Let A be a symmetric matrix and also it is real.

$A=A\prime$ (1) and $A=\bar A$ (2)

Let $\lambda$ be any eigen value of A

To prove that $\lambda$ is real, i.e $\lambda= \bar \lambda$

As $\lambda$ is eigen value of A there exists $X\neq0$ such that

$(A-\lambda I) X=0$

$AX-\lambda X=0$

$AX=\lambda X$ (3)

Taking $\theta$ on both sides,

$(AX)^\theta=(\lambda X)^\theta$

$X^\theta A\theta=X^\theta \lambda^\theta$

$X^\theta(\bar A) \prime= (\bar \lambda)\prime X^\theta$

$X^\theta( A) \prime= (\bar \lambda) X^\theta$ from (2)

$X^\theta( A) = (\bar \lambda) X^\theta$ from (1)

Post multiplying by X on both sides,

$X^\theta( A) X = (\bar \lambda) X^\theta X $

$X^\theta( \lambda) X = (\bar \lambda) X^\theta X $ from (3)

$X^\theta( \lambda) X = (\bar \lambda) X^\theta X $

$(\lambda- \bar \lambda) X^\theta X=0$ as $X\neq0$ so, $X^\theta \neq0$ so $X^\theta X\neq0$

$\lambda-\bar \lambda=0$

$\lambda=\bar \lambda$

$\lambda$ is real

Eigen values of real symmetric matrix are real.