I have a matrix for which I want to get some analytical equations of the eigenvalues. The matrix is given as \begin{align} \mathbf A &= \begin{pmatrix} \epsilon_a & 0 & 0\\ 0 & \epsilon_b & 0 \\ 0 & 0 & \epsilon_c \end{pmatrix} \\ \mathbf B &= \begin{pmatrix} \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \\ \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \\ \Sigma_{31} & \Sigma_{32} & \Sigma_{33} \\ \end{pmatrix}\\ \mathbf M &= \begin{pmatrix} \mathbf A &\mathbf B \\ \mathbf B^{\dagger} & -\mathbf A \end{pmatrix} = \begin{pmatrix} \epsilon_a & 0 & 0 & \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \\ 0 & \epsilon_b & 0 & \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \\ 0 & 0 & \epsilon_c & \Sigma_{31} & \Sigma_{32} & \Sigma_{33} \\ \Sigma_{11}^*& \Sigma_{21}^*& \Sigma_{31}^* & -\epsilon_a & 0 & 0\\ \Sigma_{12}^*& \Sigma_{22}^*& \Sigma_{32}^* & 0 & -\epsilon_b & 0\\ \Sigma_{13}^*& \Sigma_{23}^*& \Sigma_{33}^* & 0 & 0 & -\epsilon_c\\ \end{pmatrix} \end{align} The diagonal elements of the matrix are real elements, other elements are complex. By looking at the dimension one can guess that one need to solve a polynomial of order 6 to get the eigenvalues. So it seems impossible to get an exact answer for the eigenvalues. But there are some symmetries followed by the matrix $\mathbf M$, \begin{align} \mathcal C^{-1} \mathbf M \mathcal C &= -\mathbf M \\ \mathcal C &= K \sigma_x \otimes \mathbb{I}_3 \end{align} $K$ is the complex conjugation operator. I'm guessing This symmetry might give some constraints (like particle hole symmetry) which might lead to eigen values like $$ \{w_1,w_2,w_3,-w_1,-w_2,-w_3\} $$ I have computed the characteristic equation of the matrix but am unable to solve it. I'm not sure how to impose this anti-unitary kind of symmetry.
Is there any way I can get the Eigenvalues for this matrix?
I summarize below the results of the processing of an answer, results that I could prove if the space for an answer allows it. In the end I'll give a numerical example to clarify some important details.
$\boldsymbol{\S}\color{blue}{\textbf{A. Hermiticity and real eigenvalues}}$
From the very beginning the $\,6\times6\,$ matrix
\begin{align} \mathbf M & \boldsymbol{=} \begin{bmatrix} \epsilon_1 & 0 & 0 & \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \vphantom{\dfrac{a}{b}}\\ 0 & \epsilon_2 & 0 & \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \vphantom{\dfrac{a}{b}}\\ 0 & 0 & \epsilon_3 & \Sigma_{31} & \Sigma_{32} & \Sigma_{33} \vphantom{\dfrac{a}{b}}\\ \Sigma_{11}^*& \Sigma_{21}^*& \Sigma_{31}^* & \boldsymbol{-}\epsilon_1 & 0 & 0\vphantom{\dfrac{a}{b}}\\ \Sigma_{12}^*& \Sigma_{22}^*& \Sigma_{32}^* & 0 & \boldsymbol{-}\epsilon_2 & 0\vphantom{\dfrac{a}{b}}\\ \Sigma_{13}^*& \Sigma_{23}^*& \Sigma_{33}^* & 0 & 0 & \boldsymbol{-}\epsilon_3\vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \mathbf A \hphantom{^{\dagger}} & \hphantom{\boldsymbol{-}} \mathbf B \vphantom{\dfrac{a}{b}}\\ \mathbf B^{\dagger} & \boldsymbol{-} \mathbf A\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-01a}\label{A-01a}\\ &\text{where} \nonumber\\ \mathbf A & \boldsymbol{=} \begin{bmatrix} \hphantom{a} \epsilon_1 & \hphantom{a} 0 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} \epsilon_2 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} 0 & \hphantom{a} \epsilon_3 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}\: \epsilon_1,\epsilon_2,\epsilon_3 \in \mathbb{R} \quad \text{and} \quad \mathbf B \boldsymbol{=} \begin{bmatrix} \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \vphantom{\dfrac{a}{b}}\\ \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \vphantom{\dfrac{a}{b}}\\ \Sigma_{31} & \Sigma_{32} & \Sigma_{33}\vphantom{\dfrac{a}{b}} \end{bmatrix}\: \Sigma_{ij} \in \mathbb{C} \tag{A-01b}\label{A-01b} \end{align} is hermitian $\,\left(\mathbf M^{\dagger} \boldsymbol{=}\mathbf M\right)$, so it has real eigenvalues $\,\lambda_k \in \mathbb{R}\left(k=1,\cdots,6\right)$. Moreover we have zero sum of these eigenvalues \begin{equation} \sum\limits_{k\boldsymbol{=}1}^{k\boldsymbol{=}6}\lambda_k\boldsymbol{=}\rm{Trace}\left(\mathbf M\right)\boldsymbol{=}0 \tag{A-02}\label{A-02} \end{equation}
$\boldsymbol{\S}\color{blue}{\textbf{B. Property imposed on the matrix $\,\mathbf B$}}$
The $\,6\times6\,$ matrix $\,\mathcal C\,$ defined by \begin{equation} \mathcal C \boldsymbol{=} K \sigma_x \boldsymbol{\otimes} \mathbf I_3 \tag{B-01}\label{B-01} \end{equation} is anti-unitary with property \begin{equation} \mathcal C^{\boldsymbol{-}1} \boldsymbol{=}\mathcal C \quad \text{or} \quad \mathcal C^2 \boldsymbol{=} \mathbf I_6 \tag{B-02}\label{B-02} \end{equation} while the relation \begin{equation} \mathcal C^{\boldsymbol{-}1} \mathbf M \mathcal C\boldsymbol{=-} \mathbf M \tag{B-03}\label{B-03} \end{equation} imposes to the $\,3\times3\,$ matrix $\,\mathbf B\,$ of equation \eqref{A-01b} the property \begin{equation} \mathbf B\boldsymbol{=-}\mathbf B^{\boldsymbol{\top}} \tag{B-04}\label{B-04} \end{equation} that is it equals to the negative of its transpose. So we write formally \begin{equation} \mathbf B\boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_3 & \hphantom{\boldsymbol{-}}\omega_2 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}\omega_3 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\omega_2 & \hphantom{\boldsymbol{-}}\omega_1 & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=\omega\times}\,, \quad \boldsymbol{\omega}\boldsymbol{=}\left(\omega_1,\omega_2,\omega_3\right) \in \mathbb{C}^3 \tag{B-05}\label{B-05} \end{equation} In case of a real 3-vector $\,\boldsymbol{\omega}\,$ the matrix $\,\mathbf B\,$ is anti-symmetric.
These properties of the matrices $\,\mathcal C,\mathbf B \,$ are proved in $\boldsymbol{\S}$E.
Note also that from equation \eqref{B-04} \begin{equation} \mathbf B^{\boldsymbol{\dagger}}\boldsymbol{=-}\overline{\mathbf B}\boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}0 \hphantom{_3}& \hphantom{\boldsymbol{-}}\overline{\omega}_3 & \boldsymbol{-}\overline{\omega}_2 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\overline{\omega}_3 & \hphantom{\boldsymbol{-}} 0\hphantom{_3} & \hphantom{\boldsymbol{-}}\overline{\omega}_1 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}\overline{\omega}_2 &\boldsymbol{-}\overline{\omega}_1 & \hphantom{\boldsymbol{-}}0\hphantom{_3}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=-}\overline{\boldsymbol{\omega}}\boldsymbol{\times} \tag{B-06}\label{B-06} \end{equation}
$\boldsymbol{\S}\color{blue}{\textbf{C. The Characteristic Equation of $\,\mathbf M\,$ and its solution}}$
The final expression of matrix $\,\mathbf M\,$ of equation \eqref{A-01a} satisfying the condition \eqref{B-03} is \begin{align} \mathbf M & \boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}\epsilon_1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_3 & \hphantom{\boldsymbol{-}}\omega_2 \vphantom{\dfrac{a}{b}}\\ \hphantom{-}0 & \hphantom{\boldsymbol{-}}\epsilon_2 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}\omega_3 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_1 \vphantom{\dfrac{a}{b}}\\ \hphantom{-}0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}\epsilon_3 & \boldsymbol{-}\omega_2 & \hphantom{\boldsymbol{-}}\omega_1 & \hphantom{\boldsymbol{-}}0 \vphantom{\dfrac{a}{b}}\\ \hphantom{-}0 & \hphantom{\boldsymbol{-}}\overline{\omega}_3 & \boldsymbol{-}\overline{\omega}_2 & \boldsymbol{-}\epsilon_1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\overline{\omega}_3 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}\overline{\omega}_1 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\epsilon_2 & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}\overline{\omega}_2 &\boldsymbol{-}\overline{\omega}_1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\epsilon_3\vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \mathbf A \hphantom{^{\dagger}} & \hphantom{\boldsymbol{-}} \mathbf B \vphantom{\dfrac{a}{b}}\\ \mathbf B^{\dagger} & \boldsymbol{-} \mathbf A\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{C-01a}\label{C-01a}\\ &\text{where} \nonumber\\ \mathbf A & \boldsymbol{=} \begin{bmatrix} \hphantom{a} \epsilon_1 & \hphantom{a} 0 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} \epsilon_2 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} 0 & \hphantom{a} \epsilon_3 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \quad \text{and} \quad \mathbf B \boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_3 & \hphantom{\boldsymbol{-}}\omega_2 \vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}\omega_3 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\omega_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\omega_2 & \hphantom{\boldsymbol{-}}\omega_1 & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{C-01b}\label{C-01b} \end{align} So the matrix $\,\mathbf M\,$ is a function of two 3-vectors, the real $\,\boldsymbol{\epsilon}\,$ and the complex in general $\,\boldsymbol{\omega}$ \begin{equation} \boldsymbol{\epsilon}\boldsymbol{=} \begin{bmatrix} \epsilon_1 \vphantom{\dfrac{a}{b}}\\ \epsilon_2 \vphantom{\dfrac{a}{b}}\\ \epsilon_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathbb{R}^3 \quad \text{and} \quad \boldsymbol{\omega}\boldsymbol{=} \begin{bmatrix} \omega_1 \vphantom{\dfrac{a}{b}}\\ \omega_2 \vphantom{\dfrac{a}{b}}\\ \omega_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathbb{C}^3 \tag{C-02}\label{C-02} \end{equation} The characteristic equation of $\,\mathbf M\,$ is of 6th order with respect to $\,\lambda\,$ \begin{equation} a\lambda^6\boldsymbol{+}b\lambda^4\boldsymbol{+}c\lambda^2\boldsymbol{+}d\boldsymbol{=}0 \tag{C-03}\label{C-03} \end{equation} or a cubic equation with respect to $\,\mathrm x\boldsymbol{=}\lambda^2\,$ \begin{equation} a\mathrm x^3\boldsymbol{+}b\mathrm x^2\boldsymbol{+}c\mathrm x\boldsymbol{+}d\boldsymbol{=}0 \tag{C-04}\label{C-04} \end{equation} The coefficients $\:a,b,c,d\:$ are real and their expressions as functions of the components of the vectors $\,\boldsymbol{\epsilon}\,$ and $\,\boldsymbol{\omega}$ are given in $\boldsymbol{\S}$D.
Now, the analytic solution of the cubic equation \eqref{C-04} is well-known. First the change of variable \begin{equation} \mathrm x\boldsymbol{=}\mathrm t\boldsymbol{-}\dfrac{b}{3a} \tag{C-05}\label{C-05} \end{equation} leads to a cubic that has no term in $\,\mathrm t^2$. After dividing by $\,a\,$ one gets the depressed cubic equation \begin{equation} \mathrm t^3\boldsymbol{+}p\mathrm t\boldsymbol{+}q\boldsymbol{=}0 \tag{C-06}\label{C-06} \end{equation} with \begin{align} \mathrm t & \boldsymbol{=}\mathrm x\boldsymbol{+}\dfrac{b}{3a} \tag{C-07a}\label{C-07a}\\ p & \boldsymbol{=}\dfrac{3ac\boldsymbol{-}b^2}{3a^2} \tag{C-07b}\label{C-07b}\\ q & \boldsymbol{=}\dfrac{2b^3\boldsymbol{-}9 a b c\boldsymbol{+}27a^2 d}{27a^3} \tag{C-07c}\label{C-07c} \end{align} At this point we make use of the trigonometric solution for three real roots with respect to $\,\mathrm t$ \begin{equation} \mathrm t_n\boldsymbol{=}2\sqrt{\boldsymbol{-}\dfrac{p}{3}}\cos\left[\dfrac{1}{3}\arccos\left(\dfrac{3q}{2p}\sqrt{\boldsymbol{-}\dfrac{3}{p}}\right)\boldsymbol{-}\dfrac{2\pi\left(n-1\right)}{3}\right]\quad \left(n=1,2,3\right) \tag{C-08}\label{C-08} \end{equation} This formula is due to François Viète. It is purely real when the equation has three real roots (that is when $\,4p^3\boldsymbol{+}27q^2<0$).
From the roots \eqref{C-08} we have the roots of \eqref{C-04} with respect to $\,\mathrm x$ \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \mathrm x_n\boldsymbol{=}\boldsymbol{-}\dfrac{b}{3a}\boldsymbol{+}\mathrm t_n\boldsymbol{=}\boldsymbol{-}\dfrac{b}{3a}\boldsymbol{+}2\sqrt{\boldsymbol{-}\dfrac{p}{3}}\cos\left[\dfrac{1}{3}\arccos\left(\dfrac{3q}{2p}\sqrt{\boldsymbol{-}\dfrac{3}{p}}\right)\boldsymbol{-}\dfrac{2\pi\left(n-1\right)}{3}\right]\quad \left(n=1,2,3\right) \tag{C-09}\label{C-09} \end{equation} The roots $\:\mathrm x_n\left(n=1,2,3\right)\:$ are non-negative real numbers so the eigenvalues of $\,\mathbf M\,$ are \begin{equation} \boxed{\:\lambda_{1}\boldsymbol{=}\boldsymbol{-}\lambda_{4}\boldsymbol{=}\sqrt{\mathrm x_1}\,,\quad \lambda_{2}\boldsymbol{=}\boldsymbol{-}\lambda_{5}\boldsymbol{=}\sqrt{\mathrm x_2}\,,\quad \lambda_{3}\boldsymbol{=}\boldsymbol{-}\lambda_{6}\boldsymbol{=}\sqrt{\mathrm x_3}\:\vphantom{\dfrac{a}{b}}} \tag{C-10}\label{C-10} \end{equation} There is no need to prove that $\:\mathrm x_n\left(n=1,2,3\right)\:$ are non-negative real numbers since in any other case (such as one at least being negative real or complex) some of the eigenvalues $\:\lambda_k\left(k=1,2,\cdots,6 \right)\:$ would be complex in contradiction to the fact that an hermitian matrix as $\,\mathbf M\,$ has only real eigenvalues.
Viète's trigonometric expression of the roots of the cubic equation in the three-real-roots case lends itself to an elegant geometric interpretation in terms of a circle, see Figure-01. On the $\,\mathrm x\boldsymbol{-}\mathrm y\,$ graph of the function-characteristic polynomial with respect to $\,\mathrm x\boldsymbol{=}\lambda^2\,$ \begin{equation} \mathrm y\left(\mathrm x\right)\boldsymbol{=}a\mathrm x^3\boldsymbol{+}b\mathrm x^2\boldsymbol{+}c\mathrm x\boldsymbol{+}d \tag{C-11}\label{C-11} \end{equation} we choose a point \begin{equation} \mathrm K\left(\mathrm x_{_\mathrm K}\boldsymbol{=}\boldsymbol{-}\dfrac{b}{3a},\mathrm y_{_\mathrm K}\boldsymbol{=}\text{ any}\right) \tag{C-12}\label{C-12} \end{equation} Note that this point has the same abscissa as the inflection point of the curve \begin{equation} \dfrac{\mathrm d^2 \mathrm y }{\mathrm d \mathrm x^2}\boldsymbol{=}0 \quad \boldsymbol{\Longrightarrow}\quad 6a\mathrm x\boldsymbol{+}2b\boldsymbol{=}0 \quad \boldsymbol{\Longrightarrow}\quad \boxed{\:\mathrm x_{\mathtt{inflection}}\boldsymbol{=}\boldsymbol{-}\dfrac{b}{3a}\:} \tag{C-13}\label{C-13} \end{equation} For our case this abscissa is positive since from equations \eqref{D-04a}, \eqref{D-04b} \begin{equation} \mathrm x_{_\mathrm K}\boldsymbol{=}\mathrm x_{\mathtt{inflection}}\boldsymbol{=}\boldsymbol{-}\dfrac{b}{3a}\boldsymbol{=}\dfrac{\left(\boldsymbol{\Vert\epsilon\Vert}^2\boldsymbol{+}2\boldsymbol{\Vert\omega\Vert}^2\vphantom{\dfrac{a}{b}}\right)}{3}>0 \tag{C-14}\label{C-14} \end{equation}
With center the point $\:\mathrm K\:$ and radius \begin{equation} R\boldsymbol{=}2\sqrt{\boldsymbol{-}\dfrac{p}{3}} \tag{C-15}\label{C-15} \end{equation} we draw a circle and inscribe in it an equilateral triangle $\,\mathrm A_1\mathrm A_2\mathrm A_3\,$ inclined with respect to the$\,\mathrm x\boldsymbol{-}$axis by an angle \begin{equation} \theta\boldsymbol{=}\dfrac{1}{3}\arccos\left(\dfrac{3q}{2p}\sqrt{\boldsymbol{-}\dfrac{3}{p}}\right) \tag{C-16}\label{C-16} \end{equation} as shown in Figure-01. The abscissas of the vertices $\,\mathrm A_1,\mathrm A_2,\mathrm A_3\,$ are the three non-negative real roots $\,\mathrm x_1,\mathrm x_2,\mathrm x_3\,$ of equation \eqref{C-09}.
$\boldsymbol{\S}\color{blue}{\textbf{D. The coefficients of the Characteristic Equation of matrix $\,\mathbf M$}}$
The coefficients of the characteristic equations \eqref{C-03},\eqref{C-04} are \begin{align} a & \boldsymbol{=}1 \tag{D-01a}\label{D-01a}\\ b & \boldsymbol{=}\boldsymbol{-}\left[\left(\epsilon^2_1\boldsymbol{+}\epsilon^2_2\boldsymbol{+}\epsilon^2_2\right)\boldsymbol{+}2\left(\omega_1\overline{\omega}_1\boldsymbol{+}\omega_2\overline{\omega}_2\boldsymbol{+}\omega_3\overline{\omega}_3\right)\right] \tag{D-01b}\label{D-01b}\\ c & \boldsymbol{=}\hphantom{2}\left[\left(\omega_1\overline{\omega}_1\right)^2\boldsymbol{+}\left(\omega_2\overline{\omega}_2\right)^2\boldsymbol{+}\left(\omega_3\overline{\omega}_3\right)^2\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+} \nonumber\\ & \hphantom{==}2\left[\left(\omega_1\overline{\omega}_1\right)\left(\omega_2\overline{\omega}_2\right)\boldsymbol{+}\left(\omega_2\overline{\omega}_2\right)\left(\omega_3\overline{\omega}_3\right)\boldsymbol{+}\left(\omega_3\overline{\omega}_3\right)\left(\omega_1\overline{\omega}_1\right) \vphantom{\dfrac{a}{b}}\right]\boldsymbol{+} \nonumber\\ & \hphantom{==}2\left(\epsilon^2_1\omega_1\overline{\omega}_1\boldsymbol{+} \epsilon^2_2\omega_2\overline{\omega}_2\boldsymbol{+}\epsilon^2_3\omega_3\overline{\omega}_3\vphantom{\dfrac{a}{b}}\right)\boldsymbol{+}2\left(\epsilon_2\epsilon_3\omega_1\overline{\omega}_1\boldsymbol{+} \epsilon_3\epsilon_1\omega_2\overline{\omega}_2\boldsymbol{+}\epsilon_1\epsilon_3\omega_3\overline{\omega}_3\vphantom{\dfrac{a}{b}}\right)\boldsymbol{+} \nonumber\\ & \hphantom{==}\left[\left(\epsilon_2\epsilon_3\right)^2\boldsymbol{+}\left(\epsilon_3\epsilon_1\right)^2\boldsymbol{+}\left(\epsilon_1\epsilon_2\right)^2\vphantom{\dfrac{a}{b}}\right] \tag{D-01c}\label{D-01c}\\ d & \boldsymbol{=}\boldsymbol{-}\left[\left(\epsilon_1\omega_1\overline{\omega}_1\right)^2\boldsymbol{+}\left(\epsilon_2\omega_2\overline{\omega}_2\right)^2\boldsymbol{+}\left(\epsilon_3\omega_3\overline{\omega}_3\right)^2\vphantom{\dfrac{a}{b}}\right] \nonumber\\ & \hphantom{-} \boldsymbol{-}2\left[\left(\epsilon_1\omega_1\overline{\omega}_1\right)\left(\epsilon_2\omega_2\overline{\omega}_2\right)\boldsymbol{+}\left(\epsilon_2\omega_2\overline{\omega}_2\right)\left(\epsilon_3\omega_3\overline{\omega}_3\right)\boldsymbol{+}\left(\epsilon_3\omega_3\overline{\omega}_3\right)\left(\epsilon_1\omega_1\overline{\omega}_1\right) \vphantom{\dfrac{a}{b}}\right]\boldsymbol{+} \nonumber\\ & \hphantom{-} \boldsymbol{-}2\epsilon_1\epsilon_2\epsilon_3\left(\epsilon_1\omega_1\overline{\omega}_1\boldsymbol{+} \epsilon_2\omega_2\overline{\omega}_2\boldsymbol{+}\epsilon_3\omega_3\overline{\omega}_3\vphantom{\dfrac{a}{b}}\right)\boldsymbol{-}\left(\epsilon_1\epsilon_2\epsilon_3\right)^2 \tag{D-01d}\label{D-01d} \end{align}
To simplify these expressions we define first the real 3-vector \begin{equation} \boldsymbol{\zeta}\boldsymbol{=} \begin{bmatrix} \epsilon_2\epsilon_3 \vphantom{\dfrac{a}{b}}\\ \epsilon_3\epsilon_1 \vphantom{\dfrac{a}{b}}\\ \epsilon_1\epsilon_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \in \mathbb{R}^3 \tag{D-02}\label{D-02} \end{equation} and then the $\,3\times 3\,$ real matrix \begin{equation} \mathbf Z\boldsymbol{=} \begin{bmatrix} \hphantom{a} \zeta_1 & \hphantom{a} 0 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} \zeta_2 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} 0 & \hphantom{a} \zeta_3 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \hphantom{a} \epsilon_2\epsilon_3 & \hphantom{a} 0 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} \epsilon_3\epsilon_1 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 0 & \hphantom{a} 0 & \hphantom{a} \epsilon_1\epsilon_2 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{D-03}\label{D-03} \end{equation} The coefficients take the following simple expressions \begin{align} a & \boldsymbol{=}1 \tag{D-04a}\label{D-04a}\\ b & \boldsymbol{=}\boldsymbol{-}\left(\boldsymbol{\Vert\epsilon\Vert}^2\boldsymbol{+}2\boldsymbol{\Vert\omega\Vert}^2\vphantom{\dfrac{a}{b}}\right) \tag{D-04b}\label{D-04b}\\ c & \boldsymbol{=}\boldsymbol{\Vert\omega\Vert}^4\boldsymbol{+}2\boldsymbol{\Vert \mathbf A\omega\Vert}^2\boldsymbol{+} 2\boldsymbol{\langle\mathbf Z\boldsymbol{\omega},\boldsymbol{\omega}\rangle}\boldsymbol{+}\boldsymbol{\Vert\zeta\Vert}^2 \tag{D-04c}\label{D-04c}\\ d & \boldsymbol{=}\boldsymbol{-}\left(\boldsymbol{\langle\mathbf A\boldsymbol{\omega},\boldsymbol{\omega}\rangle}\boldsymbol{+}\epsilon_1\epsilon_2\epsilon_3\vphantom{\dfrac{a}{b}}\right)^2 \tag{D-04d}\label{D-04d} \end{align}
By $^{\prime\prime}\boldsymbol{\langle\hphantom{=},\hphantom{=}\rangle}^{\prime\prime}$ we mean the default inner product in the complex space $\,\mathbb C^3$.
$\boldsymbol{\S}\color{blue}{\textbf{E. Proving properties of matrices $\,\mathcal C$ and $\,\mathbf B$}}$
For the $\,6\times6\,$ matrix $\,\mathcal C\,$ defined by equation \eqref{B-01} we have \begin{equation} \mathcal C \boldsymbol{=} K \sigma_x \boldsymbol{\otimes} \mathbf I_3\boldsymbol{=}K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} 1 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} 1 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{\otimes} \mathbf I_3\boldsymbol{=}K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{E-01}\label{E-01} \end{equation} Explicitly \begin{equation} \mathcal C \boldsymbol{=} K \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \vphantom{\tfrac{a}{b}}\\ 0 & 0 & 0 & 0 & 1 & 0 \vphantom{\tfrac{a}{b}}\\ 0 & 0 & 0 & 0 & 0 & 1 \vphantom{\tfrac{a}{b}}\\ 1 & 0 & 0 & 0 & 0 & 0 \vphantom{\tfrac{a}{b}}\\ 0 & 1 & 0 & 0 & 0 & 0 \vphantom{\tfrac{a}{b}}\\ 0 & 0 & 1 & 0 & 0 & 0 \vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{E-02}\label{E-02} \end{equation} From \eqref{E-01}
\begin{equation} \mathcal C^{\boldsymbol{-}1}\boldsymbol{=} \left(K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}\right)^{\boldsymbol{-}1} \boldsymbol{=} \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}^{\boldsymbol{-}1}K^{\boldsymbol{-}1} \boldsymbol{=} \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}K \tag{E-03}\label{E-03} \end{equation}
because \begin{equation} \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}^{\boldsymbol{-}1} \boldsymbol{=} \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \text{and} \quad K^{\boldsymbol{-}1}\boldsymbol{=}K \tag{E-04}\label{E-04} \end{equation} But these two operators commute \begin{equation} \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix}K \boldsymbol{=}K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{E-05}\label{E-05} \end{equation} since the $\,6\times 6\,$ matrix is real. So finally \begin{equation} \mathcal C^{\boldsymbol{-}1}\boldsymbol{=}\mathcal C\boldsymbol{=}K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{E-06}\label{E-06} \end{equation} Although not necessary in the following we'll prove that the matrix $\,\mathcal C\,$ is anti-unitary. So, consider two complex 6-vectors $\,\mathbf x_1,\mathbf x_2 \in \mathbb{C}^6\,$ each one split to two complex 3-vectors \begin{equation} \mathbf x_1 \boldsymbol{=} \begin{bmatrix} \boldsymbol{\xi}_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\eta}_1 \vphantom{\dfrac{a}{b}} \end{bmatrix}\,, \quad \mathbf x_2 \boldsymbol{=} \begin{bmatrix} \boldsymbol{\xi}_2 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\eta}_2 \vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \Longrightarrow \quad \mathcal C\mathbf x_1 \boldsymbol{=} \begin{bmatrix} \:\overline{\boldsymbol{\eta}}_1 \vphantom{\dfrac{a}{b}} \\ \:\overline{\boldsymbol{\xi}}_1 \vphantom{\dfrac{a}{b}} \end{bmatrix}\,, \quad \mathcal C\mathbf x_2 \boldsymbol{=} \begin{bmatrix} \:\overline{\boldsymbol{\eta}}_2 \vphantom{\dfrac{a}{b}} \\ \:\overline{\boldsymbol{\xi}}_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{E-07}\label{E-07} \end{equation} then \begin{align} \boldsymbol{\langle}\mathcal C \mathbf x_1,\mathcal C \mathbf x_2\boldsymbol{\rangle}_{\mathbb{C}^6} & \boldsymbol{=} \Biggl\langle \begin{bmatrix} \:\overline{\boldsymbol{\eta}}_1 \vphantom{\dfrac{a}{b}} \\ \:\overline{\boldsymbol{\xi}}_1 \vphantom{\dfrac{a}{b}} \end{bmatrix}, \begin{bmatrix} \:\overline{\boldsymbol{\eta}}_2 \vphantom{\dfrac{a}{b}} \\ \:\overline{\boldsymbol{\xi}}_2 \vphantom{\dfrac{a}{b}} \end{bmatrix}\Biggr\rangle_{\mathbb{C}^6} \tag{E-08}\label{E-08}\\ &\boldsymbol{=} \boldsymbol{\langle}\overline{\boldsymbol{\eta}}_1,\overline{\boldsymbol{\eta}}_2\boldsymbol{\rangle}_{\mathbb{C}^3}\boldsymbol{+}\boldsymbol{\langle}\overline{\boldsymbol{\xi}}_1,\overline{\boldsymbol{\xi}}_2\boldsymbol{\rangle}_{\mathbb{C}^3}\boldsymbol{=}\overline{\left(\boldsymbol{\langle}\boldsymbol{\eta}_1,\boldsymbol{\eta}_2\boldsymbol{\rangle}_{\mathbb{C}^3}\boldsymbol{+}\boldsymbol{\langle}\boldsymbol{\xi}_1,\boldsymbol{\xi}_2\boldsymbol{\rangle}_{\mathbb{C}^3} \vphantom{\dfrac{a}{b}} \right)} \nonumber \end{align} that is \begin{equation} \boldsymbol{\langle}\mathcal C \mathbf x_1,\mathcal C \mathbf x_2\boldsymbol{\rangle} \boldsymbol{=} \overline{\boldsymbol{\langle}\mathbf x_1,\mathbf x_2\boldsymbol{\rangle}} \tag{E-09}\label{E-09} \end{equation}
Now we'll prove \eqref{B-04} starting from \eqref{B-03} and the fact that $\,\mathcal C^{\boldsymbol{-}1}\boldsymbol{=}\mathcal C$
\begin{align} & \mathcal C^{\boldsymbol{-}1} \mathbf M \mathcal C \mathbf x\boldsymbol{=}\boldsymbol{-} \mathbf M\mathbf x \: \boldsymbol{\Longrightarrow} \tag{E-10}\label{E-10}\\ & K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \mathbf A \hphantom{^{\dagger}} & \hphantom{\boldsymbol{-}} \mathbf B \vphantom{\dfrac{a}{b}}\\ \mathbf B^{\dagger} & \boldsymbol{-} \mathbf A\vphantom{\dfrac{a}{b}} \end{bmatrix} K \begin{bmatrix} \hphantom{a} 0 & \hphantom{a} \mathbf I_3 \hphantom{a} \vphantom{\dfrac{a}{b}}\\ \hphantom{a} \mathbf I_3 & \hphantom{a} 0 \hphantom{a} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \boldsymbol{\xi} \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \boldsymbol{-}\begin{bmatrix} \mathbf A \hphantom{^{\dagger}} & \hphantom{\boldsymbol{-}} \mathbf B \vphantom{\dfrac{a}{b}}\\ \mathbf B^{\dagger} & \boldsymbol{-} \mathbf A\vphantom{\dfrac{a}{b}} \end{bmatrix}\begin{bmatrix} \boldsymbol{\xi} \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\eta} \vphantom{\dfrac{a}{b}} \end{bmatrix} \quad\boldsymbol{\Longrightarrow} \nonumber\\ & K \begin{bmatrix} \mathbf B^{\dagger} & \boldsymbol{-} \mathbf A \vphantom{\dfrac{a}{b}}\\ \mathbf A & \hphantom{\boldsymbol{-}}\mathbf B\vphantom{\dfrac{a}{b}} \end{bmatrix} K \begin{bmatrix} \boldsymbol{\eta} \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\xi} \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=}\begin{bmatrix} \boldsymbol{-}\mathbf A \boldsymbol{\xi}\boldsymbol{-}\mathbf B\boldsymbol{\eta} \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\mathbf B^{\dagger} \boldsymbol{\xi} \boldsymbol{+} \mathbf A\boldsymbol{\eta}\vphantom{\dfrac{a}{b}} \end{bmatrix}\:\boldsymbol{\Longrightarrow} \nonumber\\ & \begin{bmatrix} \overline{\mathbf B^{\dagger}}\boldsymbol{\eta} \boldsymbol{-} \mathbf A \boldsymbol{\xi} \vphantom{\dfrac{a}{b}}\\ \mathbf A\boldsymbol{\eta} \boldsymbol{+}\overline{\mathbf B}\boldsymbol{\xi}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=}\begin{bmatrix} \boldsymbol{-}\mathbf A \boldsymbol{\xi}\boldsymbol{-}\mathbf B\boldsymbol{\eta} \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\mathbf B^{\dagger} \boldsymbol{\xi} \boldsymbol{+} \mathbf A\boldsymbol{\eta}\vphantom{\dfrac{a}{b}} \end{bmatrix}\:\boldsymbol{\Longrightarrow} \left. \begin{cases} \overline{\mathbf B^{\dagger}}\boldsymbol{\eta} \boldsymbol{=}\boldsymbol{-}\mathbf B\boldsymbol{\eta} \vphantom{\dfrac{a}{b}}\\ \overline{\mathbf B}\boldsymbol{\xi}\boldsymbol{=}\boldsymbol{-}\mathbf B^{\dagger} \boldsymbol{\xi}\vphantom{\dfrac{a}{b}} \end{cases}\right\} \nonumber \end{align}
so \begin{equation} \mathbf B\boldsymbol{=-}\mathbf B^{\boldsymbol{\top}} \tag{E-11}\label{E-11} \end{equation}
Note : Because of insufficient space (maximum 30000 characters) I post a numerical example as another answer.
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$ Reference : Cubic equation