A partial solution. The result is true over $M_2(\mathbb{Z})$ iff it's true over $M_2(\mathbb{Q})$. Then we may assume that the matrices are in $M_2(\mathbb{Q})$. a) Note that, if $C_1\cdots C_k=0$, then $\det(A)=0$ or $\det(B)=0$. Since $tr(A)\in\mathbb{Q}$ and $A\not= 0$ -else $k(A,B)=1$-, up to a change of basis over $\mathbb{Q}$, we may assume that i) $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ or ii) $A=diag(0,a)$, where $a\in\mathbb{Q}^*$ Since the case i) gives $k(A,B)=2$, $A$ is in the form ii). Note that $C_1\not= B,C_k\not= B$, otherwise we could remove $C_1$ or $C_k$. Thus $C_1=C_k=A$. Since $A^r=a^{r-1}A$ , necessarily, the $A$ are isolated in the sequence $(C_i)$. Then $C_1\cdots C_k=AB^{s_1}AB^{s_2}\cdots B^{s_q}A=0$, where $s_i\geq 1$. If $B^{s_1}=\begin{pmatrix}r&t\\u&v\end{pmatrix}$, then $AB^{s_1}A=avA$; if $v\not= 0$, then, in the beginning of the sequence $(C_i)$, we may replace $AB^{s_1}A$ with $A$, that is contradictory. Finally, $v=0$, $AB^{s_1}A=0$ and we are done. $\square$ b) Let $k\geq 3$ and $p=k-2$. According to the proof of a), it suffices to find $B\in GL_2(\mathbb{Q})$ s.t. $\bullet$ $B^p$ is in the form $\begin{pmatrix}r&t\\u&0\end{pmatrix}$. $\bullet$ If $1\leq l<p$, then $B^l$ is not in the above form. To be continued. cf. below post #6.

loup blanc wrote: A partial solution. The result is true over $M_2(\mathbb{Z})$ iff it's true over $M_2(\mathbb{Q})$. Then we may assume that the matrices are in $M_2(\mathbb{Q})$. a) Note that, if $C_1\cdots C_k=0$, then $\det(A)=0$ or $\det(B)=0$. Since $tr(A)\in\mathbb{Q}$ and $A\not= 0$ -else $k(A,B)=1$-, up to a change of basis over $\mathbb{Q}$, we may assume that i) $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ or ii) $A=diag(0,a)$, where $a\in\mathbb{Q}^*$ Since the case i) gives $k(A,B)=2$, $A$ is in the form ii). Note that $C_1\not= B,C_k\not= B$, otherwise we could remove $C_1$ or $C_k$. Thus $C_1=C_k=A$. Since $A^r=a^{r-1}A$ , necessarily, the $A$ are isolated in the sequence $(C_i)$. Then $C_1\cdots C_k=AB^{s_1}AB^{s_2}\cdots B^{s_q}A=0$, where $s_i\geq 1$. If $B^{s_1}=\begin{pmatrix}r&t\\u&v\end{pmatrix}$, then $AB^{s_1}A=avA$; if $v\not= 0$, then, in the beginning of the sequence $(C_i)$, we may replace $AB^{s_1}A$ with $A$, that is contradictory. Finally, $v=0$, $AB^{s_1}A=0$ and we are done. $\square$ b) Let $k\geq 3$ and $p=k-2$. According to the proof of a), it suffices to find $B\in GL_2(\mathbb{Q})$ s.t. $\bullet$ $B^p$ is in the form $\begin{pmatrix}r&t\\u&0\end{pmatrix}$. $\bullet$ If $1\leq l<p$, then $B^l$ is not in the above form. To be continued. Why can you change a basis so that A is the formi) and2)?

Tangle-Peptidoglycans wrote: loup blanc wrote: A partial solution. The result is true over $M_2(\mathbb{Z})$ iff it's true over $M_2(\mathbb{Q})$. Then we may assume that the matrices are in $M_2(\mathbb{Q})$. a) Note that, if $C_1\cdots C_k=0$, then $\det(A)=0$ or $\det(B)=0$. Since $tr(A)\in\mathbb{Q}$ and $A\not= 0$ -else $k(A,B)=1$-, up to a change of basis over $\mathbb{Q}$, we may assume that i) $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ or ii) $A=diag(0,a)$, where $a\in\mathbb{Q}^*$ Since the case i) gives $k(A,B)=2$, $A$ is in the form ii). Note that $C_1\not= B,C_k\not= B$, otherwise we could remove $C_1$ or $C_k$. Thus $C_1=C_k=A$. Since $A^r=a^{r-1}A$ , necessarily, the $A$ are isolated in the sequence $(C_i)$. Then $C_1\cdots C_k=AB^{s_1}AB^{s_2}\cdots B^{s_q}A=0$, where $s_i\geq 1$. If $B^{s_1}=\begin{pmatrix}r&t\\u&v\end{pmatrix}$, then $AB^{s_1}A=avA$; if $v\not= 0$, then, in the beginning of the sequence $(C_i)$, we may replace $AB^{s_1}A$ with $A$, that is contradictory. Finally, $v=0$, $AB^{s_1}A=0$ and we are done. $\square$ b) Let $k\geq 3$ and $p=k-2$. According to the proof of a), it suffices to find $B\in GL_2(\mathbb{Q})$ s.t. $\bullet$ $B^p$ is in the form $\begin{pmatrix}r&t\\u&0\end{pmatrix}$. $\bullet$ If $1\leq l<p$, then $B^l$ is not in the above form. To be continued. Why can you change a basis so that A is the formi) and2)? You look at the eigenvalues of $A$ and turn $A$ into it s $2$ possible Jordan matrices. $A$ s eigenvalues can either both be $0$, or one can be $0$ and the other some random value, $a$.

Here's a solution to b) that I had trouble finding. For example, take $B=\begin{pmatrix}-p-1&1\\-1&-p+1\end{pmatrix}$. EDIT. We obtain $B^k=(-p)^{k-1}\begin{pmatrix}-p-k&k\\-k&-p+k\end{pmatrix}$.