Question
Diagonalize this unitary matrix $V$ to reach $V=U \Lambda U^{\mathrm{H}}$. Again all $|\lambda|=1$ :$$V=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & 1-i \\1+i & -1\end{array}\right]$$
Step 1
The eigenvalues $\lambda$ are the roots of the characteristic equation, which is given by $\text{det}(V - \lambda I) = 0$, where $I$ is the identity matrix. Show more…
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