# submonoid of a matrix monoid with a common eigenvector

Hello, I am considering two real invertible $3\times 3$ matrices $A$ and $B$ and a nonzero vector $v\in\mathbb{R}^3$ and i am wondering if the submonoid $E$ of the monoid $(A,B)$ genererated by $A$ and $B$ with $v$ as an eigenvector

$$E:= \left( C\in (A,B), \ v \text{ is an eigenvector of } C \right)$$

is finitely generated.

Does it even hold for $2\times 2$ matrices? Thank you in advance.

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The answer is NOT NECESSARY (at least for $3\times 3$ matrices).

Let $A=\left(\matrix{2&0&1\\0&1&0\\0&0&\alpha}\right)$, $B=\left(\matrix{0&1&1\\1&0&0\\0&0&\beta}\right)$, $v=\left(\matrix{1\\1\\0}\right)$ (here $\alpha$, $\beta$ are algebraically independent over $\mathbb Q$). Then $A^nBA^nv=2^nv$, so $A^nBA^n\in E$. Next, our monoid is freely generated by $A$ and $B$. So, the element $A^nBA^n$ can be expanded as a non-trivial product of two elements in $E$ iff it can be expanded literally; this is obviously false.

So, $E$ is not finitely generated in this case.

UPDATE: Let us show that the monoid $(A,B)$ is free. Actually, we will show that every word $W=W(A,B)$ can be recovered from its action on $u=\left(\matrix{0\\0\\1}\right)$. Let $Wu=u'=\left(\matrix{u_1'\\u_2'\\u_3'}\right)$; since $A$, $B$ are invertible, it is enough to reconstruct the first letter in $W$.

So let $W=XW'$, $X$ is the first letter. Note that $u_3'=\alpha^k\beta^\ell$, where $k$ and $\ell$ are the numbers of occurences of $A$ and $B$ in $W$; hence we have reconstructed $k+\ell$. Next, on each of the previous steps, to the first two coordinates we added only the monomials in $\alpha$ and $\beta$ of lower degrees; so, $u_1'$ and $u_2'$ are polynomials in $\alpha$ and $\beta$ of degree $\leq k+\ell-1$. Moreover, the leading term in $u_1$ is either $\alpha^k\beta^{\ell-1}$ or $\alpha^{k-1}\beta^\ell$, depending on the number of letters $A$ and $B$ in $W'$, that is --- depending on $X$. Hence $X$ is reconstructible.

Surely, we used several times that $\alpha$, $\beta$ are algebraically independent over $\mathbb{Q}$. In fact, by the same lines one may prove the same for $\alpha=1/2$ and $\beta=1/3$ ;)...

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How do you prove that the monoid generated by $A$ and $B$ is free? –  Steve Richards Oct 6 '11 at 13:36
@Ilya and Steve: The subgroup generated by $A,B$ is certainly not free. For example, the commutators $[A,B]$ and $[A^2,B^2]$ generate a solvable subgroup. Still the monoid may be free but it does need a proof. –  Mark Sapir Oct 6 '11 at 15:03
I've added the proof of the fact that the monoid is free. Surely, the question about the corresponding group instead of monoid is completely different... –  Ilya Bogdanov Oct 6 '11 at 15:46
It is more clear now, but you did not define $u_1,u_2, u_3$. I assume that these are the coordinates of $W'u$. Still the fact about the leading term of $u_1$ is not proved (probably it can be proved by induction). –  Mark Sapir Oct 6 '11 at 16:19
Sorry, I was in a bit hurry. I've added some corrections; hope now it is clear enough for this site... –  Ilya Bogdanov Oct 6 '11 at 17:21