Finite Groups

F

is an invariant subspace of

Ψ : G \to U (E)

and

P : E \to F

the orthogonal projection onto

F

. Then for all

g \in G

the operators

P

and

Ψ (g)

are simultaneously diagonalizable.

We already know that $P$ commutes with $Ψ (g)$ for all $g \in G$ . Hence, it suffices to show that $P$ and $Ψ (g)$ are diagonalizable because then by Proposition, $P$ and $Ψ (g)$ are simultaneously diagonalizable. We will show that $P$ and $Ψ (g)$ are normal operators: $P$ is self-adjoint (because it is an orthogonal projection), so it commutes with $P^{*} = P$ and is thus normal. By definition, $Ψ (g)$ is a unitary operator, so we have $Ψ (g)^{*} Ψ (g) = 1_{E} = Ψ (g) Ψ (g)^{*}$ , and thus, $Ψ (g)$ is also normal. Hence, by Proposition, $P$ and $Ψ (g)$ are diagonalizable.