A geometric property of singular matrices

Question

Let $S\subset M_{n}(\mathbb{R})$ be the singular points of the equation $Det=0$. That is $S$ is the critical points of the determinant function.

What matrices belongs to $S$, precisely?

Let $M=Det^{-1}\{0\}-S$ be the codimension one submanifold of $M_{n}(\mathbb{R})$ which has a natural Riemannian metric induced by the standard metric of $M_{n}(\mathbb{R})\simeq\mathbb{R}^{n^{2}}.$

What is a linear algebraic and matrix meaning for a matrix $A\in M$ with the following property:

"The sectional curvature of $M$ at $A$ is independent of choosing a $2$-plane tangent to $M$ at $A$"

What is a precise example of this situation, for $n=2$?

Answer 1

Your first question has an easy answer. The differential of $\det$ is $$\sum_{i,j}\hat a_{ij}{\rm d}a_{ij},$$ where $\hat A$ is the cofactor matrix. Thus a singular point is such that $\hat A=0_n$, in other words, it has rank $\le n-2$.

Actually, ${\bf M}_n({\mathbb R})$ can be stratified by the sets $R_0,\ldots,R_n$ of matrices of rank $k=0,\ldots n$ respectively. Each $R_k$ is a submanifold of dimension $k(2n-k)$. $R_k$ is homogeneous, in the sense that ${\bf GL}_n({\mathbb R})\times{\bf GL}_n({\mathbb R})$ operates transitively on it by $(P,Q)\cdot A=PAQ^{-1}$. The relative boundary of $R_k$ is $R_0\cup\cdots\cup R_{k-1}$. In particular, there is no removable singularity.