A path from positive to negative determinant hits a singular matrix

Idea. The function $f(t)=\det A(t)$ is a continuous real function on an interval, so it must vanish between a positive and a negative value.

Step 1. $\det:M_n\to\mathbb{R}$ is continuous (Problem 66) and $t\mapsto A(t)$ is continuous, so $f=\det\circ A:[0,1]\to\mathbb{R}$ is continuous.

Step 2. $f(0)>0>f(1)$. By the intermediate value theorem there exists $t_0\in(0,1)$ with $f(t_0)=0$, i.e. $A(t_0)$ is singular.

Example. The path $A(t)=\operatorname{diag}(1-2t,1)$ joins $I$ ($\det=1$) to $\operatorname{diag}(-1,1)$ ($\det=-1$) and passes through the singular matrix $\operatorname{diag}(0,1)$ at $t=\frac12$. The rotation path $R_{\pi t}$ in $SO(2)$ never leaves the determinant-$1$ level set.

Consequence. Any path in $GL_n(\mathbb{R})$ keeps a constant sign of $\det$. Hence $I$ and $\operatorname{diag}(-1,1,\dots,1)$ lie in *different path-components* of $GL_n(\mathbb{R})$: $GL_n(\mathbb{R})$ is not connected (see Part 5).

Remark. The statement uses that a path is defined on a *connected* interval; it is the one-dimensional case of "the continuous image of a connected set is connected".