Let $BS(m,n)$ be the Baumslag-Solitar group defined by $B(m,n) = < a,b ~|~ b a^m b^{-1} = a^n > $, $mn \neq 0$. There is a linear representation of $BS(m,n)$ by mapping $a$ to the matrix $\left(\begin{matrix} 1&1 \cr 0&1\end{matrix}\right)$ and $b$ to the matrix $\left(\begin{matrix} \frac{n}{m}&0 \cr 0&1\end{matrix}\right)$. Denote this representation homomorphism as $f$, assume $|m| \neq |n|$, my main question is:

What is the kernel of $f$ ?

Some observations

(1) Commutator of the form $ [a, a^b], [a,a^{b^2} ], [a,a^{b^3}] \ldots$, are in the kernel. Do these elements generate the Kernel of $f$? Do they form an infinite generated free group? .

(2) If $|m| \ne |n|$ and either $|m| = 1$ or $|n| = 1$ then $f$ is known to be injective.

No, if you take $|n|=1=|m|$. – Anton Klyachko Nov 13 '12 at 23:35notinjective and the kernel isnotgenerated by the commutators (1). – Anton Klyachko Nov 13 '12 at 23:57