Let A be a given Hermitian matrix, then $A^\theta = A$

Let $\lambda$ be the characteristic root of the matrix A with corresponding characteristic vector X then $AX = \lambda X$ (1)

Taking transpose conjugate of both the sides of (1) we get

$$\therefore (AX)^\theta = (\lambda X)^\theta \\ \therefore X^\theta A^\theta = \bar{\lambda} X^\theta \\ X^\theta A = \bar{\lambda} X^\theta$$

Post - multiplying by X, we get

$$X^\theta AX = \bar{\lambda}X^\theta X \\ X^\theta \lambda X = \bar{\lambda}X^\theta X \\ \lambda X^\theta X = \bar{\lambda}X^\theta X \\ (\lambda - \bar{\lambda})X^\theta X = 0$$

Since X is the non-zero vector, $X^\theta X \neq 0$ $\hspace{1cm} \rightarrow \lambda = \bar{\lambda}$ which shows that $\lambda$ is real.