Does this change of basis exist? Let $A$ be a unital associative $\mathbb{C}$ algebra and let $M$ be a 2-dimensional $A$-module with basis $\left\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\}$.  $A$ has two generators $\left\{a_{1}, a_{2}\right\}$ and they act on the basis of $M$ by
\begin{align*}
  & a_{1} \cdot \mathbf{e}_{1} = -\mathbf{e}_{1} & a_{2} \cdot \mathbf{e}_{1} = \mathbf{e}_{1} + \mathbf{e}_{2}\\
  & a_{1} \cdot \mathbf{e}_{2} = \mathbf{e}_{1} + \mathbf{e}_{2} & a_{2} \cdot \mathbf{e}_{2} = -\mathbf{e}_{2}\\
\end{align*}
Is it possible to find a basis $\left\{\overline{\mathbf{e}_{1}}, \overline{\mathbf{e}_{2}}\right\}$ of $M$ such that the action of the $a_{i}$ are still integer values, but are non-negative, i.e.
\begin{align*}
  & a_{1} \cdot \overline{\mathbf{e}_{1}} = n_1 \overline{\mathbf{e}_{1}} + n_2 \overline{\mathbf{e}_{2}}
  & a_{2} \cdot \overline{\mathbf{e}_{1}} = m_1 \overline{\mathbf{e}_{1}} + m_{2} \overline{\mathbf{e}_{2}}\\
  & a_{1} \cdot \overline{\mathbf{e}_{2}} = n_3 \overline{\mathbf{e}_{1}} + n_4 \overline{\mathbf{e}_{2}}
  & a_{2} \cdot \overline{\mathbf{e}_{2}} = m_3 \overline{\mathbf{e}_{1}} + m_{4} \overline{\mathbf{e}_{2}}\\
\end{align*}
where the $n_{i}$ and $m_{i}$ are in $\mathbb{Z}$ and $n_{i}, m_{i} \geq 0$ for $1 \leq i \leq 4$.
 A: Assuming that a unital $\Bbb C$ algebra means an algebra over $\Bbb C$ with a unit, the answer is negative even  for $A=\Bbb C$. Indeed, assume the contrary. Put $A_1=\begin{pmatrix} –1 & 0\\ 1 & 1 \end{pmatrix}$,  $A_2=\begin{pmatrix} 1 & 1\\ 0 & -1 \end{pmatrix}$. Let $i\in\{1,2\}$ be any index. Then  $A_j$ is a matrix of a transformation $a_j$ with respect to the basis $\{\mathbf{e}_{1},\mathbf{e}_{2}\}$. Let $\overline{A_j}$ be a matrix of a transformation $a_j$ with respect to the basis $\{\overline{\mathbf{e}_{1}},\overline{\mathbf{e}_{1}}\}$. Since the matrix $\overline{A_j}$ is similar to $A_j$, it has the same characteristic polynomial $\lambda^2-1$. In particular, the trace of $\overline{A_j}$ (which is equal to $n_1+n_4$ or $m_1+m_4$, depending on $j$) is zero. Since all $n_i$ and $m_i$ are non-negative, we have  $n_1=n_4=m_1=m_4=0$. Then $-1=\det A_j$, which is equal to $-n_2n_3$ or $-m_1m_3$, depending on $j$. Since all $n_i$ and $m_i$ are non-negative integers, we have $n_2=n_3=m_2=m_3=1$. Thus $\overline{A_1}=\overline{A_2}$, and so $a_1=a_2$, a contradiction.
