3d volume of the set of $(2\times2)$-matrices of rank $\leq1$ and norm $\leq1$

Question

The problem:Let $M$ be the set of all $2\times2$ matrices $A$ with rank less than or equal to 1 and $\left|A\right|\le1$,where $|A|^2$ denotes the sum of squares of the entries of $A$.Find the three-dimensional volume of $M$.
My thought to this question is as follows:
The condition of $A$ with rank $\le1$ means that $\begin{pmatrix} a_{11} &a_{12} \\ a_{21} &a_{22} \end{pmatrix}$ can be represented as $\begin{pmatrix} t &pt \\ kt &pkt \end{pmatrix}$,$where \, t, p, k \in \mathbb{R}$. Also $\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}\le1$ since $|A|\le1$.
Let $f\colon \mathbb{R^3} \to \mathbb{R}$ be defined by the equation $f(t,p,k)=\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}$,where $t=a_{11}, pt=a_{12},kt=a_{21},pkt=a_{22}$.
In the above, let's define $g\colon \mathbb{R^3} \to \mathbb{R^4}$ by $g(t,p,k)=(a_{11},a_{12},a_{21},a_{22})=(t,pt,kt,pkt)$.
Then find the Jacobian matrix $D$ of $g$.
The ineqaulity $f(t,p,k)=\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}\le1$ define a region $S$ in $\mathbb{R^3}$,then maybe we can parametrize it .Then find $\int_{S}\sqrt{\det{D^TD}}$ to find the volume of $\mathbb{R^3}$ in $\mathbb{R^4}$.
Am I correct?Is there a better approach?

Answer 1

Your plan is correct, but there is great danger that you will run into computational difficulties. Apart from this we have to make sure that we have an essentially one-one parametrization of $M$, or else we might obtain the double of the desired volume.

I propose to use the following parametrization of $M$: $${\bf f}:\quad (t,u,v)\mapsto\quad t\ \left[\matrix{\cos u\cr\sin u}\right]\bigl[\matrix{\cos v&\sin v\cr}\bigr]\ ,$$ with parameter domain $[0,1]\times[0,2\pi]\times[0,2\pi]$. This leads to the matrix elements $$a_{11}=t\cos u\cos v,\quad a_{12}=t\cos u\sin v,\quad a_{21}=t\sin u\cos v,\quad a_{22}=t\sin u\sin v\ ,$$ hence $a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2=t^2$. It is easy to check that ${\bf f}(t,u,v)={\bf f}(t',u',v')\>$ iff $\>t=t'$ and $(u,v)=(u',v')$, or $t=t'$ and $(u,v)=(u'\pm\pi,v'\pm\pi)$. This means that ${\bf f}$ produces a double covering of $M$.

We need the matrix $$D:=\bigl[d{\bf f}(t,u,v)\bigr]=\left[\matrix{ {\partial a_{11}\over\partial t}& {\partial a_{11}\over\partial u}& {\partial a_{11}\over\partial v}\cr {\partial a_{12}\over\partial t}& {\partial a_{12}\over\partial u}& {\partial a_{12}\over\partial v}\cr {\partial a_{21}\over\partial t}& {\partial a_{21}\over\partial u}& {\partial a_{21}\over\partial v}\cr {\partial a_{22}\over\partial t}& {\partial a_{22}\over\partial u}& {\partial a_{22}\over\partial v}\cr}\right]=\ldots\quad.$$

The Gramian simplifies considerably: One computes $$D^\top D= \left[\matrix{1&0&0\cr 0&t^2&0\cr0&0&t^2\cr}\right]\ ,$$ so that $\sqrt{{\rm det}\bigl(D^\top D\bigr)}=t^2$. In this way we obtain $${\rm vol}(M)={1\over2}\int_{[0,1]\times[0,2\pi]\times[0,2\pi]}t^2\>{\rm d}(t,u,v)={2\pi^2\over3}\ .$$