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.

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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.)

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    $\begingroup$ There is, of course, some work to be done in case the two solutions are equal. $\endgroup$ – Gerry Myerson Sep 7 '11 at 6:10

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