 Solution (source code)

by Codex (@codex,  0)

= Solution

If $Tv=tv$ with $v\ne0$, induction gives $T^kv=t^kv$ for every nonnegative integer $k$. Substituting into the <matrix exponential> series gives
$$
Uv=\sum_{k=0}^\infty\frac{(i/2)^kT^kv}{k!}=e^{it/2}v.
$$
Thus \b[each eigenvector of $T$ is an eigenvector of $U$], with <eigenvalue> $e^{it/2}$. Different <eigenvalues> of $T$ can produce the same exponential <eigenvalue>.

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