Symmetry and eigen values To prove: 

A symmetric matrix has only real eigenvalues.

For this I took a symmetric matrix $A$, an eigenvalue $k$ and an eigenvector $X$. 
$AX=kX$
Taking $X$ transpose on both sides
$X'AX=X'kX$
Taking transpose
$(X'AX)'=(X'kX)'$
After solving
$X'AX=k'X'X$
$X'kX=k'X'X$
$(k-k')X'X=0$
$k-k'=0$
$k=k'$
$k$ is equal to the transpose of $k$. How does it prove that $A$ symmetric matrix has only real eigenvalues?
 A: First of all, let me fix the terminology.
The transpose of the matrix $A=[a_{ij}]$ is the matrix $A^T=[b_{ij}]$, where $b_{ij}=a_{ji}$. A (square) matrix is symmetric if $A=A^T$.
The hermitian transpose of the matrix $A$ is the matrix $A^H=[c_{ij}]$ where $c_{ij}=\overline{a_{ij}}$; that is, besides transposing, every entry is also conjugated.
A matrix is hermitian if $A=A^H$. For a real matrix, being symmetric is the same as being hermitian.

Theorem. A real symmetric matrix only has real eigenvalues.

Suppose $k$ is a complex eigenvalue, with complex eigenvector $x$ of the real symmetric matrix $A$; then we have
\begin{gather}
Ax=kx \\[4px]
x^HAx=x^H(kx) \\[4px]
x^HAx=k(x^Hx)
\end{gather}
Take the hermitian transpose of both sides:
$$
x^HA^Hx=\bar{k}(x^Hx)
$$
where $\bar{k}$ is the complex conjugate of $k$.
But $A^H=A$, since $A$ is real symmetric, so we get
$$
k(x^Hx)=\bar{k}(x^Hx)
$$
and, from $x\ne0$, we obtain $x^Hx\ne0$. Therefore $k=\bar{k}$ is real.
Actually, we see that we just need $A^H=A$, so the theorem can be extended.

Theorem. A (complex) hermitian matrix only has real eigenvalues.


Note that the result is false for matrices with non real coefficients. A simple counterexample is the matrix
$$
A=\begin{bmatrix}
i & 0 \\
0 & i
\end{bmatrix}
$$
Then we have $A=A^T$, so $A$ is symmetric, but its eigenvalues aren't real. As a matter of fact, the attribute symmetric is mostly used for real matrices.
A: One way to show that a self-adjoint matrix has real eigenvalues is to use the inner product. Let $ A $ be our self-adjoint matrix so that $ A^H = A $ ($ A^H $ is the Hermitian transpose, also known as the conjugate transpose) and let $ \lambda $ be our eigenvalue, and $ v $ a corresponding eigenvector. Define the inner product of two vectors as $ \langle v, w \rangle = v^H w $, then we have
$$ \bar{\lambda}\langle v, v \rangle = \langle Av, v \rangle = (Av)^H v = v^H A^H v = v^H Av = \langle v, Av \rangle = \langle v, \lambda v \rangle = \lambda \langle v, v \rangle$$
But an eigenvector is nonzero by definition, and the inner product is positive definite, so we have $ \lambda = \bar{\lambda} $. As $ \lambda $ is equal to its complex conjugate, it is real. To finish, simply note that $ A^T = A^H $ for matrices with real entries, so symmetric matrices with real entries are self-adjoint.
