Prove that every symmetric $2\times2$ matrix is diagonalizable over the reals 
Prove that every symmetric $2\times2$ matrix is diagonalizable over the reals.

Every symmetric matrix looks like this: $$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22}\end{pmatrix}$$
So my idea is to find the characteristic polynomial and show that it's discriminant is $\geq 0$. The characteristic polynomial is
$$|\lambda I-A| = (\lambda-a_{11})(\lambda-a_{22})-(a_{12})^2=\lambda^2+(-a_{11}-a_{22})\lambda+a_{11}a_{22}-(a_{12})^2 ,$$
so the discriminant is
$$(a_{11})^2+2a_{11}a_{22}+a_{22}^2-4(a_{11}a_{22}-(a_{12})^2)=(a_{11})^2-2a_{11}a_{22}+a_{22}^2+4(a_{12})^2 .$$
But I can't prove that this is always positive because there is the $-2a_{11}a_{22}$ element in the middle.
And hints? I think i'm missing something and that's probably not the correct way to prove this.

$\textbf{Hint}$: $(a_{11})^2-2a_{11}a_{22} + (a_{22})^2 = (a_{11}-a_{22})^2$
And you should be able to say something about the sign on that. (This leaves you with $4(a_{12})^2$ to deal with, but the same thing applies there.)
