Bourbaki

Corollary

Tag 05WN, permanent. Cite it as [Bourbaki, Tag 05WN].

Algebra, chapter II, § 10. Matrices

Let E, F, G be three right (resp. left) modules over a ring A, (ei)i∈I(e_i)_{i \in I}, (fk)k∈K(f_k)_{k \in K}, (gl)l∈L(g_l)_{l \in L} respective finite bases of E, F, G, u:E→Fu : E \to F, v:F→Gv : F \to G two linear mappings, M(u)M(u) the matrix of u relative to the bases (ei)(e_i), (fk)(f_k), M(v)M(v) the matrix of v relative to the bases (fk)(f_k), (gl)(g_l) and M(v∘u)M(v \circ u) the matrix of v∘uv \circ u relative to the bases (ei)(e_i), (gl)(g_l); then

(18 D)

M(v∘u)=M(v)M(u) M(v \circ u) = M(v) M(u)
(18 G)
tM(v∘u)=tM(u)tM(v). {}^t M(v \circ u) = {}^t M(u) {}^t M(v).

We prove for example (18 G). For all x∈Ex \in E, by (17 G):

tM(x)⋅tM(v∘u)=tM(v(u(x)))=tM(u(x))⋅tM(v)=tM(x)⋅tM(u)⋅tM(v) {}^t M(x) \cdot {}^t M(v \circ u) = {}^t M(v(u(x))) = {}^t M(u(x)) \cdot {}^t M(v) = {}^t M(x) \cdot {}^t M(u) \cdot {}^t M(v)
by associativity; the corollary then follows from no. 3, Proposition 1 since the matrix tM(x){}^t M(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∈Ex \in E the linear mapping θx:Ad→E\theta_x : A_d \to E mapping every α∈A\alpha \in A to xαx \alpha (§ 2, no. 1). It is immediate that the matrix M(θx)M(\theta_x) with respect to the basis 1 of AdA_d and the basis (ei)(e_i) of E is just the matrix M(x)M(x); similarly M(θu(x))=M(u(x))M(\theta_{u(x)}) = M(u(x)) and formula (17 D) can therefore be considered as a translation of the relation

θu(x)=u∘θx. \theta_{u(x)} = u \circ \theta_x.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed edition, 1998, Springer. Not translated.