I am trying to solve a problem in Hatcher. I reduced my problem to showing that if $f:\mathbb{R}^n\to\mathbb{R}^n$ is an invertible linear map, then $f$ is homotopic to a composition of reflections, i.e there is a path of invertible matrices between $f$ and a diagonal matrix with only $\pm 1$ on the diagonal.
The hint is to use Gaussian elimination. Please help.
Consider the matrix $I + E_{ij}$ where $E_{ij}$ has a $1$ in the $ij$ slot and a $0$ elsewhere. Then, for any matrix $X$, the product $(I+E_{ij})(X)$ is what you get by adding row $j$ of $X$ to row $i$.
For $i\neq j$, form a path $\gamma(t) = (I+tE_{ij})X$. Notice first that $\det(\gamma(t)) = \det(I+tE_{ij})\det (X) = 1\det(X)\neq 0$, so $\gamma(t)$ is invertible for all $t$.
Further, $\gamma(0) = (I+0)X = X$, and $\gamma(k) = (I+kE_{ij})(X)$, so $\gamma$ is a path which starts at $X$ and ends at the matrix obtained by starting with $X$ and adding $k$ times row $j$ to row $i$.
In other words, there is a path through invertible matrices starting at any arbitrary invertible matrix, and ending at the resulting matrix obtained by performing the "add $k$ times a row to a different row" row operation. Using these kinds row operations repeatedly, you can move any $X$ to a diagonal matrix. Further, the diagonal entries of this resulting diagonal matrix must all be non-zero, because the diagonal matrix is invertible.
In a similar fashion, for $k>0$, $\gamma(t) = (I+tkE_{ii})X$ is a path starting at $X$ and ending at the matrix obtained by multiplying row $i$ of $X$ by $k$. Using these on a diagonal matrix whose entries are all non-zero, you can move it to a diagonal matrix with $\pm 1$ on the diagonals.
(In case it matters, one can further guarantee that at most one $-1$ appears, by using paths involving the rotation matrices $\begin{bmatrix} \cos t & \sin t\\ -\sin t & \cos t\end{bmatrix}$.)