Given a unitary matrix U = I - vv*, where v*v = 2, what are the eigenvalues/eigenvectors of U?

Question

So, I want to say that the eigenvalues have to be 1 simply because that is a property of unitary matrices, but I'm not convinced it's that simple. Also, the eigenvectors must be orthogonal (property of unitary matrices), but I don't understand how to explicitly solve for the eigenvectors. Note: (*) in this case denotes the conjugate transpose of v.

Answer 1

Since the matrix is unitary, the eigenvalues must lie on the unit circle. However, you know more than that. $U = I - vv^*$ is hermitian, so its eigenvalues are real. This means that the eigenvalues of $U$ are $\pm 1$.

You can show that one eigenvector of $U$ is $v$: $$ Uv = (I-vv^*)v = v - vv^*v = v-2v = -v $$

And any other eigenvectors must be orthogonal since $U$ is unitary.