Bourbaki

Proposition 19

Tag 0779, permanent. Cite it as [Bourbaki, Tag 0779].

Algebra, chapter VII, § 5. Endomorphisms of vector spaces

— Let E be a finite dimensional vector space over a commutative field K, and let S\mathscr{S} be a set of commuting triangularisable endomorphisms of E. Then there exists a basis of E with respect to which the matrix of each element u of S\mathscr{S} is lower triangular and the matrix of usu_s is diagonal, with the same diagonal elements as that of u.

By Cor. 3 of VII, p. 4.5, the set Ts\mathfrak{T}_s of absolutely semi-simple components of elements of T\mathfrak{T} consists of diagonalisable elements which commute with one another, so is diagonalisable (VII, p. 41, Prop. 13), the set Tn\mathfrak{T}_n of nilpotent components of elements of S\mathscr{S} consists of nilpotent elements which commute with one another, and each element of Tn\mathfrak{T}_n commutes with each element of Ts\mathfrak{T}_s. Arguing as in the proof of Prop. 13 (VII, p. 41), we see that there exists a decomposition of E as a direct sum of subspaces EiE_i, which are invariant under Ts\mathfrak{T}_s and Tn\mathfrak{T}_n, and such that the restriction of each element of S\mathscr{S}, to each EE, is a homothety. Replacing EE by each EE, in turn, we may assume that the elements of S\mathscr{S}, are homotheties; it is enough to prove that there exists a basis of E with respect to which the elements of Tn\mathfrak{T}_n are represented by lower triangular matrices with zero diagonals; we are thus reduced to the case where S\mathscr{S} consists of nilpotent elements.

Now suppose E≠0E \neq 0, and let F be a nonzero subspace of EE, invariant under T\mathfrak{T}, of minimum dimension. Then for each u∈Tu \in \mathfrak{T}, the kernel of the restriction of uu to F is nonzero and invariant under T\mathfrak{T} (VII, p. 41, Lemma 3); by the choice of F the restriction of uu to F is thus zero for all u∈Tu \in \mathfrak{T}. Let x∈F, x≠0x \in F,\ x \neq 0; then u(x)=0u(x) = 0 for all u∈Su \in \mathscr{S}; arguing by induction on the dimension of E, we may suppose that there exists a basis (eˉ1,...,eˉn−1)(\bar{e}_1, ..., \bar{e}_{n-1}) of the quotient E′=E/KxE' = E/Kx such that, for all u∈Tu \in \mathfrak{T}, the endomorphism uˉ\bar{u} of E′E' induced by uu has a matrix with respect to this basis which is lower triangular with zero diagonal; if ei∈Ee_i \in E projects onto a, for i=1,...,n−1i = 1, ..., n-1, then the basis (e1,...,en−1,x)(e_1, ..., e_{n-1}, x) satisfies the required conditions.

Exercises

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

Cited by

Cites

Nothing.

Transcribed from the printed edition, 2003, Springer. Not translated.