diagonal of the principal $m\times
m$-submatrix.

The following fact is obvious.

\[le:full rank\] Let $(B,\Lambda)$ be a compatible pair. Then $B\cdot \Lambda$ has full rank.

Let $(B,\Lambda)$ be a compatible pair and let $k\in [1,m]$. We define for $\varepsilon\in \{+1,-1\}$ a $n\times n$ matrix $E_{k,\varepsilon}$ via

-   $(E_{k,\varepsilon})_{ij}=\delta_{ij}$ if $j\ne k$,

-   $(E_{k,\varepsilon})_{ij}= -1$ if $i=j= k$,

-   $(E_{k,\varepsilon})_{ij}= max(0,-\varepsilon b_{ik})$ if $i\ne j= k$.

Similarly, we define a $m\times m$ matrix $F_{k,\varepsilon}$ via

-   $(F_{k,\varepsilon})_{ij}=\delta_{ij}$ if $i\ne k$,

-   $(F_{k,\varepsilon})_{ij}= -1$ if $i=j= k$,

-   $(F_{k,\varepsilon})_{ij}= max(0,\varepsilon b_{kj})$ if $i= k\ne j$.

We define a new pair $(B_k,\Lambda_k)$ as $$\label{eq:mutation matrix and Poisson}
   B_k=      F^T_{k,\varepsilon}   B E_{k,\varepsilon}^T \ , \quad
\Lambda_k=E_{k,\varepsilon} \Lambda E_{k,\varepsilon}^T\ ,$$ where $X^T$ denotes the transpose of $X$. We chose this rather non-straightforward way of defining $E_{k\varepsilon}$ and $F_{k,\varepsilon}$ in order to show how our definition relates to that of [@bz-qclust]. We will not need it in what follows. The motivation for the definition is the following fact.

[@bz-qclust Prop. 3.4] \[pr:comp under mutation\] The pair $(B_k,\Lambda_k)$ is compatible. Moreover, $\Lambda_k$ is independent of the choice of the sign $\varepsilon$.

The following fact is clear.

Let $\AA$ be a cluster algebra given by an initial seed 