T normal operator has splitting characteristic polynomial - eigenvectors form basis?
Question: is a normal operator on a finite-dimensional real inner product space with a characteristic polynomial that splits. Prove that has an orthonormal basis of eigenvectors of .
By Schur's theorem, because the characteristic polynomial of splits, there exists an orthonormal basis for such that is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of ? Thanks!