# Exercise 2.17

In this exercise we need to prove

$$
\text{A normal matrix is Hermitian} \iff \text{A normal matrix has real eigenvalues}
$$(eqn:2.17.1)

---

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 $A$, 

$$
A = \sum_{i}\lambda_i|i\rangle\langle i|
$$(eqn:2.17.2)

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 

$$
A = A^{\dagger} \iff \sum_{i}\lambda_i|i\rangle\langle i| = \left(\sum_{i}\lambda_i|i\rangle\langle i|\right)^{\dagger} = \sum_{i}\lambda^{*}_i|i\rangle\langle i|
$$(eqn:2.17.3)

where $\lambda^*_i$ is the conjugate of $\lambda_i$. From eq. {eq}`eqn:2.17.3` 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

$$
A = \sum_{i}\lambda_i|i\rangle\langle i|
$$(eqn:2.17.4)

From eq. {eq}`eqn:2.17.4` we could directly calculate $A^{\dagger}$ as

$$
A^{\dagger} = \left(\sum_{i}\lambda_i|i\rangle\langle i|\right)^{\dagger} = \sum_{i}\lambda^{*}_i|i\rangle\langle i| =\sum_{i}\lambda_i|i\rangle\langle i| = A 
$$(eqn:2.17.5)

where $\lambda^*_i$ is the conjugate of $\lambda_i$. From eq. {eq}`eqn:2.17.5` 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.