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.

Here is a possible solution (although perhaps not the best). Let 
$$A=\begin{pmatrix} a & c \\ c & b \end{pmatrix}.$$
Then $p(\lambda)= |A-\lambda I| = \lambda^2 - (a+b) \lambda + ab -c^2$. You can easily check that this has two distinct real roots (your eigenvalues $\lambda_1$ and $\lambda_2$). Thus you have $2$ linearly independent eigenvectors $e_1$ and $e_2$. These, if you put them as columns of your matrix $P$, will diagonalize your matrix.
