# Calculating a $9$-dimensional volume

I'd like to calculate the $9$-dimensional volume of the set of all $3\times 3$ matrices whose eigenvalues have negative real part, bounded by the unit $9$-ball.

More precisely, if we define $\lambda_{1,2,3}$ to be the three solutions to the characteristic equation $p_3(z)=0$ for a $3 \times 3$ matrix $A$ with entries $a_{ij}$, I'd like the calculate the volume of the region

$$\mathcal{S}:=\{(a_{ij})_{i,j=1}^3\in\mathbb{R}^9:\Re[\lambda_{1,2,3}]<0\; \wedge\; \sum_{i,j=1}^3 a_{ij}^2\leq1\}.$$

I suspect (and hope) that there is a nice way to utilize the symmetries of $\mathcal{S}$ (like for the $2\times 2$ case), because setting up the integral(s) seems almost impossible. Nevertheless, below is what is needed to do so (not required reading, it doesn't really lead anywhere).

Any help with regards to either setting up the integral or (preferably) showing how some symmetry can be used would be greatly appreciated.

The characteric polynomial is $$p_3(z)=c_3z^3+c_2z^2+c_1z+c_0,$$ where \begin{align} c_0 &= \det A\\ c_1 &= a_{12} a_{21}-a_{11} a_{22}+a_{13} a_{31}+a_{23} a_{32}-a_{11} a_{33}-a_{22} a_{33}\\ c_2 &= \mathrm{tr}A\\ c_3 &= -1, \end{align} and has the roots

\begin{align} \lambda_1&= -\frac{1}{3}a_2+(S+T)\\ \lambda_2&= -\frac{1}{3}a_2+\frac{1}{2}(S+T)+\frac{1}{2}i\sqrt{3}(S-T)\\ \lambda_3&= -\frac{1}{3}a_2+\frac{1}{2}(S+T)-\frac{1}{2}i\sqrt{3}(S-T),\\ \end{align} following the notation here (see Eq.s 22-23 and 49-51).

Now, we want to see what happens when we take the real part (so we can get some inequalities to integrate over). While we cannot say anything about which of them are complex, we can say that when the determinant $D$ is positive, there is one real root, while the two others are complex conjugates of each other. When $D$ is negative, all three roots are real and distinct. Lastly, when $D$ is zero, all three are real, but at least two are equal. However, I don't think we need to bother about this last case, since it covers a zero $9$-dimensional volume. I think using $D$ to split up the integrals in this way could be useful.

• Huh, you're right! I'd only need to look at the first two principal minors then, since my $H(p)=\begin{pmatrix}c_2 & c_0 \\ 1 & c_1 \end{pmatrix},$ right? If you would be so kind as to show how to set up the integral (or just so I get the idea of it, I'm a bit confused about how to approach it), I'd happily mark this an answer. – Bobson Dugnutt Oct 14 '17 at 14:08