Unusual decomposition of 3x3 real symmetric matrices - is this possible?

If $M$ is a 3x3, real symmetric matrix, then I know there are a few ways to decompose $M$ as

$M = A^T D A$,

where $D$ is a real diagonal matrix: e.g., this can always be done for some $A \in SO(3)$, or for some lower triangular $A$. Can it always be done for some $A \in SO(2,1)$?

-Jeanne

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Doesn't same argument as for $SO(n)$ (that is, the argument that a symmetric matrix has an orthogonal basis of eigenvectors, to be found in every linear algebra text)? – Igor Rivin Aug 14 '12 at 23:07
I'm not sure - if I remember correctly, the argument works for SO(n) in part because inverse and transpose are the same thing, which isn't true for SO(2,1). I'll have to think about it some more. – Jeanne Clelland Aug 14 '12 at 23:15

This happens already in dimension $2$. You can't simultaneously diagonalize $x^2$ and $xy$, for example. I think you cannot simultaneously diagonalize $-x^2 + y^2 + z^2$ and $2xy + z^2$ (if I remember the example correctly).
@Igor: $M$ represents a quadratic form $Q_1=x^TMx$ and $\mathrm{SO}(2,1)$ is defined to be the subgroup of $\mathrm{GL}(3,\mathbb{R})$ that preserves a quadratic form $Q_2={x_1}^2+{x_2}^2-{x_3}^2$. Jeanne's problem is really to find a basis of $\mathbb{R}^3$ in which $Q_1=y^TDy$ (where $D$ is diagonal) and $Q_2={y_1}^2+{y_2}^2-{y_3}^2$, in particular, a basis (dual to the coordinates $y_i$) in which both $Q_1$ and $Q_2$ are diagonal. This cannot always be done, as I point out above. If one of the $Q_i$ (or some linear combination) were positive definite, then, of course, it could be done. – Robert Bryant Aug 15 '12 at 0:45