Suppose M(m,n) is the vector space which is composed of all $n \times m$ real matrices. Let $F(m,n)$, $n>m$, be the set of all $n \times m$ matrices and, for each element $u \in F(m,n)$, the rank of $u$ is $m$. Note that $F(m,n)$ is a subset in vector space $M(m,n)$. Now we can define a equivalence relation on $F(m,n)$. That is，for $u,v \in F(m,n)$, we say $u\sim v$ if there exists an $m \times m$ invertible matrix $A$ such that $u=vA$. Then, for $u\in F(m,n)$, we define a subset of $F(m,n)$ as follows: $$ G(u)=\{ v \in F(m,n) : v \sim u \}. $$ Finally, suppose that $\left< G(u) \right>$ is the linear space generated by $G(u)$, that is, $\left< G(u) \right>$ is the intersection of all linear spaces that containing $G(u)$.
My question is:
What's the dimension of the vector space $\left< G(u) \right>$ ?
Supplement:
It is easy to get an answer for $m=1$. In this case, we have $\dim(\left< G(u) \right>)=1=m$. For $m>1$ and $u_1 \in F(m,n)$, I have found $m$ linearly independent vectors $u_1, \cdots, u_m$ in $G(u_1)$. Hence, the dimension of $\left< G(u) \right>$ is at least $m$.
Edit $1$.
Inspired by amomin's comments, I have found a way to attack this problem, but not completely. Let $u \in F(m,n)$ and suppose that $u_1, \cdots, u_k$ are $k$ elements in $G(u)$. Then, for $i=1, \cdots, k$, there exist $A_i \in \text{GL}(m)$ such that $u_i = uA_i$. Considering the following equation $$ \lambda_1 uA_1 + \lambda_2 uA_2 + \cdots + \lambda_k uA_k =0. \tag{1} $$ Easily, $(1)$ can be written as an equivalent form $$ u(\lambda_1 A_1 + \lambda_2 A_2 + \cdots + \lambda_k A_k)=0. \tag{2} $$ Together with the rank $r(u)=m$, we can choose $m$ linearly independent rows in $u$. Meanwhile, the $m$ row vectors can also form an $m \times m$ invertible matrix $u'$. Hence, by $(2)$ we have $$ u'(\lambda_1 A_1 + \lambda_2 A_2 + \cdots + \lambda_k A_k)=0, $$ and then $$ \lambda_1 A_1 + \lambda_2 A_2 + \cdots + \lambda_k A_k =0. $$ Hence, if $A_1, \cdots, A_k$ are linearly independent in $\text{GL}(m)$, then we have that $u_1, \cdots, u_k$ are $k$ linearly independent vectors in $G(u)$. More specifically, $r(u_1, \cdots, u_k) \geqslant r(A_1, \cdots, A_k)$. Taking this into consideration, let us suppose that the rank $r(\text{GL}(m))=s$ and choose $s$ linearly independent vectors $A_1, \cdots, A_s$ in $\text{GL}(m)$. Set $u_i = uA_i$, for $i=1, \cdots, s$, then, we get $s$ linearly independent vectors $u_i \in G(u)$. Therefore, we have $$ \text{dim}(\left< G(u) \right>) = r(G(u)) \geqslant r(\text{GL}(m)) = s. $$ Here, I have two questions:
$1.$ What's the rank of $\text{GL}(m)$ ?
$2.$ Is this equation $r(G(u)) = r(\text{GL}(m))$ valid ?
Yes. Let $A_1, \dots, A_s\in \text{GL}(m)$ be such that any element of $\text{GL}(m)$ can be expressed as their linear combination. Let $v=uA\in G(u)$, and let $A=\lambda_1A_1+\dots+\lambda_sA_s$. Then $v = \lambda_1uA_1+\dots+\lambda_suA_s$, i.e., any element of $G(u)$ is a linear combination of $uA_1,\dots,uA_s$. Thus $r(G(u)) \leq r(\text{GL}(m))$; together with the opposite inequality, it means that these two values are equal.
It's $m^2$. Let $E^{i,j}$, where $1\leq i,j\leq m$, be $m\times m$ matrix with $1$ at the position $(i,j)$ and $0$ at all other positions, and let $I$ be the unit $m\times m$ matrix. Then $I\in\text{GL}(m)$, and for any $i,j$ $I+E^{i,j}\in\text{GL}(m)$, so $E^{i,j}\in\left<\text{GL}(m)\right>$, and the matrices $E^{i,j}$ span the whole $M(m,m)$.