Bourbaki

Exercise 17, § 10

Algebra, II

Bourbaki marks this one as harder than the rest.

Let K be a field (commutative or not), E a right vector space of dimension n over K, u an endomorphism of E. a) Now prove that if n>1n > 1 and if u is not a central homothety, there exists a basis of E such that the matrix of u with respect to this basis is of the form (αij)(\alpha_{ij}) with αij=0\alpha_{ij} = 0 for j>i+1j > i + 1 and ∑i=1n−1αi,i+1=1\sum_{i=1}^{n-1} \alpha_{i, i+1} = 1 (reason by induction on n). b) Now prove that if n>2n > 2, there exists a basis of E such that the matrix of u with respect to this basis is of the form (αij)(\alpha_{ij}) with αij=0\alpha_{ij} = 0 for j>i+1j > i + 1 and ∑i=1n−1αi,i+1=0\sum_{i=1}^{n-1} \alpha_{i, i+1} = 0. (Reason by induction on n using a), noting that it amounts to the same thing to prove the proposition for u or for u+γ1Eu + \gamma 1_E, where γ\gamma belongs to the center of K, and by studying separately the case where u is of rank 1 or 2.) c) Let A=(αij)A = (\alpha_{ij}) be a matrix having the properties described in b). Let, on the other hand, S be the matrix (εij)(\varepsilon_{ij}) of order n such that εij=0\varepsilon_{ij} = 0, except for i=j+1i = j + 1 where εj+1,j=1\varepsilon_{j+1, j} = 1. Now prove that there exists a matrix B=(βij)B = (\beta_{ij}) of order n such that A=λEnn+(BS−SB)A = \lambda E_{nn} + (BS - SB) with λ∈K\lambda \in K. (Take B such that βi1=0\beta_{i1} = 0 for 1≤i≤n1 \leq i \leq n and βij=0\beta_{ij} = 0 for j>i+2j > i + 2). d) Suppose K commutative. Deduce from c) that every endomorphism u of E such that Tr(u)=0\mathrm{Tr}(u) = 0 can be written vw−wvvw - wv, where v and w are endomorphisms of E.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.