A proof that eigenvalues of unitary matrix must be complex

Question

I have a problem with unitary matrix. I have to proof that its eigenvalues are complex and it lies on a circle.

So, I know how to do it.

I have to start with eigenfunction

$ \hat{\textbf{U}} | \psi \rangle = u | \psi \rangle . $

So

$ \langle \psi | \hat{\textbf{U}}^\dagger = \langle \psi | u^*. $

When I calculated the inner product of these two, I got

$ \langle \psi | \psi \rangle = |u|^2 \langle \psi | \psi \rangle. $

So, I know that

$|u| = 1.$

That means, that eigenvalues lies on a circle. But, I wonder where in this proof is proof that $u$ is complex? Or how can I proof that?

I mean, I see here only that I suppose, that $u$ is complex, not a proof.

Answer 1

Unitary matrices in general have complex entries, so that the eigenvalues are also complex numbers, and as you have shown, they must have modulus equal to $1$.

Answer 2

Recall that the eigenvalues of a matrix are precisely the roots of its characteristic polynomial. Since $\mathbb{C}$ is algebraically closed, all of these roots are in $\mathbb{C}$.

Answer 3

Your proof is correct. To button up your understanding, here are a couple of ideas for you to try:

1. Identify an example of a unitary matrix $U$ whose eigenvalues have no imaginary component.
2. Find an example of a unitary matrix $U$ whose eigenvalues have no real component.
3. Given a countable subset $S$ of the unit circle, can you find a unitary operator whose eigenvalues correspond with $S$?

You've already shown that the eigenvalues of a unitary operator must lie on the unit circle. These examples will help you understand which subsets of the unit circle can be the spectrum of a unitary operator.