Let \(\theta\) be a real number. Prove that the following two matrices are similar over the field of complex numbers: \[ \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \qquad \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix} \] (Hint: Let \(T\) be the linear operator on \(C^2\) which is represented by the first matrix in the standard ordered basis. Then find vectors \(\alpha_1\) and \(\alpha_2\) such that \(T\alpha_1 = e^{i\theta}\alpha_1\), \(T\alpha_2 = e^{-i\theta}\alpha_2\), and \(\{\alpha_1, \alpha_2\}\) is a basis.)
Solution
Let \[ R=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}. \] Over \(\mathbb C\), set \[ \alpha_1=\begin{bmatrix}1\\-i\end{bmatrix},\qquad \alpha_2=\begin{bmatrix}1\\ i\end{bmatrix}. \] They are linearly independent. Direct calculation gives \[ R\alpha_1=e^{i\theta}\alpha_1, \qquad R\alpha_2=e^{-i\theta}\alpha_2. \] Thus, with \[ P=\begin{bmatrix}1&1\\-i&i\end{bmatrix}, \] we have \[ P^{-1}RP=\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{bmatrix}. \] Hence the two displayed matrices are similar over \(\mathbb C\).