[proofplan] We use the defining property of the change-of-basis matrix: it sends $\mathcal B$-coordinates to $\mathcal C$-coordinates. For each basis vector $b_i$, its coordinate vector in the basis $\mathcal B$ is the $i$th standard coordinate vector $e_i \in k^n$. Multiplying a matrix by $e_i$ extracts its $i$th column, so applying $P_{\mathcal C \leftarrow \mathcal B}$ to $[b_i]_{\mathcal B}=e_i$ identifies the $i$th column with $[b_i]_{\mathcal C}$. [/proofplan] [step:Evaluate the change-of-basis matrix on each standard coordinate vector] For each $i \in \{1,\ldots,n\}$, let $e_i \in k^n$ denote the $i$th standard coordinate vector, whose $j$th entry is $1$ if $j=i$ and $0$ if $j \neq i$. Since $\mathcal B=(b_1,\ldots,b_n)$ is ordered, the coordinate vector of $b_i$ with respect to $\mathcal B$ is \begin{align*} [b_i]_{\mathcal B}=e_i. \end{align*} By the defining property of $P_{\mathcal C \leftarrow \mathcal B}$, applied to the vector $v=b_i \in V$, we have \begin{align*} [b_i]_{\mathcal C}=P_{\mathcal C \leftarrow \mathcal B}[b_i]_{\mathcal B}. \end{align*} Substituting $[b_i]_{\mathcal B}=e_i$ gives \begin{align*} [b_i]_{\mathcal C}=P_{\mathcal C \leftarrow \mathcal B}e_i. \end{align*} [guided] Fix $i \in \{1,\ldots,n\}$. Let $e_i \in k^n$ be the $i$th standard coordinate vector, meaning that its $j$th entry is $1$ if $j=i$ and $0$ if $j \neq i$. The reason this vector appears is that $b_i$ is the $i$th element of the ordered basis $\mathcal B=(b_1,\ldots,b_n)$. In $\mathcal B$-coordinates, the vector $b_i$ has coefficient $1$ on $b_i$ and coefficient $0$ on every other $b_j$. Therefore \begin{align*} [b_i]_{\mathcal B}=e_i. \end{align*} Now use the defining convention for the change-of-basis matrix. The matrix $P_{\mathcal C \leftarrow \mathcal B}$ is characterized by the identity \begin{align*} [v]_{\mathcal C}=P_{\mathcal C \leftarrow \mathcal B}[v]_{\mathcal B} \end{align*} for every $v \in V$. This convention says that the matrix converts $\mathcal B$-coordinate columns into $\mathcal C$-coordinate columns. Applying this identity to the particular vector $v=b_i$ gives \begin{align*} [b_i]_{\mathcal C}=P_{\mathcal C \leftarrow \mathcal B}[b_i]_{\mathcal B}. \end{align*} Since $[b_i]_{\mathcal B}=e_i$, substitution yields \begin{align*} [b_i]_{\mathcal C}=P_{\mathcal C \leftarrow \mathcal B}e_i. \end{align*} Thus the image of the $i$th standard coordinate vector under the change-of-basis matrix is exactly the $\mathcal C$-coordinate vector of the basis vector $b_i$. [/guided] [/step] [step:Identify multiplication by $e_i$ with extraction of the $i$th column] Write $P_{\mathcal C \leftarrow \mathcal B}=(p_{ji})_{1 \leq j,i \leq n}$, where $p_{ji} \in k$ is the entry in row $j$ and column $i$. For fixed $i$, the $j$th entry of the column vector $P_{\mathcal C \leftarrow \mathcal B}e_i$ is \begin{align*} \sum_{\ell=1}^{n} p_{j\ell}(e_i)_\ell=p_{ji}. \end{align*} Hence $P_{\mathcal C \leftarrow \mathcal B}e_i$ is precisely the $i$th column of $P_{\mathcal C \leftarrow \mathcal B}$. [/step] [step:Assemble the columns into the full matrix identity] From the first step, \begin{align*} P_{\mathcal C \leftarrow \mathcal B}e_i=[b_i]_{\mathcal C} \end{align*} for every $i \in \{1,\ldots,n\}$. From the second step, $P_{\mathcal C \leftarrow \mathcal B}e_i$ is the $i$th column of $P_{\mathcal C \leftarrow \mathcal B}$. Therefore the $i$th column of $P_{\mathcal C \leftarrow \mathcal B}$ is $[b_i]_{\mathcal C}$ for every $i$, and consequently \begin{align*} P_{\mathcal C \leftarrow \mathcal B}=\begin{pmatrix} [b_1]_{\mathcal C} & \cdots & [b_n]_{\mathcal C} \end{pmatrix}. \end{align*} This proves the theorem. [/step]