If \(\theta\) is a real number, prove that the following matrices are unitarily equivalent \[ \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}, \qquad \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}. \]
Solution
Let \[ R_\theta= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}. \]
1. Find an orthonormal eigenbasis over \(\mathbb C\).
Put \[ v_+=\frac1{\sqrt2}\binom{1}{-i}, \qquad v_-=\frac1{\sqrt2}\binom{1}{i}. \] Then \[ R_\theta v_+ =\frac1{\sqrt2} \binom{\cos\theta+i\sin\theta} {\sin\theta-i\cos\theta} =e^{i\theta}v_+, \] and similarly \[ R_\theta v_-=e^{-i\theta}v_-. \] Moreover \[ v_+^*v_+=v_-^*v_-=1, \qquad v_+^*v_-=0. \] Hence \((v_+,v_-)\) is an orthonormal basis of \(\mathbb C^2\).
2. Use that eigenbasis to obtain a unitary equivalence.
Let \[ U=\frac1{\sqrt2} \begin{pmatrix} 1&1\\ -i&i \end{pmatrix}. \] Its columns are the orthonormal vectors from step 1, so \(U\) is unitary. The two eigenvector equations are exactly \[ R_\theta U =U \begin{pmatrix} e^{i\theta}&0\\ 0&e^{-i\theta} \end{pmatrix}. \] Multiplying on the left by \(U^*\) gives \[ \boxed{ U^*R_\theta U =\begin{pmatrix} e^{i\theta}&0\\ 0&e^{-i\theta} \end{pmatrix}.} \] Thus the two matrices are unitarily equivalent.