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Let $E(c)$ be an elementary matrix of the type to add $c$ times a row to another row when applied to another matrix on the left (with $c$ in some off-diagonal position $(i, j)$), and, with the usual notation, let $F_i(c) := \operatorname{diag}(1, \ldots, c, \ldots, 1)$, with the $c$ in position $i$, be another type of elementary matrix. Then, as these two types span the elementary matrix that switches rows, only applying them from the left, they generate $GL_n(\mathbb{R}) := G$. Let $G^+$ be those $A \in G$ with $|A| \gt 0$, and define $G^-$ similarly.

It is well-known that $G^+$ and $G^-$ are the path-components of G. I am stuck with a proof using only the above, and only basic algebra and matrix algebra (no linear algebra (like spectral theorem) results even, and certainly no topology beyond the definition of "path-connected" - using only basic topology makes this fairly trivial).

It is obvious that all $E(c)$ are path-connected to $I$. So they can by and large be ignored, as long as we remember that they could appear in any position relative to the $F_i$.

The $F_i(c)$ are path-connected to either $F_i(1) = I$, or $F_i(-1)$. Note that, obviously, no $F_i(a)$ is path-connected to $F_i(-a)$ ($a \ne 0$), nor, for $i \ne j$, $F_i(c)$ to $F_j(d)$ ($c, d \lt 0$). But only looking at $\operatorname{diag}(-1,-1) = F_1(-1) F_2(-1) \in G^+$, we see that the matrices in $G^+$ also include those whose elementary matrix decomposition includes some which do not path-connect to $I$.

There is an old question in which the answer using this approach simply claims that all matrices $A \in G$ are either path-connected to $I$, or to $F_n(-1)$, which I fail to see (eg, look at the example, or at $F_1(-1)$). It is clear that all $A \in G^+$ have an even number of $F_i(c)$ with $c \lt 0$, and that you get the same determinant if you exchange those pairs with $F_i(-c)$ both, but this would usually result in a different $\bar{A}$.

Assuming a clever way along this last thought could be found, then the $A \in G^-$ could still be connected to any $F_i(-1)$, with $i \in \{1, \ldots, n\}$, and more is needed to show that they are in one path-component.

Stuff I've shown using only similar methods, and that could be used:

1) $SL_n(\mathbb{R})$ is path-connected.

2) $SL_n(\mathbb{R})$ is normal in $G$

I feel that I'm overlooking something easy, and would appreciate feedback.

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gnometorule
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    One way of making progress might be to observe that $-I_2$ is path connected to $I_2$, because $-I_2$ is the matrix of a 180 degree rotation, and the rotation group of the plane is surely path connected. Therefore the product of any pair of $F_i(-1)$s is in the path component of the identity. – Jyrki Lahtonen Apr 29 '15 at 22:08
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    Tyvm, @Jyrki! This made me go back to it, and finish it I think. I forgot that as part of 1) at the bottom of my question, I had already shown that $F_1(-1)F_2(-1)$ can be generated by 4 $E(c)$ (which all p.c to Id). This generalizes to correcting the sign for a pair of higher dimensional matrices in the obvious way. Call this fact (X). But as we already know that all $A \in G^+$ are p.c to "an identity matrix with an even number of pairs $(-1,-1)$ on the diagonal", and p.c is an ER, then by (X), all $A \in G^+$ are p.c to the Id. – gnometorule Apr 30 '15 at 00:54
  • Similarly, we find matrices $E(c)$ that when applied in the right order, transform $F_i(-1)$ to $F_n(-1)$, which handles $G^-$ (correct the pairs as above, then apply this sequence of matrices). If no one else does, I'll write this up when I'll have time as an answer. – gnometorule Apr 30 '15 at 00:55

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