Prove that a normal operator on a complex inner product space is selfadjointif and only

Step 1 of 2

Consider a normal operator  on a finite-dimensional complex inner product space .

The objective is to show that the operator  is self-adjoint if and only if all its eigenvalues are real.

The theorem can be used here are:

     If  is an eigenvector of a normal operator  on an inner product space , then  is also an eigenvector of  . Moreover, if  then .

     Suppose  is a linear operator on a finite-dimensional complex inner product space . Then  is a normal operator if and only if there exists an orthonormal basis for  consisting of eigenvectors of .

     If  and  are two linear operators on an inner product space  with a basis   such that   . Then  for all .

First, let  is a self-adjoint operator. It implies that, , where  is the adjoint of  . Also, if  is an eigenvector of  corresponding to an eigenvector  , then using result (1),  is an eigenvector of  corresponding to an eigenvalue . Since . It implies the following:

                                                                …… (1)

X cannot be zero as it is an eigenvector.

Equation (1) implies that the eigenvalue  is real since  is an arbitrary eigenvalue of the self-adjoint operator . 

It implies that every eigenvalue of a self-adjoint operator is real.