to commute.) The nullspace of that matrix will then give you a nontrivial solution of your original problem.

Answer 2

I post this as a comment to Ted's post, rather than just posting the final result as a comment so that I can show all my work in case I've made a mistake. From what I have below, it seems that the upper right should be $-B-I\nu$ instead of $-B+I\nu$. $$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}} \overbrace{(A+iB)}^H\overbrace{\,(x+iy)\,}^z=\overbrace{\,(\mu+i\nu)\,}^\lambda\overbrace{\,(x-iy)\,}^{\bar z}\tag1 $$ $$ \overbrace{(Ax-By)}^{\Re(Hz)}+i\overbrace{(Bx+Ay)}^{\Im(Hz)}=\overbrace{(\mu x+\nu y)}^{\Re(\lambda\bar z)}+i\overbrace{(\nu x-\mu y)}^{\Im(\lambda\bar z)}\tag2 $$ Putting the real part on top and the imaginary part on the bottom, we get $$ \begin{bmatrix}A&-B\\B&A\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\mu&\nu\\\nu&-\mu\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\tag3 $$ $$ \begin{bmatrix}A-I\mu&-B-I\nu\\B-I\nu&A+I\mu\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=0\tag4 $$