A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms

By [L3], $L_A$ and $L_B$ are linear. If $AB=BA=I_n$, then $L_AL_B=L_{AB}=\operatorname{id}$ and $L_BL_A=L_{BA}=\operatorname{id}$, so $L_A$ is a linear isomorphism with inverse $L_B$. This includes $n=0$, where the unique empty matrix and the unique zero-space map are their own inverses.

Conversely, if $L_A$ has a linear inverse $S$, [L4] represents $S$ by a matrix $B$ in the standard coordinate basis. The two identity composites and [L2] give $AB=I_n=BA$, so $A$ is invertible.

For a general linear isomorphism $T$, the list $T(\mathcal B)$ is an ordered basis of $W$: $T^{-1}$ transfers both linear independence and spanning back to $\mathcal B$. Hence [L5] shows that $\mathcal B$ and $\mathcal C$ have the same length, so the two displayed representation matrices are square. Representing $T^{-1}\circ T$ and $T\circ T^{-1}$ in the chosen bases, [L2] gives both inverse equations for $[T]_{\mathcal B}^{\mathcal C}$ and $[T^{-1}]_{\mathcal C}^{\mathcal B}$.