EDIT: At this point, geometric interpretations of conditions 2-4 would qualify as an answer. This can include symmetries of the region.

I have a real $3 \times 3$ matrix $A$ with entries $a_{ij},$ and I want to find out how much of the unit $9$-ball the $9$-dimensional volume of the region in which $A$ is stable (meaning all its eigenvalues have negative real part) takes up. However, since this region is a cone, it suffices to look at how much of the surface of the $9$-ball the region takes up, so we can look at

  1. $\sum_{i,j=1}^3 a_{ij}^2=1.$

We can prove (see below) that $A$ is stable iff

  1. $\mathrm{tr}(A)<0$
  2. $\det(A)<0$
  3. $\mathrm{tr}^3(A)<\mathrm{tr}(A^3)$

are all met.

To be more precise, if we let $\mathcal{C}$ denote the set of all $3 \times 3$ matrices satisfying all four of the above conditions and let $\mathcal{S}^8$ denote the $8$-sphere (i.e. the surface of the unit $9$-ball), I'd like to find $$R=\frac{\mathrm{vol}(\mathcal{C})}{\mathrm{vol}(\mathcal{S}^8)}.$$

But how would I set up the integrals for actually finding this? Or, better than messing with integrals, can we perhaps infer this ratio from looking at the symmetries of conditions 2-4?

From simulations, $R\sim 0.1045.$

Any help is much appreciated. For instance, what would the geometric interpretations of conditions 2-4 be?

Proof: (Not necessary reading)

The (monic) characteristic polynomial for $A$ is $$p_3(z)=z^3+c_2z^2+c_1z+c_0,$$ where \begin{align} c_0 &= -\det A\\ c_1 &= \frac{1}{2}\left[\mathrm{tr}^2(A)-\mathrm{tr}(A^2)\right]\\ c_2 &= -\mathrm{tr}A.\\ \end{align}

$A$ is stable iff the characteristic polynomial satisfies The Hurwitz Stability Criterion. Written out for the present case, it reduces to

\begin{align} \Delta_1&=c_2 && \mkern-18mu \mkern-18mu >0 \\ \Delta_2&=c_1c_2-c_0 && \mkern-18mu \mkern-18mu >0 \\ \Delta_3&=c_0\Delta_2 && \mkern-18mu \mkern-18mu >0, \end{align}

of which the two first can be written as $\mathrm{tr}(A)<0$ and $\mathrm{tr}^3(A)<\mathrm{tr}(A^3).$

Now, we'd like to reduce the third inequality. The idea is that we know that $\det(A)<0$ is necessary for $A$ being stable (since the determinant of a matrix is the product of its eigenvalues). This means we can divide with $c_0$ without changing the inequality, at which point the third inequality reduces to the second. Note that we're ignoring singular matrices, since their contribution to the volume is zero anyway. $\Box$

  • $\begingroup$ @MrYouMath Thank you for the reference, I didn't know about that theorem. When you write the eigenvalue equation, do you then mean the solved characteristic equation? If you simply mean the characteristic function, I've already written that in the question. Also, the final three conditions I list are the inequalities resulting from using the Hurwitz Stability Criterion, and I unfortunately don't think the Liénard–Chipart Criterion results in any simpler inequalities. I'm not sure exactly what your suggestion is. Could you perhaps elaborate? Thanks. $\endgroup$ – Bobson Dugnutt Oct 18 '17 at 8:18
  • $\begingroup$ @MrYouMath I have written them out explicitly. The result are the listed conditions 2.-4. $\endgroup$ – Bobson Dugnutt Oct 18 '17 at 8:43
  • $\begingroup$ Have you tried solving the simpler of $A\in\mathbb{R}^{2\times2}$ so it only requires solving for the area of a 4-ball, since often solving a simpler problem can give you insight into the more difficult problem. Another thought I had was solving for the contour of the area, which only requires that $\det(A)=0$. $\endgroup$ – Kwin van der Veen Oct 20 '17 at 18:13
  • $\begingroup$ @KwinvanderVeen Yes, I've solved the problem in that case. The result is $R=\frac{1}{4}.$ You can see a nice proof for it here. The inequalities were much simpler in that case, however. W.r.t. to the contour-idea, what made you put $\det(A)=0,$ and not for instance $\mathrm{tr}(A)=0$? Also, by contour, do you mean boundary? $\endgroup$ – Bobson Dugnutt Oct 20 '17 at 18:35
  • 1
    $\begingroup$ @Lovsovs With contour indeed meant the boundary between the regions where $A$ is and is not Hurwitz. For this you need the determinant and not the trace. Because the determinant is the product of eigenvalues, while the trace is the sum. In order for a matrix pass from Hurwitz to not Hurwitz one of the eigenvalues has to become zero and therefore its determinant as well. Also using spherical coordinates might help (removes one constraint and coordinate). $\endgroup$ – Kwin van der Veen Oct 20 '17 at 18:58

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