Is there a $2\times 2$ symmetric matrix that can't be diagonalized by a special orthogonal matrix? The spectral theorem guarantees an orthogonal matrix, but both the algebra and the geometry suggest this fact degenerates, in the two-dimensional case, to the existence of a special orthogonal matrix that will do the trick. (Geometrically I can't see why we would need a reflection to bring a quadratic form into standard form; algebraically there appears to always be a solution for a rotation by $\theta$, at least if I did my algebra right.)
In higher dimensions, what's the (right way to think about the) geometric obstruction to an orientation-preserving change-of-basis for the diagonalization? (I'm assuming that for $n>2$ there are indeed $n\times n$ symmetric matrices that require a reflection, even though I can't quickly write one down.) I'm thinking over the reals, if that wasn't clear.