$GL_n(\mathbb{R})$ is the union of two path connected subsets (Michael Artin's Algebra). I have that $SL_n(\mathbb{R})$ is path connected and generated by elementary matrices of the form $I + a e_{i,j} (i \neq j)$, where $e_{i,j}$ is the matrix with $1$ at position $(i,j)$ and zero elsewhere.  I also have that $GL_n(\mathbb{R})$ is path connected and generated by elementary matrices of the above type together with ones of the form $I + c e_{i,i}$.  So clearly I already have a union of two path connected subsets, but I was wondering if there is a smaller second set, for instance, matrices that have at least one factor of the second type.  Hints are welcome.
 A: At first i wanted to mention that $GL_n(\mathbb{R})$ is not connected, the determinant is continuous and the image of a connected space under a continuous function is a connected space . 
But the image is $\mathbb{R}\setminus \{0\}$ which is surely not connected..
On the other hand 
$GL_n(\mathbb{R})$ is surely
$$GL_n(\mathbb{R})=GL_n^- (\mathbb{R}) \cup  GL_n^+ (\mathbb{R})$$
Now think of Row reduced Echolon form and you will find continuous generators, for each of them.
For the generating system of $GL_n^+ (\mathbb{R})$ you make the following: 
\begin{align*}
B_{b}^\lambda=(b_{k,l})_{1\leq k,l\leq n}=
\left\{
\begin{array}{ll}
\lambda & k=i \text{ and } l=j\\
0 & \text{ else }\\
\end{array}
\right.
\end{align*}
\begin{align*}
\gamma_{b}(t)=E_n+t\cdot B_{b}^\lambda
\end{align*}
\begin{align*}
C(t)=(c_{k,l}(t))_{1\leq k,l \leq n}&=\left\{
\begin{array}{cl}
1 & k=l \text{ with } k,l\neq i,j,\\
-\sin{(\pi \cdot t)} & k=i,\ l=j\\
\sin{(\pi \cdot t)} & k=j, \ l=i\\
\cos{(\pi \cdot t)} & k=l=i \\
\cos{(\pi \cdot t)} & k=l=j \\
0 & \text{else}\\
\end{array}
\right.\\
\gamma_{c)}(t)&=C(t)
\end{align*}
\begin{align*}
A(t)=(a_{kl})_{1\leq k,l\leq n}=
\left\{
\begin{array}{ll}
1 & k=l\neq i\\
1+t\cdot (\lambda-1) & k=l=i\\
0 & \text{else}\\
\end{array}
\right.
\end{align*}
with $\lambda > -1$
$t$ is always in $[0,1]$.
A: 
Exercise 2.M.8(b) (Artin's Algebra, 2nd edition). Show that $GL_n(\mathbb{R})$ is a union of two path-connected subsets, and describe them.

Let $\sim_1$, $\sim_2$ be the binary operations corresponding to path-connectivity in $SL_n(\mathbb{R})$, $GL_n(\mathbb{R})$, respectively; by my answer here, $\sim_1$, $\sim_2$ are equivalence relations.
In the same spirit as my answer here, we consider the map$$M\to f(c,M)$$that multiplies the first row of $M$ by $c$, and note that $f(c,M)$ is continuous in $c$ for fixed $M$ (but multiplies the determinant by $c$). If $c>0$ and $M\in GL_n(\mathbb{R})$, the continuous function$$X(t) = f(1 + (c-1)t\,,M)$$on $[0,1]$ takes$$X(0)=M \to X(1) = f(c,\,M),$$ while keeping$$1+(c-1)t\ne 0$$at all times (we only hit the interval $[1,c]$) and thus remaining inside $GL_n(\mathbb{R})$. Hence,$$f(c,M)\sim_2 M.$$Yet any $B\in GL_n(\mathbb{R})$ can be written as $f(\epsilon\det{B},\,\epsilon A)$ for some $\epsilon\in\{-1,1\}$ and $A\in \epsilon SL_n$ such that$$\epsilon\det{B}>0,$$so$$B\sim_2 \epsilon A\sim_2 \epsilon I_n$$by here and the trivial observation$$P\sim_1 Q\iff dP\sim_2 dQ$$ for nonzero scalars $d$. Thus,$$S_+ = \{B\in GL_n(\mathbb{R}): \det(B)>0\},\text{ }S_- = \{B\in GL_n(\mathbb{R}): \det(B)<0\}$$are path-connected subsets covering $GL_n(\mathbb{R})$.
On the other hand, $B_+\in S_+$ and $B_-\in S_-$ can not be connected in $GL_n(\mathbb{R})$. Suppose otherwise, so some continuous function$$X:[0,1]\to\mathbb{R}^{n\times n}$$takes$$X(0)=B_+,\text{ }X(1)=B_-$$while staying inside $GL_n(\mathbb{R})$. Then $\det{X(t)}$, a polynomial in the $n^2$ continuous component functions of $X$, is continuous itself. However, $$\det{X(0)}>0>\det{X(1)},$$so $\det{X(t)}$ must vanish somewhere by the Intermediate Value Theorem. This is clearly absurd, so we indeed have $$B_+\not\sim_2 B_-,$$and that $\sim_2$ partitions $GL_n(\mathbb{R})$ into $S_+$ and $S_-$.
A: $GL_n(\mathbb R)$ is not path connected. Think about determinants.
