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.
 A: 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.
A: 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.
A: Any symmetric matrix is normal; i.e. $A^T A = A A^T = A^2$. Hence, according to Corollary 7.1.4 of "Matrix Computations" by Golub & Van Loan, 3rd ed. (page 314) it is diagonalizable.
A: Here is a constructive proof. Let $c=\cos\theta$ and $s=\sin\theta$. Then
$$
\pmatrix{c&s\\ -s&c} \pmatrix{x&y\\ y&z} \pmatrix{c&-s\\ s&c}
=\pmatrix{
c^2x + 2scy + s^2z
&(c^2-s^2)y - sc(x-z)\\
(c^2-s^2)y - sc(x-z)
&c^2z - 2scy + s^2x}.
$$
Hence the RHS is diagonal if we pick a $\theta$ such that
$$
(c^2-s^2)y - sc(x-z) = y\cos(2\theta) - \frac{x-z}{2}\sin(2\theta) = 0.
$$
