Prove that if $A \in M_{2\times2}\mathbb {(R)}$ is symmetric then A is diagonalizable Given that: $$A \in M_{2\times2} \mathbb {(R)}$$
we have to prove that $A$ is diagonalizable.
As in:

$$\text{There exists a turnable matrix } P \; (\text{det(P) != 0 }) \; \text{such that}:$$
  $$P^{-1} \cdot A \cdot P = D$$

When $D$ is:
$$\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \; , \; \lambda_{1,2} \; \text{are the eigenvalues of A} $$
Any help will be appreciated.

As $A$ is symmetric, it looks like 
$$ A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$$
has trace $a+d$, determinant $ad-b^2$, and eigenvalues are the soluitons of 
$$X^2-(a+d)X+(ad-b^2)=0. $$
Since $(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2$ is nonnegative, all eigenvalues are real. And unless $a=d, b=0$, they are distinct, thus making $A$ diagonalizable. In the exceptional case $a=d, b=0$, matrix $A$ is already in diagonal form.
