The singular matrices of M_2(R) form a cone over a torus

Idea. The singular matrices are the null cone of the quadratic form $|z|^2-|w|^2$. A nonzero point of it has two nonzero phases $(\alpha,\beta)$ and a radius $\rho$.

Step 1: description. A matrix is singular iff $\det A=|z|^2-|w|^2=0$, i.e. $|z|=|w|$. It is zero iff $z=w=0$. A matrix of rank one is singular and nonzero: $|z|=|w|=\rho>0$.

Step 2: rank-one matrices. Consider
$$\Psi:T^2\times(0,\infty)\to\{\text{rank one}\},\qquad(e^{i\alpha},e^{i\beta},\rho)\mapsto(\rho e^{i\alpha},\rho e^{i\beta}).$$
It is continuous, bijective, and its inverse $(z,w)\mapsto(z/|z|,\,w/|w|,\,|z|)$ is continuous on $\{|z|=|w|>0\}$. So $\Psi$ is a homeomorphism: the rank-one matrices are homeomorphic to $T^2\times(0,\infty)$.

Step 3: the whole singular set is the cone. The cone over $T^2$ is $C(T^2)=T^2\times[0,\infty)/(T^2\times\{0\})$. The extended map $\overline\Psi:T^2\times[0,\infty)\to\{|z|=|w|\}$, $(e^{i\alpha},e^{i\beta},\rho)\mapsto(\rho e^{i\alpha},\rho e^{i\beta})$ is a continuous surjection that is proper (it satisfies $\|(z,w)\|=\sqrt2\rho$, so preimages of bounded sets are bounded, and $T^2$ is compact), hence a closed map. It identifies exactly the points of $T^2\times\{0\}$ (they all go to $0$). A closed continuous surjection is a quotient map, so it induces a homeomorphism $C(T^2)\cong\{\text{singular matrices}\}$.

Check. Consistency with the algebra: every rank-one matrix is $uv^T$ with $u,v\in\mathbb{R}^2\setminus\{0\}$, determined up to $(u,v)\sim(\lambda u,\lambda^{-1}v)$. Normalising $u$ to a unit vector (using the freedom $\lambda=1/|u|$), the pair $(u,v)$ becomes a point of $S^1\times(\mathbb{R}^2\setminus\{0\})\cong S^1\times S^1\times(0,\infty)$, and the only remaining ambiguity is $(u,v)\sim(-u,-v)$, a free involution on this space. The quotient of the torus factor by this free involution is again a torus, in agreement with Step 2. A numerical test on random $uv^T$ confirms $|z|=|w|$.

Remark. The singular set is *not* a manifold at $0$: removing the vertex disconnects a small neighbourhood into pieces that do not look like a ball of $\mathbb{R}^3$. The rank-one set is a smooth $3$-manifold, a "tube" $T^2\times(0,\infty)$, that is, the light cone of a quadratic form of signature $(2,2)$ minus its vertex.

Lesson. A homogeneous equation $q(z,w)=0$ defines a cone: study it through its "link" (the intersection with the unit sphere), here a torus.