Fix an integer $d \ge 2$ and let $M\_d$ be the space of real $d \times d$ matrices. Let $E$ be a vector subspace of $M\_d$. We say that $E$ is *transitive* if $E \cdot \mathbb{R}^d\_\* = \mathbb{R}^d$, that is, for all vectors $v \in \mathbb{R}^d\_\* = \mathbb{R}^d-\{0\}$ and $w \in \mathbb{R}^d$ there exists a matrix $A \in E$ such that $A \cdot v = w$.


The question is **how to determine *algebraically* if a space of matrices is transitive or not**.


More precisely, which algebraic (ie, polynomial) conditions on the entries of matrices $A\_1,...,A\_k$ express the fact that the space $E$ spanned by them is non-transitive?




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Remarks:


1) Fix the number $k$ of generators of $E$. Let $Z$ be the subset of $\mathbb{R}^{kd^2}$ corresponding to the $k$-tuples of matrices that generate a non-transitive set. That $Z$ is the projection of an algebraic set, and therefore by Tarski-Seidenberg theorem, is a *semi-algebraic* set.


2) Consider the analogous problem with complex matrices and vectors in $\mathbb{C}^d$, and let $Z\_C$ be the set corresponding to $Z$ above. Then $Z$ is algebraic (projectivize everything and apply the theorem that says that pprojection of algebraic is algebraic). Anyway what I'd like to see are the explicit equations for this algebraic set.



Algebraic characterization of transitive spaces of matrices
Here is the real answer:


A space of matrices is transitive iff its "orthogonal complement" contains no matrix of rank one.


The idea was not mine; it is I found in Sec. 4 from the paper below (See also some more modern and more readable papers that cite it):


Azoff, E.A. On finite rank operators and preannihilators. Mem. Amer. Math. Soc. 64, no. 357 (1986).


