Answer to Question #214229 in Complex Analysis for SID

Question

Let θ be a real number. Then check whether the matrices 

cos θ − sin θ

sin θ cos θ

and 

e

iθ −0

0 e

−iθ 

are similar over the field of complex numbers.

Answer 1

$$A=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

The characteristic polynomial is

$$\det(A-\lambda I)=0$$

$$\begin{vmatrix} \cos \theta-\lambda & -\sin \theta \\ \sin \theta & \cos \theta-\lambda \end{vmatrix}=0$$

$$\cos^2\theta-2\lambda \cos\theta+\lambda^2+\sin^2\theta=0$$

$$\lambda^2-2\lambda \cos\theta+1=0$$

$$\lambda=\cos\theta\pm\sqrt{\cos^2\theta-1}$$

$$\lambda=\cos\theta\pm i \sin\theta=e^{\pm i\theta}$$

$\lambda_1=e^{i \theta}=\cos\theta+ i \sin\theta$

$$\begin{bmatrix} -i\sin\theta & -\sin \theta \\ \sin \theta & -i\sin \theta \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$-i\sin\theta\begin{bmatrix} 1 & -i \\ i & 1 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$\sin\theta\begin{bmatrix} 1 & -i \\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=\begin{bmatrix} i \\ 1 \end{bmatrix}, \sin \theta\not=0$$

$\lambda_2=e^{-i \theta}=\cos\theta- i \sin\theta$

$$\begin{bmatrix} i\sin\theta & -\sin \theta \\ \sin \theta & i\sin \theta \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$i\sin\theta\begin{bmatrix} 1 & i \\ - i & 1 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$\sin\theta\begin{bmatrix} 1 & i \\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0$$

$$\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=\begin{bmatrix} - i \\ 1 \end{bmatrix}, \sin \theta\not=0$$

$$T\alpha_1=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} i \\ 1 \end{bmatrix}=\begin{bmatrix} i\cos \theta-\sin \theta \\ i\sin \theta+\cos \theta \end{bmatrix}$$

$$=( \cos \theta+i\sin \theta)\begin{bmatrix} i\\ 1 \end{bmatrix}=e^{i\theta}\alpha_1$$

$$T\alpha_2=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} - i \\ 1 \end{bmatrix}=\begin{bmatrix} -i\cos \theta-\sin \theta \\ - i\sin \theta+\cos \theta \end{bmatrix}$$

$$=( \cos \theta-i\sin \theta)\begin{bmatrix} -i\\ 1 \end{bmatrix}=e^{-i\theta}\alpha_2$$

Therefore the following two matrices are similar over the field of complex numbers:

$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \text{and} \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}$$