Prove if $A$ is a symmetric matrix with real entries, then the eigenvalues of $A$ are real. Given a matrix 
$$A=\begin{bmatrix}a & b \\ b& c \end{bmatrix} $$
then let $\lambda = p+qi$ be a complex eigenvalue of $A$. 
I used the characteristic equation to get a factorizacion of
$i(2pq - aq - qc)$ because I thought I would get zero. But it didn´t work.
 A: Let $\langle \cdot, \cdot \rangle$ denote the complex inner product. Suppose $v$ is a nonzero eigenvector of $A$ with eigenvalue $\lambda$. Then $$\lambda \langle v, v \rangle = \langle Av, v \rangle = \langle v, A^{T} v \rangle = \langle v, Av \rangle = \overline{\lambda} \langle v, v \rangle$$ Hence $\lambda = \overline{\lambda} \implies \lambda \in \mathbb{R}$.
Alternative proof: Compute the characteristic polynomial $$\det(x\cdot \text{Id} - A) = (x-a)(x-c)-b^2 = x^2 - (a+c)x + ac-b^2$$ This quadratic has discriminant $$(a+c)^2 - 4(ac-b^2) = (a-c)^2 + 4b^2 \ge 0$$ Therefore, its roots (i.e., the eigenvalues of $A$) are in $\mathbb{R}$.
A: Let $A$ be the real symmetric matrix, $x$ the eigen vector and $\lambda$ the eigen value. Then 
$$Ax=\lambda x \implies (Ax)^*=Ax^*=(\lambda x)^*=\lambda^*x^* $$
Also
$$(Ax).x^*=(\lambda x).x^*=\lambda||x||^2 $$
Similarly $$(Ax^*).x=(\lambda^* x^*).x=\lambda^*||x||^2 $$
LHS are equal since: $$(Ax).x^*=(A^Tx^*).x=(Ax^*).x$$
Now equate RHS to give:
$$\lambda^*||x||^2=\lambda||x||^2\implies \lambda=\lambda^*$$
