If \(T\) is a normal operator, prove that characteristic vectors for \(T\) which are associated with distinct characteristic values are orthogonal.
Solution
Let \[ Tv=\lambda v, \qquad Tw=\mu w, \] with \(v,w\ne0\) and \(\lambda\ne\mu\).
1. An eigenvector of a normal operator is an eigenvector of the adjoint with conjugate eigenvalue.
Since \(T\) is normal, so is \(T-\lambda I\), and therefore \[ \|(T-\lambda I)x\| =\|(T^*-\overline\lambda I)x\| \] for every \(x\). Taking \(x=v\) gives \[ 0 =\|(T-\lambda I)v\| =\|(T^*-\overline\lambda I)v\|, \] hence \[ T^*v=\overline\lambda v. \] Similarly, \[ T^*w=\overline\mu w. \]
2. Distinct eigenvalues force orthogonality.
Using the adjoint identity and the eigenvector equations, \[ \lambda(v\mid w) =(Tv\mid w) =(v\mid T^*w). \] Since \(T^*w=\overline\mu w\) and the inner product is conjugate-linear in its second slot, \[ (v\mid T^*w) =(v\mid\overline\mu w) =\mu(v\mid w). \] Therefore \[ (\lambda-\mu)(v\mid w)=0. \] Because \(\lambda\ne\mu\), we conclude \[ \boxed{(v\mid w)=0}. \] Thus characteristic vectors belonging to distinct characteristic values are orthogonal.