Let \(T\) be a normal operator on a finite-dimensional complex inner product space. Prove that \(T\) is self-adjoint, positive, or unitary according as every characteristic value of \(T\) is real, positive, or of absolute value 1. (Use Theorem 22 to reduce to a similar question about diagonal matrices.)
Solution
Since \(T\) is normal on a finite-dimensional complex inner product space, the spectral theorem gives a unitary operator \(U\) and scalars \(\lambda_1,\ldots,\lambda_n\) such that \[ U^*TU=D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \] The \(\lambda_i\) are exactly the characteristic values of \(T\).
1. Real characteristic values give a self-adjoint operator.
If every \(\lambda_i\) is real, then \[ D^*=\operatorname{diag}(\overline{\lambda_1},\ldots,\overline{\lambda_n})=D. \] Hence \[ T^*=(UDU^*)^*=UD^*U^*=UDU^*=T. \] Thus \(T\) is self-adjoint.
2. Positive characteristic values give a positive operator.
Assume every \(\lambda_i>0\). For any nonzero vector \(x\), write \(y=U^*x\). Since \(U\) is unitary, \(y\ne0\), and \[ (Tx\mid x) =(Dy\mid y) =\sum_{i=1}^n\lambda_i|y_i|^2>0. \] Therefore \(T\) is positive.
3. Characteristic values of absolute value one give a unitary operator.
If \(|\lambda_i|=1\) for every \(i\), then \[ D^*D =\operatorname{diag}(|\lambda_1|^2,\ldots,|\lambda_n|^2) =I. \] Consequently \[ T^*T =UD^*U^*UDU^* =U(D^*D)U^* =I. \] Thus \(T\) is unitary.