Let $R$ be a commutative ring with identity, let $n \in \mathbb{Z}$ with $n \geq 0$, and let $M$ be a finite free left $R$-module of rank $n$. Let $\mathcal{B} = (b_1,\dots,b_n)$ and $\mathcal{C} = (c_1,\dots,c_n)$ be ordered bases of $M$, with the convention that if $n = 0$ both bases are empty. Define the change-of-basis matrix $P_{\mathcal{C} \leftarrow \mathcal{B}} \in R^{n \times n}$ by requiring that, for every $m \in M$,

\begin{align*} [m]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow \mathcal{B}}[m]_{\mathcal{B}}, \end{align*}

where $[m]_{\mathcal{B}}$ and $[m]_{\mathcal{C}}$ denote coordinate columns in the ordered bases $\mathcal{B}$ and $\mathcal{C}$. Then $P_{\mathcal{C} \leftarrow \mathcal{B}}$ is invertible in $R^{n \times n}$, and its inverse is $P_{\mathcal{B} \leftarrow \mathcal{C}}$.

Conversely, if $\mathcal{B} = (b_1,\dots,b_n)$ is an ordered basis of $M$ and $A = (a_{ij}) \in R^{n \times n}$ is invertible, then the ordered family $\mathcal{C} = (c_1,\dots,c_n)$ defined by

\begin{align*} c_j = \sum_{i=1}^{n} a_{ij} b_i \end{align*}

for each $j \in \{1,\dots,n\}$ is an ordered basis of $M$. Equivalently, an invertible matrix sends an ordered basis to another ordered basis by using its columns as the coordinates of the new basis vectors.