$T\mapsto [T{]}_{\mathcal{B}}^{\mathcal{C}}$ is a vector-space isomorphism $\mathcal{L}(V,W)\cong M_{m\times n}(F)$

For every basis vector $b_j$, coordinate uniqueness in [L2] gives $[(S+T)(b_j){]}_{\mathcal{C}}=[S(b_j){]}_{\mathcal{C}}+[T(b_j){]}_{\mathcal{C}}$ and $[(\lambda T)(b_j){]}_{\mathcal{C}}=\lambda [T(b_j){]}_{\mathcal{C}}$, so $\mathrm{\Phi }$ is linear column by column.

If $\mathrm{\Phi }(S)=\mathrm{\Phi }(T)$, then [L2] gives $S(b_j)=T(b_j)$ for every $j$; linearity and the unique expansion of every vector in $\mathcal{B}$ give $S=T$, so $\mathrm{\Phi }$ is injective.

Given $A=(a_{ij})\in M_{m\times n}(F)$, prescribe $T(b_j):={\sum }_{i<m}{a}_{ij}{c}_i$ and, for the unique expansion $v={\sum }_{j<n}{x}_j{b}_j$ from [L2], define $T(v):={\sum }_{j<n}{x}_{j}T(b_j)$. This is well defined and linear, and the $j$-th matrix column is the $j$-th column of $A$; hence $\mathrm{\Phi }(T)=A$. Together with steps 1.1 and 2.1, $\mathrm{\Phi }$ is a linear bijection. Its set-theoretic inverse is linear: if $A=\mathrm{\Phi }(S)$ and $B=\mathrm{\Phi }(T)$, then injectivity and linearity give ${\mathrm{\Phi }}^{-1}(A+B)=S+T$ and ${\mathrm{\Phi }}^{-1}(\lambda A)=\lambda S$. Thus $\mathrm{\Phi }$ is a linear isomorphism by [L3].

If $n=0$, then $V$ is the zero space and both sides contain only their zero element; if $m=0$, then $W$ and $M_{m\times n}(F)$ are zero spaces and the only map is the zero map. Thus the construction also proves the isomorphism in every zero-dimensional case.