Let E, F, G be three right (resp. left) modules over a ring A, (ei)i∈I, (fk)k∈K, (gl)l∈L respective finite bases of E, F, G, u:E→F, v:F→G two linear mappings, M(u) the matrix of u relative to the bases (ei), (fk), M(v) the matrix of v relative to the bases (fk), (gl) and M(v∘u) the matrix of v∘u relative to the bases (ei), (gl); then
(18 D)
M(v∘u)=M(v)M(u) (18 G)
tM(v∘u)=tM(u)tM(v).
We prove for example (18 G). For all x∈E, by (17 G):
tM(x)⋅tM(v∘u)=tM(v(u(x)))=tM(u(x))⋅tM(v)=tM(x)⋅tM(u)⋅tM(v) by associativity; the corollary then follows from no. 3, Proposition 1 since the matrix
tM(x) with one row is arbitrary.
Remark (1). Formula (17 D) can be considered as a special case of (18 D). For there corresponds canonically to every x∈E the linear mapping θx:Ad→E mapping every α∈A to xα (§ 2, no. 1). It is immediate that the matrix M(θx) with respect to the basis 1 of Ad and the basis (ei) of E is just the matrix M(x); similarly M(θu(x))=M(u(x)) and formula (17 D) can therefore be considered as a translation of the relation
θu(x)=u∘θx.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.