Real symmetric matrix similar to diagonal matrix

Let $A$ be a $2\times 2$ matrix with real entries, which is symmetric. Prove that $A$ is similar over $\Bbb{R}$ to a diagonal matrix.

-
wande: welcome to MathSE. I see that this is your first question. So I wanted to let you know a few things about MathSE. We like to know the sources of questions - if it's homework, please add the [homework] tag. People will still help, so don't worry. We also like to know what you've tried on a problem. These sort of pleasantries usually result in more and better answers. Finally, I should add that posting questions in the imperative (i.e. Compute all such, Prove that...) is considered rude by some of the members, so it would be nice of you to change that wording. Thank you. –  Arturo Magidin Sep 6 '11 at 19:13

You want to find an invertible matrix $S$ such that $S^{-1} A S = \Lambda$, where $\Lambda$ is a diagonal matrix. So, you need to solve $AS=S\Lambda$ for both $S$ and $\Lambda$, and you want $S$ invertible and $\Lambda$ diagonal. Looking at the columns, you see that you need to solve the system $Ax= \lambda x$ for $x \in \mathbb R^2, x\ne0$ and $\lambda \in \mathbb R$. When $A$ is symmetric, this system gives rise to a quadratic equation for $\lambda$ whose discriminant is non-negative and so has real solutions. If there are two solutions, the corresponding $x$ form a basis of $\mathbb R^2$ and the linear transformation defined by $A$ has a diagonal matrix with respect to that basis. This should get you started.
(As hinted above, the key to understanding this problem is to consider the linear transformation $T$ defined by $A$ in $\mathbb R^2$. Similarity then corresponds to a change of basis. The matrix of $T$ with respect to the canonical basis is $A$. You want to find a basis such that the matrix of $T$ with respect to that basis is diagonal.)