I am presently giving tutorials on linear algebra for first-year mathematics students, and there seems to be some ambiguity in the terminology related to the change of basis matrix. I aim to provide a clear and persuasive explanation on this topic to the students.
Consider a vector space $V$ over a field $K$ having two bases $\mathcal B=(b_1,\dots,b_n)$ and $\mathcal C$. Then we define the change of basis matrix $$P_{\mathcal C}^{\mathcal B}:=\Big([b_1]_\mathcal C\mid\cdots\mid[b_n]_\mathcal C\Big)\in K^{n\times n},$$ where for a vector $v\in V$, we write $[v]_\mathcal C$ for the coordinates of $v$ in the basis $\mathcal C$. For any $v\in V$, we then have the following identity: $$P_\mathcal C^\mathcal B\cdot[v]_\mathcal B=[v]_\mathcal C.$$
Because of this property, it makes sense to me to say that $P_\mathcal C^\mathcal B$ is the change of basis matrix from $\mathcal B$ to $\mathcal C$. However, it seems to be common to call this matrix $P_\mathcal C^\mathcal B$ the change of basis matrix from $\mathcal C$ to $\mathcal B$. For instance, have a look at the French Wikipedia page.
Why is this? I would like to think that this is just a typo, but considering that many sources have the roles of $\mathcal B$ and $\mathcal C$ reversed, there must be a different reason for this.
Besides changing the coordinates relative to the $\mathcal{B}$-basis into the coordinates relative to the $\mathcal{C}$-basis, the same matrix also carries out the change of the basis vectors themselves but in the other direction from $\mathcal{C}$ to $\mathcal{B}$.
To see this, note that if $\mathcal{C}=(c_1,...,c_n)$ and the coordinates of a vector in the $\mathcal{B}$-basis are $(v_1,...,v_n)$, then the $\mathcal{C}$-coordinates are $\sum_{j=1}^nP_{ij}v_j$ (where I have suppressed the $\mathcal{B}$ and $\mathcal{C}$ sub/superscripts on $P$ for now) so that the vector may be written as
$$\sum_{i=1}^n\left(\sum_{j=1}^nP_{ij}v_j\right)c_i.$$
Re-order the summation as
$$\sum_{j=1}^nv_j\left(\sum_{i=1}^nP_{ij}c_i\right)$$
So now this says that the $\mathcal{B}$-basis is $(\sum_{i=1}^nP_{i1}c_i,...,\sum_{i=1}^nP_{in}c_i)$.