For $T\colon V\to V$ every anti self-adjoint $T$ is orthogonally diagonalizable.

Question

I know how to prove that every self-adjoint operator is orthogonally diagonalizable (using spectral the and Gram Schmidt process). But I am not sure how to apply to an anti self-adjoint operator. Can someone please help me write a proof for this?

Answer 1

A skew-Hermitian (anti self-adjoint) operator $T$ is always normal, meaning that it commutes with its adjoint: $T\, T^*=T^*\, T$ (just like a self-adjoint operator). A variant of the spectral theorem says exactly that an operator on a finite-dimensional inner product space is normal if and only if it is orthogonally diagonalizable.