Exercise 2.17#

In this exercise we need to prove

(1)#\[ \text{A normal matrix is Hermitian} \iff \text{A normal matrix has real eigenvalues} \]

To check whether a Hermitian operator \(A\) as normal operator has real eigenvalues (from the left to the right), we first notice that any normal operator is diagonalizable, so could write down the diagonal representation of normal operator \(A\),

(2)#\[ A = \sum_{i}\lambda_i|i\rangle\langle i| \]

where \(\lambda_i\) is eigenvalue of \(A\) and \(|i\rangle\) is corresponding eigenvector of \(\lambda_i\). Note that \(A\) is also Hermitian, it means that

(3)#\[ A = A^{\dagger} \iff \sum_{i}\lambda_i|i\rangle\langle i| = \left(\sum_{i}\lambda_i|i\rangle\langle i|\right)^{\dagger} = \sum_{i}\lambda^{*}_i|i\rangle\langle i| \]

where \(\lambda^*_i\) is the conjugate of \(\lambda_i\). From eq. (3) we could conclude that \(\lambda_i = \lambda^*_i\) and thus \(A\) has real eigenvalues if \(A\) is a normal operator and Hermitian operator.

To check whether a normal operator \(A\) that has real eigenvalues is a Hermitian operator (from the right to the left), since any normal operator is diagonalizable, we could still have

(4)#\[ A = \sum_{i}\lambda_i|i\rangle\langle i| \]

From eq. (4) we could directly calculate \(A^{\dagger}\) as

(5)#\[ A^{\dagger} = \left(\sum_{i}\lambda_i|i\rangle\langle i|\right)^{\dagger} = \sum_{i}\lambda^{*}_i|i\rangle\langle i| =\sum_{i}\lambda_i|i\rangle\langle i| = A \]

where \(\lambda^*_i\) is the conjugate of \(\lambda_i\). From eq. (5) we could conclude that \(A=A^{\dagger}\) and thus \(A\) is a Hermitian operator if \(A\) is a normal operator with all eigenvalues as real.