Show for any matrix $A$ with positive determinant, there is a continuous path in $GL_+(n,R)$ from $A$ to the identity. Can someone please help me prove the following?
Prove $GL_+(n,R)$ is path connected. That is, show that for any matrix $A$ with positive determinant, there is a continuous path in $GL_+(n,R)$ from $A$ to the identity.
$\textbf{My idea:}$ Start with diagonal matrices, then upper/lower-triangular matrices, then arbitrary matrices.
 A: There are several ways to do this depending on which matrix decompositions you like the best. Using QR decomposition it suffices to prove the result for matrices in $SO(n)$ and for upper triangular matrices with positive diagonal.

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*For upper triangular matrices $U$ with positive diagonal it is very easy and a linear interpolation $(1 - t) U + t I$ works (every intermediate step is another upper triangular matrix with positive diagonal; the space of such matrices is convex).


*For matrices in $SO(n)$ you can use the spectral theorem to show that the exponential map $\exp : \mathfrak{so}(n) \to SO(n)$ is surjective, then linearly interpolate in $\mathfrak{so}(n)$ (every element of $SO(n)$ is a direct sum of rotations and you are interpolating the degrees of the rotations), or you can use Givens rotations.
Similarly, using singular value decomposition it suffices to prove the result for matrices in $SO(n)$ and for diagonal matrices with positive diagonal (also convex), and using polar decomposition it suffices to prove the result for matrices in $SO(n)$ and for positive-definite symmetric matrices (also convex).
The appearance of $SO(n)$ in all of these results can be explained abstractly as follows: every connected Lie group $G$ has a maximal compact subgroup $K$, and as a manifold is diffeomorphic to $K \times \mathbb{R}^n$. (So $G$ deformation retracts onto $K$, meaning the two are homotopy equivalent and share e.g. the same homotopy groups and (co)homology, and in particular the same $\pi_0$.) And the maximal compact subgroup of $GL_n^{+}(\mathbb{R})$ is $SO(n)$.
A: You can also use the Gauss decomposition
$$g = l\cdot u$$
where $l$ is lower triangular, and $u$ is upper triangular. It works for $g$ in a dense open subset defined by "all principal minors non-zero".
So: take $g'$ with positive determinant.  Get $g$ close by, with positive determinant and all principal minors non-zero. Now join $g'$ to $g$ by a segment ( $g$ was chosen close enough to $g'$), and reduce the problem to connecting $g$.  Connect $l$ and $u$ to diagonal matrices. Now reduce to a diagonal matrix ( with elements $\pm 1$).  Pair the $-1$ if any, look at them as rotations by $\pi$. You are mostly done.
