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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?


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.

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Transitive Lie algebras are a classically studied subject. (Google it to check.) However my space $E$ is not assumed to be a Lie subalgebra of $M_d$. – Jairo Bochi Feb 18 '10 at 13:11
up vote 3 down vote accepted

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).

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The condition for $E$ to be intransitive is that the determinant form is the $0$ form somewhere other than the origin. That is, every vector $v \in \mathbb R^d$ gives you an alternating $d$-form on $M_d$ and on $E$ by the determinant of the images of $v$. This form is nonzero on a vector if and only if the images of the vector by $E$ are all of $\mathbb R^d$.

Edit: You are right that the above was not a complete answer.

To be more explicit with bases for everything: Let $E$ have dimension $D$, with $d \le D \le d^2$. Let $E$ have a basis $\{E_1, ... E_D \}$ so that every element of $E$ is represented by a vector $(a_1,...,a_D)$. Represent every element of $\mathbb R^d$ by a vector $(b_1,....,b_d)$.

Then for any vector $(b_1,....,b_d)$, the determinant form is an alternating $d$-form on $E$ identified with $\mathbb R^D$. These forms have a basis of size ${D \choose d}$ given by the determinants of $d\times d$ minors, that is, project to a given $d$ coordinates, and take the determinant. To check whether the determinant form is the $0$ $d$-form, express it in terms of the basis, and see if all ${D \choose d}$ coefficients are $0$. That is, check if the ${D \choose d}$ determinants $\det [E_{f(1)}v, ..., E_{f(d)}v]$ are all $0$ for each integer-valued function $f$ with $1 \le f(1) \lt f(2) \lt ... \lt f(d) \le D$.

As we let $v$ vary but fix a basis for $E$, the coefficients of the determinant form are homogeneous polynomials of degree $d$ in the coordinates $\{b_i\}$. The variety of intersections of the zeros of those polynomials on $\mathbb R^d$ is the origin if and only if $E$ is transitive.

This gives you a test for a particular $E$ in terms of recognizing whether a variety is just a point. It still leaves the condition on $E$ in the Grassmannian as a projection of a variety.

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OK, but this does not answer the question. You suggest to take each $v\neq 0$ and then compute some associated determinant. So in theory you would have to perform infinitely many calculations: one for each $v$. __________ What I need is some computable test on the generators of $E$. – Jairo Bochi Feb 17 '10 at 9:56

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