Suppose $G$ is a semigroup (i.e., closed under matrix multiplication) of invertible $2\times 2$ real matrices. Suppose also that $G$ is transitive i.e., for any two nonzero vectors $u$ and $v$ there exists a matrix in $G$ that maps $u$ to $v$. Are there any nontrivial examples of such a $G$?

$\begingroup$ What upper triangular matrix takes (1,0) to (1,5)? Gerhard "Ask Me About System Design" Paseman, 2010.11.11 $\endgroup$– Gerhard PasemanNov 11, 2010 at 18:20

$\begingroup$ You are right Gerhard, my bad. $\endgroup$– HejNov 11, 2010 at 22:32
1 Answer
Here is a complete answer:
Every semigroup $S$ of invertible $2\times 2$*matrices which is transitive on* $\mathbb R^2$ is either conjugate to $SO_2(\mathbb R) \times \mathbb R^+$ or $SO_2(\mathbb R) \times \mathbb R$ or it is a product of $SL_2(\mathbb R)$ and a multiplicative subgroup of $\mathbb R$*.*
Proof: Let S be such a semigroup. Then the intersection $S_0$ with $SL_2(\mathbb R)$ is a subsemigroup of $SL_2(\mathbb R)$. By a theorem of Hilgert and Hofmann (see their beautiful paper on "Old and new on $SL_2$") there are only three choices for $S_0$: Either $S_0$ is all of $SL_2(\mathbb R)$, a circle group or contained in a conjugate of the elements of $SL_2(\mathbb R)$ with only positive entries. If $S_0$ is a circle group, then $S$ will be conjugate to $SO_2(\mathbb R) \times \mathbb R^+$ or $SO_2(\mathbb R) \times \mathbb R$. If $S_0$ happens to be all of $SL_2(\mathbb R)$, then we have $SL_2(\mathbb R)\subset S \subset GL_2(\mathbb R)$, so $S$ is a product of $SL_2(\mathbb R)$ and a multiplicative subgroup of $\mathbb R$. In the third case, we may assume that $S_0$ is actually contained in the semigroup described above. Then $S_0$ maps every vector with two positive entries to a vector with two posiitve entries, hence $S$ maps the upper right quadrant to a subset of itself and the lower left quadrant. In particular, $S$ cannot be transitive.