Spectral theorem for unitary operators T o F

Question

Si $T$ es unitaria y $B$ es una base de $V$ formada por vectores propios de $T$ entonces $B$ es un conjunto ortogonal.

If $T$ is unitary and $B$ is a basis for $V$ consisting of eigenvectors of $T$ then $B$ is an orthogonal set.

Why is this statement false? Does this not contradict the spectral theorem for unitary operators?

Answer 1

Spectral theorem implies that $\ker(T-\lambda I) \perp \ker(T-\mu I)$ for every distinct eigenvalues $\lambda$ and $\mu$. (Indeed, every normal operator has this property)

Hence if you have eigenvectors corresponding to distinct eigenvalues respectively, then these are orthogonal.

However, if those eigenvectors correspond to the same eigenvalue, then those need not be orthogonal. You need to apply Gram-Schmidt process to make them orthogonal.

Answer 2

The problem is multiplicity of eigenvalues. The eigenspaces are orthogonal, but within an eigenspace of dimension greater than one, you can find a non-orthogonal basis.