Show that the set of $m\times n$ matrices with full rank is connected A subset $S\subset X$ is connected if for every $x$ and $y$ in $S$, there exists a continuous curve $r:[0,1]\rightarrow S$ such that $r(0)=x$ and $r(1)=y$.
I want to show that the set of all $m\times n$ matrices with $m>n$ with rank $n$ is connected.
This paper shows that the set of all $m\times n$ matrices with $m>n$ with rank $n$ is connected by analytic regular arcs, which is more than what I need. I am struggling to follow all the arguments presented in the paper. I image that if the analytic arc part is removed, then the argument becomes much easier, but I am not sure.
The argument I was thinking is as follows: Let $V=\{v_1,v_2,...,v_n\}$ be an orthonormal basis, then all full rank matrices can be generated by placing the vectors $v_1,...,v_n$ into a matrix with some scalar factor distinct than $0$, namely, $[a_1v_1,a_2v_2,...,a_nv_n]$ (the only restriction for the $a_i$ is that they are nonzero). If we have two full rank matrices given by $m=[a_1v_1,a_2v_2,...,a_nv_n]$ and $n=[b_1v_1,b_2v_2,...,b_nv_n]$, then the curve $r$ is $r(\lambda)=(1-\lambda) m+\lambda n$ is clearly continuous and $r(0)=m$ and $r(1)=n$.
I don't think this argument is enough though. For example, if $n=-m$ then $r(0.5)=0$ which is outside of $S$.
Does anyone know how to show only the result I want? or is the analytic arc part unavoidable to show the connectedness?
 A: This is true more generally for matrices of rank $r$, where $ 0\leq  r\le \min(m,n)$.
I am assuming here that the matrices have real entries.
Theoritical proof.
Two rectangular matrices $M_1,M_2$ of same size $m\times n$ have same rank if and only if there exist two invertible matrices $U,V$ of appropriate sizes such that $M_2=UM_1V$.
Notation. if $\varepsilon\in\{\pm\}$, let $ GL_k^{\varepsilon}(\mathbb{R})$ be the set of $k\times k$ matrices whose determinant has sign $\varepsilon$.
Let $\varepsilon$ be the sign of $\det(U)$, and let $\varepsilon'$ be the sign of $\det(V)$.
The map $f:(U,V)\in GL_m^{\varepsilon}(\mathbb{R})\times GL_n^{\varepsilon'}(\mathbb{R})\mapsto UM_1V\in M_{m\times n}(\mathbb{R})$ is continuous, and its image contains $M_1$ and $M_2$ by choice of $U$ and $V$.
Now  $ GL_m^{\varepsilon}(\mathbb{R})\times GL_n^{\varepsilon'}(\mathbb{R})$ is (path) connected, since it is the product of two (path) connected spaces, so the image of $f$ is (path) connected. In particular, you can join $M_1$ and $M_2$ by a continuous curve lying in the space of matrices of rank $r$.
If you  want a more explicit version of the proof. You need first to prove that $GL_k^\varepsilon (\mathbb{R})$ is path connected. For it is enough to prove that $GL_k^+ (\mathbb{R})$ is path connected (since any element of $GL_k^- (\mathbb{R})$ maybe written as $D_{-1}P$, where $D_{-1}=diag(1,\ldots,1,-1)$ and $P\in GL_k^+(\mathbb{R}))$.
Now, it is enough to show that you can connect an invertible matrix $P$ of positive determinant with the identity matrix $I_k$ using a path on $GL_k^+(\mathbb{R})$.
Let $\lambda=\det(P)>0$, so that $P=D_\lambda Q$, where $D_\lambda=diag (1,\ldots,1,\lambda)$, and $\det(Q)=1$.
Since $Q$ has determinant $1$, it is a product of transvection matrices. Hence, it is enough to convince ourselves that:

*

*we can connect $I_k$ and $D_\lambda$ using a path on $GL_k^+(\mathbb{R})$.


*we can connect $I_k$ and a transvection matrix using a path on $GL_k^+(\mathbb{R})$.
Both statements are obvious: for the first one, take $t\in [0,1]\mapsto diag (1,\ldots, 1, (1-t)+t\lambda)$, and for the second one, if $T=I_k+c E_{ij}$ is a transvection matrix, take $t\in [0,1]\mapsto I_k+ c tE_{ij}$.
Now to conclude the explicit argument, take a path $r:[0,1]\to GL_m^{\varepsilon}(\mathbb{R})$ such that $r(0)=I_m$  and $r(1)=U$, and a path $s:[0,1]\to GL_n^{\varepsilon'}(\mathbb{R})$ such that $s(0)=I_n$ and $s(1)=V$.
Then $\rho: t\in [0,1]\mapsto r(t)M_1s(t)\in M_{m\times n}(\mathbb{R})$ is a path such that $\rho(0)=M_1$ and $\rho(1)=M_2$, whose image consists of matrices of rank $r$, since $r(t)$ and $s(t)$ are invertible for all $t\in [0,1]$.
