Problem 20 Goal — Transpose injectivity semiring

For every natural number $n$ (including $0$), every type $R$ equipped with a chosen unital commutative-semiring structure, and every matrix $A=(A_{i,j})$ with entries in $R$ and row and column indices $i,j\in\operatorname{Fin}(n)=\{k\in\mathbb N\mid k<n\}$, the following two conditions are equivalent. The first condition is that left multiplication by $A$ is injective on column vectors: for all functions $u,v:\operatorname{Fin}(n)\to R$, if $\sum_{j\in\operatorname{Fin}(n)}A_{i,j}u_j=\sum_{j\in\operatorname{Fin}(n)}A_{i,j}v_j$ for every $i\in\operatorname{Fin}(n)$, then $u_j=v_j$ for every $j\in\operatorname{Fin}(n)$. The second condition is that left multiplication by the transpose of $A$ is injective: for all functions $u,v:\operatorname{Fin}(n)\to R$, if $\sum_{j\in\operatorname{Fin}(n)}A_{j,i}u_j=\sum_{j\in\operatorname{Fin}(n)}A_{j,i}v_j$ for every $i\in\operatorname{Fin}(n)$, then $u_j=v_j$ for every $j\in\operatorname{Fin}(n)$. All sums, products, zeros, and ones are those of the chosen commutative-semiring structure, and transposition means exactly that the entry in position $(i,j)$ becomes $A_{j,i}$. No nontriviality, finiteness of $R$, additive inverses, cancellation, integral-domain, or field assumption is imposed on $R$, and no condition is imposed on $A$. In particular, semirings with $0=1$ are included; when $R$ is a subsingleton, every such vector type is a subsingleton and both maps are automatically injective. When $n=0$, the index set is empty, every displayed sum is the empty sum $0$, there is exactly one vector $\operatorname{Fin}(0)\to R$, and both maps are injective. The assertion is a biconditional, so each injectivity condition implies the other; it does not assert that the two matrix-vector maps are equal or that either one is injective without the other being injective.