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EDIT 1: I've made a major mistake in the calculations leading up to this question. The question should be regarded as "on hold" until further notice. Sorry for any inconvenience!

EDIT 2: I've updated my question. It it hopefully correct and more transparent now.

I have a real $3 \times 3$ matrix $A$ with entries $a_{ij}.$ I want to find the $9$-dimensional volume of the region satisfying the following three constraints:

  1. $\mathrm{tr} (A)<0,$
  2. $\det (A) > b \,\mathrm{tr}(A),$ and
  3. $\mathrm{tr}(A^TA)\leq 1,$

where $$b=a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33}-a_{12}a_{21}-a_{13}a_{31}-a_{23}a_{32},$$ and where the third constraint can also be written as $\sum_{i,j=1}^3 a_{ij}^2\leq1,$ i.e., the volume is bounded by the unit $9$-ball.

How should I set up the appropriate integrals for finding this volume? I'm at a bit of a loss here.

A numerical approach would also be of interest, but as I don't know how to set the integrals up in the first place, I don't know how to approach the problem numerically either.

My previous question provides some context.

Thank you.

Bobson Dugnutt
  • 10,912
  • Why not integrate over $S^8$? If you scale $A$ by $\lambda>0$, then the first condition is preserved, in the second condition, both sides are scaled by $\lambda^3$ and the third inequality is scaled by $\lambda^2$, so it is true for all matrices on the line between $0$ and $A$ inside the sphere). Therefore, you could drop the third condition. The first condition means that you're integrating over a hemisphere, so you're left with the third condition. How does that simplify (if you write out the determinant and the product)? – Michael Burr Oct 14 '17 at 15:19
  • In general, if you want the surface area measure of a unit hemisphere $S^{d-1}$, then you can integrate over the disk $D^{d-1}$ and surface area measure is $\frac{1}{x_d}dx_1\dots dx_{d-1}$. – Michael Burr Oct 14 '17 at 15:24
  • @MichaelBurr What you are saying is that the region defined by the two first constraints is a cone, right? That is certainly true. I included the third constraint in order to bound the region (otherwise the volume is infinite). So I don't understand how can I drop the third condition? Are you saying that I can do this is I instead integrate over the unit $9$-sphere (i.e., the surface of a unit $9$-ball)? You, however, also write "so you're left with the third condition". I'm must admit, I'm a bit confused. I'd appreciate it if you were to write this whole thing out as an answer. – Bobson Dugnutt Oct 14 '17 at 15:48
  • I meant second condition. Yes, I'm thinking that you might integrate over the sphere instead of the cone. I'll try to write out an answer when I have time. – Michael Burr Oct 14 '17 at 17:29
  • @MichaelBurr Ah, I just now realized what you meant in your first comment, it all makes more sense now! Btw, you asked how the second constraint simplifies. It actually simply becomes $(1+\mathrm{tr}(A))\det(A)>0.$ Your idea about only integrating over one hemisphere of the unit sphere sounds intriguing, but I don't quite understand how this is equivalent with the original question. Either way, thanks for the ideas so far. – Bobson Dugnutt Oct 16 '17 at 00:44
  • @MichaelBurr Hmm, I don't think the second condition actually simplifies to what I wrote in my previous comment. Either that or I've made a mistake in an earlier calculation. I'll look at it first thing tomorrow. – Bobson Dugnutt Oct 16 '17 at 01:21
  • I tried a million random matrices, and found 104699 with all eigenvalues' real parts positive; 395881 with two positive; 395569 with one positive and two negative; and 103851 with all three negative. So 10.4 percent or so. The matrices nine entries were all Gaussian normal random real numbres, and the matrices were not symmetric. – Empy2 Oct 16 '17 at 01:48
  • @Michael Yep, I have simulated it as well, with a billion matrices, and found 0.10450(1). When I wrote about how to approach the problem numerically, I meant the integral. I should have been clearer about that. But that is exactly the probability I'm looking to calculate exactly. Nice catch! – Bobson Dugnutt Oct 16 '17 at 01:59
  • @MichaelBurr I've now completely rewritten my question, which is hopefully clearer. I'm interested in what you make of the updated (correct) conditions. Link to question. Thanks. – Bobson Dugnutt Oct 17 '17 at 14:21

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