TITLE: Prove that the set of $n$-by-$n$ real matrices with positive determinant is connected
QUESTION [35 upvotes]: Math people:
In the fourth edition of Strang's "Linear Algebra and its Applications", page 230, he poses the following problem (I have changed his wording): show that if $A \in \mathbf{R}^{n \times n}$ with $\det(A) >0$, then there exists a continuous function $f:[0,1] \to \mathbf{R}^{n \times n}$ with $f(0) = I$, $f(1) = A$, and $\det(f(t)) > 0$ for all $t \in [0,1]$.  He says "the problem is not so easy, and solutions are welcomed by the author" (direct quote from the book).  Since the author welcomes solutions, I am not 100% sure even he has solved it.  Does anyone know if this is true or false, and may I have a solution or a hint?  I assigned this problem to a graduate student in a linear algebra class, because I wanted to give him a challenge.  Neither one of us has solved it.  I have a lot of other things to do, but I'd like to give him a hint.
STEFAN (Stack Exchange FAN)

REPLY [10 votes]: A proof using action of groups:
Let $GL_n(\mathbb{R})_+= \{ M \in GL_n(\mathbb{R}) \mid \det(M)>0 \}$ act on $\mathbb{R}^n \backslash \{0\}$ in the canonical way; notice that the action is transitive. Let $e_1=(1,0,...,0)$.
Introduce the subgroups $H$ and $K$ defined by $H= \left\{ \left( \begin{array}{cc} 1 & 0 \dots 0 \\ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} & A \end{array} \right) \mid A \in GL_{n-1}(\mathbb{R})_+ \right\}$ and $G= \left\{ \left( \begin{array}{cc} 1 & a_1 \dots a_{n-1} \\ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} & I_{n-1} \end{array} \right) \mid (a_1,...,a_{n-1}) \in \mathbb{R}^{n-1} \right\}$. Then the stabilizer of $e_1$ is $HG$, homeomorphic to $G \times H \simeq \mathbb{R}^{n-1} \times GL_{n-1}(\mathbb{R})_+$.
You deduce that $\mathbb{R}^{n}\backslash \{0\}$ is homeomorphic to $GL_n(\mathbb{R})_+ /( \mathbb{R}^{n-1} \times GL_{n-1}(\mathbb{R}_+))$. Finally, you can conclude by induction using the following lemma:

Lemma: Let $G$ be a topological group and $H$ be a subgroup of $G$. If $H$ and $G/H$ are connected, then $G$ is connected.

Proof: Let $f : G \to \{0,1\}$ be a continuous function. Since $H$ is connected, $f$ is constant on the classes of $G$ modulo $H$, hence a continuous function $\tilde{f} : G/H \to \{0,1\}$. But $G/H$ is connected, so $\tilde{f}$ is constant. You deduce that $f$ is constant, hence $G$ is connected. $\square$