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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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  • $\begingroup$ 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)? $\endgroup$
    – Igor Rivin
    Commented Aug 14, 2012 at 23:07
  • $\begingroup$ 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. $\endgroup$
    – Jeanne Clelland
    Commented Aug 14, 2012 at 23:15

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Unfortunately, the answer is 'no'. You are basically asking whether you can simultaneously diagonalize two quadratic forms in three variables, and the answer is that, 'generically' you can (and you always can if some linear combination of the two is definite), but there are special pairs that cannot be simultaneously diagonalized.

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).

Generally, if the (real) null cones of the two indefinite quadratic forms are tangent, then they can't be simultaneously diagonalized.

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  • $\begingroup$ @Robert: why TWO forms? $\endgroup$
    – Igor Rivin
    Commented Aug 14, 2012 at 23:53
  • $\begingroup$ @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. $\endgroup$
    – Robert Bryant
    Commented Aug 15, 2012 at 0:45
  • $\begingroup$ Thanks, Robert! The case I have in mind is when both quadratic forms have signature (1,2), so it sounds like I'm probably out of luck. $\endgroup$
    – Jeanne Clelland
    Commented Aug 15, 2012 at 1:23
  • $\begingroup$ @Jeanne: But is there any chance that some linear combination of them is definite? Also, remember that simultaneous diagonalizability is an open property on pairs of quadratic forms over the reals, so you have to be somewhat unlucky to be in the case that you can't do it. $\endgroup$
    – Robert Bryant
    Commented Aug 15, 2012 at 4:07
  • $\begingroup$ @Robert: I've been playing with precisely the question of whether some linear combination of them is definite; I don't think there is such a linear combination, but I haven't tried to write out a rigorous proof of that yet. (The restriction on these forms is that the sum of these two forms of signature (1,2) has signature (2,1).) Is the simultaneously diagonalizable property dense as well as open? If so, I'd settle for that; I can live with knowing that it's true generically. $\endgroup$
    – Jeanne Clelland
    Commented Aug 15, 2012 at 4:45

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