GL_n is dense and its boundary is the singular set

Idea. A matrix can be made invertible by adding a tiny multiple of the identity: this only fails for finitely many values of the multiple.

Step 1: density. Fix $A\in M_n(\mathbb{R})$. The function $p(t)=\det(tI+A)$ is a *monic* polynomial of degree $n$ in $t$ (it equals $(-1)^n\chi_A(-t)$, where $\chi_A$ is the characteristic polynomial). A non-zero polynomial has finitely many roots, so there is $\delta>0$ such that $p(t)\neq0$ for $0<|t|<\delta$. For such $t$, $A+tI\in GL_n(\mathbb{R})$ and $\|A+tI-A\|=|t|\to0$. Hence $A\in\overline{GL_n(\mathbb{R})}$.

(Equivalently: $A+tI$ is singular iff $-t$ is an eigenvalue of $A$, and $A$ has at most $n$ eigenvalues.)

Step 2: the boundary. For any set $E$, $\partial E=\overline E\setminus\operatorname{int}E$. Here $E=GL_n$ is open (previous problem), so $\operatorname{int}E=E$, and by Step 1 $\overline E=M_n(\mathbb{R})$. Therefore
$$\partial GL_n(\mathbb{R})=M_n(\mathbb{R})\setminus GL_n(\mathbb{R})=\{A:\det A=0\}.$$

Remark.
- Equivalently, the singular matrices form a closed set with empty interior (they are the zero set of a non-zero polynomial). This is the key reason "generic matrices are invertible".
- A typical use: to prove an identity between continuous functions of $A$ (such as $\det(AB)=\det(A)\det(B)$ or Cayley-Hamilton) it suffices to prove it on the dense set $GL_n$, or even on the dense set of matrices with distinct eigenvalues.