This is a math question, but is based on a physics topic I'm trying to develop an intuition for. Let me know if this question would be better placed in physics.
What I know: The "characteristic matrix" in thin film optics relates the E and H field amplitudes before and after passing through a thin film of material. For a single layer on a substrate, the equation is:
$$ \begin{pmatrix} E_b \\ H_b \end{pmatrix} = \begin{pmatrix} \cos(\delta) & i(1/y_1)\sin(\delta) \\ iy_1 \sin(\delta) & \cos(\delta) \\ \end{pmatrix} \begin{pmatrix} E_a \\ H_a \end{pmatrix} $$
where $E_a,H_a$ are the field amplitudes at the film/substrate interface, $E_b,H_b$ are the field amplitudes at the air/film interface, $\delta$ = the phase thickness of the material, $y_1$ = characteristic admittance of the thin film. It can also be written in terms of the substrate admittance by normalizing to $E_a$:
$$ \begin{pmatrix} B \\ C \end{pmatrix} = \begin{pmatrix} \cos(\delta) & i(1/y_1)\sin(\delta) \\ iy_1 \sin(\delta) & \cos(\delta) \\ \end{pmatrix} \begin{pmatrix} 1 \\ y_{sub} \end{pmatrix} $$
I was playing around with the matrix, when I realized: if you view $M$ as a linear transformation of $\mathbf{C}$, it corresponds to a pure rotation and scaling of the basis vectors (1,0) and (0,$ i$). In other words, for vector $\mathbf{v}$ in the complex plane, $M\mathbf{v}$ would rotate $\mathbf{v}$ by $\delta$ and then it's components would both be scaled by $y_1$ and $(1/y_1)$. See below for how the basis vectors rotate while staying normal to one another for $\delta = \frac{\pi}{2}$:
$$ \begin{pmatrix} 0 \\ iy_1 \end{pmatrix} = \begin{pmatrix} 0 & i(1/y_1) \\ iy_1 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$
$$ \begin{pmatrix} -(1/y_1) \\ 0 \end{pmatrix} = \begin{pmatrix} 0 & i(1/y_1) \\ iy_1 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 \\ i \end{pmatrix} $$
I thought this was a really interesting geometric interpretation of what this matrix was doing, until I realized that the vectors $(E_a, H_a)$ or $(1, Y_{sub})$ have complex components themselves and therefore don't exist in $\mathbf{C}$, but actually in $\mathbf{C}^2$ or $\mathbf{R}^4$.
What I don't know: So my questions now are, what does a linear transformation look like in $\mathbf{C}^2$? Does my pure rotation + scaling idea still make sense for this matrix? I'm not sure I even understand what $\mathbf{C}^2$ looks like much less how it would be transformed by a complex matrix...
$\mathbf{C}^2$ is a bit of a mirage. You may change your basis to $$ \begin{pmatrix} E_b \\ H_b \end{pmatrix} = \begin{pmatrix} \cos(\delta) & i(1/y_1)\sin(\delta) \\ iy_1 \sin(\delta) & \cos(\delta) \\ \end{pmatrix} \begin{pmatrix} E_a \\ H_a \end{pmatrix} \qquad \mapsto \qquad \\ \begin{pmatrix} E_b \\ iH_b \end{pmatrix} = \begin{pmatrix} \cos(\delta) & (1/y_1)\sin(\delta) \\ -y_1 \sin(\delta) & \cos(\delta) \\ \end{pmatrix} \begin{pmatrix} E_a \\ iH_a \end{pmatrix}, $$ so the transition matrix is , in fact, a real rotation and scaling, unimodular, $$ \begin{pmatrix} 1/\sqrt{ y_1 } & 0 \\ 0 & \sqrt{y_1} \\ \end{pmatrix} \begin{pmatrix} \cos(\delta) & \sin(\delta) \\ - \sin(\delta) & \cos(\delta) \\ \end{pmatrix} \begin{pmatrix} \sqrt{y_1 } & 0 \\ 0 & 1/ \sqrt{y_1 } \\ \end{pmatrix} . $$