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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.

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Every symmetric matrix can be diagonalized by a special orthogonal matrix. To say that $A$ is diagonalized by $O$ is to say that $AO = OD$ where $D$ is diagonal. This is equivalent to the statement that the columns of $O$ consist of an orthonormal basis of eigenvectors of $A$, and you can permute such an orthonormal basis if necessary to ensure that $\det O = 1$.

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    $\begingroup$ oh, yes, so stupid of me -- swapping rows, if necessary, flips the determinant back to 1. thanks. $\endgroup$ – symplectomorphic Nov 28 '12 at 23:05

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