Proposition 2

Suppose I and K are finite. For every element $x \in E$, the matrix $M(u(x))$ with respect to the basis $(f_k)$ is given by the formula

(17 D)

(17 G)

We verify for example (17 G). Let $x = \sum_i x_{i0} e_i, \quad u(x) = \sum_k y_{k0} f_k$ with $x_{i0} \in A, \ y_{k0} \in A$; then $u(x) = u\left( \sum_i x_{i0} e_i \right) = \sum_i x_{i0} u(e_i) = \sum_{i, k} x_{i0} u_{ki} f_k$; whence y_{k0} = \sum_i x_{i0} u_{ki}. In order to bring the two indices $i$ along side one another, we consider the transpose matrices ${}^t M(x) = (x'_{0i})$, where $x'_{0i} = x_{i0}$ and

where $u'_{ik} = u_{ki}$; then $y_{k0} = \sum_i x'_{0i} u'_{ik}$ and the right hand side is the element of index $k$ of the matrix with one row ${}^t M(x) \cdot {}^t M(u)$, whence (17 G).

When A is commutative, (17 G) reduces to (17 D) by formula (4) of no. 2.