Proposition 1

Let $\mathbf{A}, \mathbf{B}$ be two rings and $M = (m_{ik})_{(i,k) \in I \times K}$ and

two matrices with finite indexing sets over an $(\mathbf{A}, \mathbf{B})$*-bimodule* $G$. Suppose that for every matrix unit $L = (a_i)_{i \in I}$ with one row and elements in $\mathbf{A}$ and every matrix unit $C = (b_k)_{k \in K}$ with one column and elements in $\mathbf{B}$, $L.M.C = L.M'.C$ (the products being calculated via the external laws of the $(\mathbf{A}, \mathbf{B})$*-module* $G$); then* $M = M'$.

If $L$ is taken to be the matrix unit $(a_s)$ with $a_i = 1, a_s = 0$ for $s \neq i$, and $C$ the matrix unit $(b_t)$ with $b_k = 1, b_t = 0$ for $t \neq k$, the products $L.M.C$ and $L.M'.C$ are matrices with a single element respectively equal to $m_{ik}$ and $m'_{ik}$.

Let $\mathbf{A}, \mathbf{B}$ be two rings and $\sigma : \mathbf{A} \to \mathbf{B}$ a homomorphism.

For every matrix $M = (m_{ik})$ over $\mathbf{A}$, we shall denote by $\sigma(M)$ the matrix $(\sigma(m_{ik}))$ over $\mathbf{B}$; clearly $\sigma(aM) = \sigma(a)\sigma(M)$, $\sigma(Ma) = \sigma(M)\sigma(a)$ for $a \in \mathbf{A}$, also $\sigma(^tM) = (^t(\sigma(M)))$ and

when the operations considered are defined, the products on the left and right hand sides of (12) being calculated in $\mathbf{A}$ and $\mathbf{B}$ respectively. When $\sigma$ is denoted by $x \mapsto x^\sigma$, we write $M^\sigma$ instead of $\sigma(M)$.

Consider in particular an anti-endomorphism $\sigma$ of $\mathbf{A}$, that is a homomorphism of $\mathbf{A}$ to the opposite ring $\mathbf{A}^0$, or a mapping of $\mathbf{A}$ into itself such that

for all $a, a'$ in $\mathbf{A}$; then, for two matrices $M, M'$ over $\mathbf{A}$ whose product $MM'$ is defined, where the products on the two sides are calculated in $\mathbf{A}$; this follows immediately from (10) and (12).