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We are given that $C$ is an $n\times n$ complex matrix, and $C$ is Hermitian ($C = C^*$). Now we define $C = A + Bi$ where $A$ and $B$ are real matrices.

Let $M$ be defined by

\begin{bmatrix} A & -B \\ B & A \end{bmatrix}

Given that $\lambda$ is an eigenvalue of $C$ with eigenvector $z = x + iy$, we want to find two independent orthogonal eigenvectors of $M$ with eigenvalue $\lambda$.

I found that

\begin{bmatrix} z \\ -iz \end{bmatrix}

is one eigenvector. How do I find the other one?

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2 Answers 2

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Given a Hermitian matrix

$C = C^\dagger \in M_{n \times n}(\Bbb C), \tag 1$

which has an eigenvalue $\lambda$ with associated eigenvector $z$:

$Cz = \lambda z, \tag 2$

it is well-known that

$C^\dagger = C \Longrightarrow \lambda \in \Bbb R; \tag 3$

since the entries of $C$ lie in $\Bbb C$, with real $\lambda$, we will generally have

$z \in \Bbb C^n; \tag 4$

that is,

$z = x + iy, \tag 5$

with

$x, y \in \Bbb R^n; \tag 6$

thus we may write

$C(x + iy) = \lambda(x + iy); \tag 7$

now (1) implies we also have

$C = A + iB, \; A, B \in M_{n \times n}(\Bbb R); \tag 8$

thus (7) becomes

$(A + iB)(x + iy) = \lambda x + i \lambda y, \tag 9$

or

$(Ax - By) + i(Ay + Bx) = \lambda x + i \lambda y; \tag{10}$

equating the real and imaginary parts of either side yields

$Ax - By = \lambda x, \tag{11}$

$Ay + Bx = \lambda y. \tag{12}$

We can now in fact find two real orthogonal eigenvectors for

$M = \begin{bmatrix} A & -B \\ B & A \end{bmatrix}, \tag{13}$

each corresponding to the eigenvalue $\lambda$; set

$w = \begin{pmatrix} x \\ y \end{pmatrix}; \tag{14}$

$Mw = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} Ax - By \\ Bx + Ay \end{pmatrix} = \begin{pmatrix} \lambda x \\ \lambda y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix} = \lambda w; \tag{15}$

now set

$v = \begin{pmatrix} -y \\ x \end{pmatrix}; \tag{16}$

$Mv = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} \begin{pmatrix} -y \\ x \end{pmatrix} = \begin{pmatrix} -Ay - Bx \\ -By + Ax \end{pmatrix} = \begin{pmatrix} -\lambda y \\ \lambda x \end{pmatrix} = \lambda \begin{pmatrix} -y \\ x \end{pmatrix} = \lambda v; \tag{17}$

furthermore,

$\langle v, w \rangle = \begin{pmatrix} -y \\ x \end{pmatrix}^T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y^T & x^T \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = -y^Tx + x^Ty; \tag{18}$

$y^T x, x^Ty \in \Bbb R \Longrightarrow y^Tx = (y^Tx)^T = x^Ty, \tag{19}$

thus,

$\langle v, w \rangle = 0. \tag{20}$

Note: It may be worth pointing out that the condition on $C$

$C^\dagger = (\bar C)^T = C; \tag{21}$

has implitions for $A$ and $B$ as in (8), for then

$\bar C = A - iB, \tag{22}$

$(\bar C)^T = A^T - iB^T; \tag{23}$

thus,

$A = A^T, \tag {24}$

$B = -B^T; \tag{25}$

we see that $A$, the real part of $C$, must be symmetric, whilst $B$, the imaginary part, must be skew-symmetric. Finally we observe that the matrix $M$ (13) is in fact symmetric as well:

$M^T = \begin{bmatrix} A & -B \\ B & A \end{bmatrix}^T = \begin{bmatrix} A^T & B^T \\ -B^T & A^T \end{bmatrix} = \begin{bmatrix} A & -B \\ B & A \end{bmatrix} = M, \tag{26}$

by virtue of (24)-(25). End of Note.

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Try $\ v_1=\begin{bmatrix}x\\y\end{bmatrix}\ $ and $\ v_2=\begin{bmatrix}y\\-x\end{bmatrix}\ $. Your $\ \begin{bmatrix}z\\-iz\end{bmatrix}\ $ is $\ v_1 + iv_2\ $, and $\ v_1-iv_2 =\begin{bmatrix}\overline{z}\\i\overline{z}\end{bmatrix}\ $ is an eigenvector of $\ M\ $ orthogonal to it, and with the same eigenvalue $\ \lambda\ $.

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