[proofplan] We compute the matrix of $\beta \circ \alpha$ by evaluating the composition on each basis vector of $U$, expanding using the matrix representations of $\alpha$ and $\beta$, and recognising the resulting expression as the standard matrix product $BA$. The computation reduces to exchanging the order of two finite sums and identifying the inner sum as the definition of matrix multiplication. [/proofplan] [step:Express $\alpha(e_i)$ and $\beta(f_j)$ in terms of their matrices] Let $(e_1, \dots, e_p)$, $(f_1, \dots, f_m)$, and $(g_1, \dots, g_n)$ be the fixed ordered bases for $U$, $V$, and $W$ respectively. Let $A = (a_{ji}) \in \mathrm{Mat}_{m,p}(\mathbb{F})$ be the matrix of $\alpha$ and $B = (b_{kj}) \in \mathrm{Mat}_{n,m}(\mathbb{F})$ be the matrix of $\beta$, so that \begin{align*} \alpha(e_i) = \sum_{j=1}^{m} a_{ji}\, f_j, \qquad \beta(f_j) = \sum_{k=1}^{n} b_{kj}\, g_k. \end{align*} [/step] [step:Evaluate $(\beta \circ \alpha)(e_i)$ by substituting and exchanging summation order] For each $i \in \{1, \dots, p\}$, apply $\beta$ to the expansion of $\alpha(e_i)$, using linearity of $\beta$: \begin{align*} (\beta \circ \alpha)(e_i) &= \beta\!\left(\sum_{j=1}^{m} a_{ji}\, f_j\right) = \sum_{j=1}^{m} a_{ji}\, \beta(f_j) = \sum_{j=1}^{m} a_{ji} \sum_{k=1}^{n} b_{kj}\, g_k. \end{align*} Exchanging the order of the finite sums: \begin{align*} (\beta \circ \alpha)(e_i) = \sum_{k=1}^{n} \left(\sum_{j=1}^{m} b_{kj}\, a_{ji}\right) g_k. \end{align*} [guided] We want to find the coordinates of $(\beta \circ \alpha)(e_i)$ in the basis $(g_1, \dots, g_n)$. Starting from $\alpha(e_i) = \sum_{j=1}^{m} a_{ji} f_j$, we apply $\beta$ and use its linearity to pass it through the sum: \begin{align*} (\beta \circ \alpha)(e_i) = \beta\!\left(\sum_{j=1}^{m} a_{ji}\, f_j\right) = \sum_{j=1}^{m} a_{ji}\, \beta(f_j) = \sum_{j=1}^{m} a_{ji} \sum_{k=1}^{n} b_{kj}\, g_k. \end{align*} To read off the coefficient of $g_k$, we exchange the order of summation (both sums are finite, so this is valid without any convergence considerations): \begin{align*} (\beta \circ \alpha)(e_i) = \sum_{k=1}^{n} \left(\sum_{j=1}^{m} b_{kj}\, a_{ji}\right) g_k. \end{align*} The coefficient of $g_k$ is $\sum_{j=1}^{m} b_{kj}\, a_{ji}$, which is the $(k,i)$-entry of the matrix of $\beta \circ \alpha$. [/guided] [/step] [step:Identify the coefficients as the entries of the matrix product $BA$] The $(k,i)$-entry of the matrix product $BA$ is, by definition of matrix multiplication, \begin{align*} (BA)_{ki} = \sum_{j=1}^{m} b_{kj}\, a_{ji}. \end{align*} Comparing with the expansion from the previous step, we have $(\beta \circ \alpha)(e_i) = \sum_{k=1}^{n} (BA)_{ki}\, g_k$ for each $i \in \{1, \dots, p\}$. Therefore the matrix of $\beta \circ \alpha$ with respect to the bases $(e_i)$ and $(g_k)$ is $BA$. [/step]