Prove if $U$ unitary, $U^k$ has convergent subsequence to $I$ Let $U$ be a unitary matrix in $M_{n}(\mathbb{C})$. Prove $\{U^{k}\}_{k \in\Bbb N}$ has a subsequence that converges to $I$ (identity matrix). How to prove that?
I tried using the Spectral theorem.
 A: It suffices to apply the spectral theorem, noting that $U$ can be diagonalized, i.e. there is a unitary $V$ such that $VUV^* = D$ is diagonal.  From there, one must apply an analysis result:

Lemma: Let $\xi \in \Bbb C$ with $|\xi| = 1$.  Then $\{\xi^k\}$ has a subsequence converging to $1$.  That is, $1$ is an accumulation point of the sequence $\{\xi^k\}$.

Proof: Let $\epsilon > 0$ be arbitrary.  Let $N$ be such that $\epsilon > 2\pi/N$. 
By the pigeonhole principle (hopefully you can figure out what I mean here), there must exist $k_1 < k_2 \leq N$ such that $|\xi^{k_2} - \xi^{k_1}|<\epsilon$.  So, we have
$$
|\xi^{k_2 - k_1} - 1| = |\xi^{k_1}||\xi^{k_2 - k_1} - 1| < \epsilon
$$
Thus, setting $n = k_2 - k_1$, we may now state that for any $\epsilon > 0$, there exists an $n \in \Bbb N$ for which $|\xi^{n} - 1| < \epsilon$.  The conclusion follows.

I've left a lot of blanks for you to fill in here.  Let me know if there's something you'd like clarified.
A: Observation 1: The set of unitary matrices is compact.
Proof: The operator norm of a unitary is 1, so the set of unitary matrices is bounded.  The defining relation $U^{\dagger}U = 1$ shows that the set of unitaries is the preimage of a closed set $\{1\}$ of a continuous map, so the set of unitaries is closed.  Closed & bounded implies compact.
Observation 2: A sequence in a compact set has a convergent (Cauchy) subsequence.  Call the limit of the subsequence $L$.
So for any $\epsilon>0$ we can find a $q$ s.t. for any element $U^p$ of the subsequence with $p\geq q$, $||L-U^p||<\epsilon/2$.  WLOG pick $q$ s.t. $U^q$ is itself in the subsequence.  For any $n\in \mathbb{N}$ such that $U^{n+q}$ is in the Cauchy subsequence, $$\epsilon > ||L-U^{n+q}|| + ||L-U^q|| \geq ||U^{n+q}-U^q|| = ||U^q (U^n-1)|| = ||U^n-1||.$$  So there exist arbitrarily high powers of $U$ arbitrarily close to 1.  Any Cauchy subsequence of these must converge to 1. 
No spectral theorem required!
(Thanks to A. Barron's comment on the OP for suggesting this approach.)
