# Find two independent orthogonal eigenvectors

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?

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.

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\$$.