The general linear group is path connected

Idea. Every invertible matrix factors as "rotation $\times$ stretch". The rotation part lives in the connected group $SO(n)$ and the stretch part lives in a convex cone.

Step 1: polar decomposition. For $A\in GL_n(\mathbb{R})$, the matrix $A^TA$ is symmetric positive definite (since $x^TA^TAx=\|Ax\|^2>0$ for $x\ne0$). Let $P=(A^TA)^{1/2}$ be its positive definite square root (obtained from the spectral theorem by taking square roots of the eigenvalues), and $U=AP^{-1}$. Then
$$U^TU=P^{-1}A^TAP^{-1}=P^{-1}P^2P^{-1}=I,$$
so $U\in O(n)$ and $A=UP$.

Step 2: sign. $\det P>0$ (product of positive eigenvalues), so $\det A=\det U\det P$ has the sign of $\det U$. If $A\in GL_n^+$, then $\det U=1$: $U\in SO(n)$.

Step 3: a path to $I$. Let $U_t$ be a path in $SO(n)$ from $I$ to $U$ (it exists by assumption) and $P_t=(1-t)I+tP$, $t\in[0,1]$. Each $P_t$ is symmetric positive definite (convexity of the cone of positive definite matrices, Problem 1). Then
$$A_t=U_tP_t$$
is continuous, lies in $GL_n^+$ (as $\det U_t=1$ and $\det P_t>0$), and goes from $A_0=I$ to $A_1=UP=A$.

Conclusion. Every $A\in GL_n^+$ is joined to $I$ inside $GL_n^+$, so $GL_n^+$ is path connected. Together with Problem 2 and the map $A\mapsto DA$ ($D=\operatorname{diag}(-1,1,\dots,1)$), this shows that $GL_n(\mathbb{R})$ has exactly two components.

Remark. The decomposition $GL_n^+\cong SO(n)\times\{\text{positive definite matrices}\}$ is a homeomorphism (polar decomposition depends continuously on $A$). It is the source of the deformation retraction of Problem 25.