Timeline for Unusual decomposition of 3x3 real symmetric matrices - is this possible?
Current License: CC BY-SA 3.0
12 events
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| Aug 16, 2012 at 18:10 | comment | added | Robert Bryant | @Jeanne: In the case in which $p(x,y)$ has a linear factor and an irreducible quadratic factor (and $Q_2$ has type $(2,1)$), I think you can put them in the form $$ Q_1 = \lambda_1\ x^2 + \lambda_2\ (y^2 - z^2) + 2\lambda_3\ yz\ , $$ where $\lambda_3 > 0$ and $$ Q_2 = x^2 + y^2 - z^2\ . $$ | |
| Aug 15, 2012 at 16:19 | comment | added | BS. | Assuming $Q_1$ non-degenerate, the two forms are sim diag iff $Q_^{-1}Q_2$ is diagonalizable over $\mathbb{R}$. On the other hand, in the "generic" case (no tangency), the criterion for sim diag seems to be that the intersection of real nullcones is either zero or four lines, but I only quickly checked. | |
| Aug 15, 2012 at 14:54 | comment | added | Jeanne Clelland | @Robert: Related question: in the case where they can't be simultaneously diagonalized, is there some other normal form that they can be put into? | |
| Aug 15, 2012 at 13:07 | comment | added | Jeanne Clelland | @Robert: Thanks much; that really helps! | |
| Aug 15, 2012 at 12:10 | comment | added | Robert Bryant | @Jeanne: It's not dense. However, consider the homogeneous cubic $p(x,y) = \det(x Q_1 + y Q_2)$. If $p$ has three distinct, real linear factors (which is an open condition on the pair $(Q_1,Q_2)$), then $Q_1$ and $Q_2$ are simultaneously diagonalizable. For $p$ to have distinct factors is an open, dense condition, but 'distinct and real' is open but not dense. This condition on $p$ is not an if-and-only-if, so my above claim that the set of $(Q_1,Q_2)$ that are simultaneously diagonalizable is 'open' is not correct. After all, $(0,0)$ is simultaneously diagonal, but not all 'small' pairs are. | |
| Aug 15, 2012 at 4:45 | comment | added | Jeanne Clelland | @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. | |
| Aug 15, 2012 at 4:07 | comment | added | Robert Bryant | @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. | |
| Aug 15, 2012 at 1:23 | vote | accept | Jeanne Clelland | ||
| Aug 15, 2012 at 1:23 | comment | added | Jeanne Clelland | 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. | |
| Aug 15, 2012 at 0:45 | comment | added | Robert Bryant | @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. | |
| Aug 14, 2012 at 23:53 | comment | added | Igor Rivin | @Robert: why TWO forms? | |
| Aug 14, 2012 at 23:16 | history | answered | Robert Bryant | CC BY-SA 3.0 |