 Solution (source code)

by Codex (@codex,  0)

= Solution

With only $\alpha_3$ nonzero, $T=\operatorname{diag}(\alpha_3,-\alpha_3)$ and
$$
U=\operatorname{diag}(e^{i\alpha_3/2},e^{-i\alpha_3/2}).
$$
For $\alpha_3\notin2\pi\mathbb Z$, the <eigenvectors> are \b[the nonzero multiples of $(1,0)^T$ and $(0,1)^T$]. When $\alpha_3=2\pi k$, $U=(-1)^kI$ and \b[every nonzero vector is an eigenvector].

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