I was studying some kind of linear representations of the form
$\mathbb{R} ^m \ni \xi \mapsto A(\xi) \in gl(n, \mathbb{R}), \quad$ i.e $\quad a_{ij} = \xi_k, \ \forall i,j =1, \dots, n,\ $ for some $k \in \{ 1, \dots, m\}$.
Let us denote $A_{\xi_k}=\frac{\partial A(\xi)}{\partial \xi_k}=A(0,\dots,1,\dots,0)\ $, with the $k$-th entry equal to one, and with $A_{\xi_k}^l$ its $l$-th column. I was able to prove, basically by chain differentiation, the following lemma:
$\textit{Given a generic $x \in \mathbb{R}^n$, the dimension of its orbit under the action $x \mapsto A(\xi)x\ $ is given by}$ \begin{equation*} dim~\mathcal{O}(x)= {max} \{ r \ |\ \exists\ A_{\xi_{k_1}}^{l_1}, \dots, A_{\xi_{k_r}}^{l_r} \text{ independent, with $k_1, \dots{}, k_r$ all different}\}. \end{equation*}
I wanted to know if this property of the matrix A can be rephrased in a more mathematical or at least known way. Basically it says that the dimension of a generic orbit is given by the maximum number of independent column I can find picking at most one from each matrix $A_{\xi_k}$.
Dropping our initial problem, given for example two matrices $A$ and $B$, I can find two such columns if and only if they are not of the form $A=(\lambda_1 v | \dots | \lambda_n v)$ and $B=(\mu_1 v | \dots | \mu_n v)$ for some $v \in \mathbb{R}^n \setminus \{0\}$ and coefficients $\lambda_i, \mu_j$. Given three matrices, if not all the columns belong to the same plane and so on and so forth.
Or, we could try to describe $r$ in a more compact way, looking at all the possible wedge products $A_{\xi_{k_1}}^{l_1} \wedge \dots \wedge A_{\xi_{k_r}}^{l_r}$, as elements in $\bigwedge^r \mathbb{R}$, the dimension of the orbit is at least $r$ if at least one of these is non null. Denoting its entries as $(c_{j_1 \dots j_N})_{k_1 \dots k_r}^{l_1 \dots l_r}$ we can then define the map $\psi_r$ from $\mathbb{R}^m$ to $\bigwedge^r \mathbb{R}^n \otimes \mathbb{1}_{\mathbb{R}^{n^r}} \otimes \mathbb{1}_{\bigwedge^k \mathbb{R}^m}$, which correspond, respectively, to the entries of the above wedge product, to all the possible choices of columns, and all the possible choice of variables, as \begin{equation} A(\xi) \mapsto \psi_r(A(\xi)) = \sum_{\substack{l_1, \dots, l_r =1, \dots, n \\ 1 \le j_1 < \dots < j_r \le n \\1 \le k_1 < \dots < k_r \le m}} (c_{j_1 \dots j_r})_{k_1 \dots k_r}^{l_1 \dots l_r}\ (e_{j_1 \dots j_r})_{k_1 \dots k_r}^{l_1 \dots l_r}. \end{equation} Then, if the rhs in this big big vector space is not the zero element, the dimension of a generic orbit for the action $A(\xi)$ is at least $r$, and it is exactly equal to the maximum of such an $r$.
The question is, am I rephrasing in a weird way some basic properties and/or relation between matrices? Could you point me at it? It seems to me as a kind of "independence degree" between sets of matrices, even though I am just writing it cause it sounds good not because of any explicit relation.
Thank you for your time!