[proofplan] We express $\alpha$ as the tautological composition $\mathrm{id}_V^{-1} \circ \alpha \circ \mathrm{id}_U$, where the identity maps serve to translate between the old and new bases. The matrices of $\mathrm{id}_U$ and $\mathrm{id}_V$ in mixed bases are the change-of-basis matrices $P$ and $Q$, and two applications of [Composition Corresponds to Matrix Multiplication](/theorems/383) yield $B = Q^{-1}AP$. [/proofplan] [step:Identify $P$ and $Q$ as matrices of identity maps in mixed bases] The matrix $P \in \mathrm{GL}_m(\mathbb{F})$ has $i$th column equal to the coordinates of $u_i$ in the basis $(e_1, \dots, e_m)$: $u_i = \sum_{k=1}^m P_{ki}\, e_k$. This is the matrix of $\mathrm{id}_U: U \to U$ with respect to the bases $(u_i)$ (domain) and $(e_i)$ (codomain). Similarly, $Q \in \mathrm{GL}_n(\mathbb{F})$ has $j$th column equal to the coordinates of $v_j$ in the basis $(f_1, \dots, f_n)$: $v_j = \sum_{l=1}^n Q_{lj}\, f_l$. This is the matrix of $\mathrm{id}_V: V \to V$ with respect to the bases $(v_j)$ (domain) and $(f_j)$ (codomain). [guided] The change-of-basis matrix $P$ records how to express each new basis vector $u_i$ in terms of the old basis vectors $e_k$. Since $\mathrm{id}_U(u_i) = u_i = \sum_{k=1}^m P_{ki}\, e_k$, the matrix of $\mathrm{id}_U$ with domain basis $(u_i)$ and codomain basis $(e_i)$ is precisely $P$. The $i$th column of $P$ contains the coordinates of $u_i$ in the basis $(e_k)$. The same reasoning applies to $Q$. Since $\mathrm{id}_V(v_j) = v_j = \sum_{l=1}^n Q_{lj}\, f_l$, the matrix of $\mathrm{id}_V$ with domain basis $(v_j)$ and codomain basis $(f_j)$ is $Q$. Both $P$ and $Q$ are invertible because the identity map is an isomorphism, and an isomorphism between finite-dimensional spaces has an invertible matrix representation. [/guided] [/step] [step:Express $\alpha$ as a three-fold composition translating between bases] The map $\alpha: U \to V$ factors as \begin{align*} \alpha = \mathrm{id}_V^{-1} \circ \alpha \circ \mathrm{id}_U, \end{align*} where we read the composition as \begin{align*} (U, (u_i)) \xrightarrow{\mathrm{id}_U} (U, (e_i)) \xrightarrow{\;\alpha\;} (V, (f_j)) \xrightarrow{\mathrm{id}_V^{-1}} (V, (v_j)). \end{align*} The matrix of $\mathrm{id}_U$ (from $(u_i)$ to $(e_i)$) is $P$. The matrix of $\alpha$ (from $(e_i)$ to $(f_j)$) is $A$. The matrix of $\mathrm{id}_V^{-1}$ (from $(f_j)$ to $(v_j)$) is $Q^{-1}$. [/step] [step:Apply matrix multiplication to obtain $B = Q^{-1}AP$] By the [Composition Corresponds to Matrix Multiplication](/theorems/383) theorem applied twice, the matrix of the composition $\mathrm{id}_V^{-1} \circ \alpha \circ \mathrm{id}_U$ is \begin{align*} B = Q^{-1} A P. \end{align*} This is the matrix of $\alpha$ with respect to the bases $(u_i)$ and $(v_j)$. [/step]