Cours de géométrie algébrique. 2 : précis de géométrie by Jean Alexandre Dieudonné

By Jean Alexandre Dieudonné

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0 0 · · · Bn where N := k1 + · · · + kn . We consider a matrix A ∈ MN (C) with spectrum σ(A) = {λ1 , . . , λp }. The main goal of this section consists in, for each 1 ≤ j ≤ p, constructing a basis Bj := {ej1 , . . , ejma (λj ) } 20 Chapter 1. The Jordan Theorem of the ascent generalized eigenspace N [(A − λj I)ν(λj ) ] so that the matrix Aj,Bj of the operator from N [(A − λj I)ν(λj ) ] to itself defined by restriction of A is blockdiagonal Aj,Bj = diag{Jj,i1 , Jj,i2 , . . , Jj,inj } for some integer number nj ≥ 1, where ih ∈ {1, .

Given a linear subspace Y ⊂ CN such that CN = N [A0 − λ0 I] ⊕ Y, use the implicit function theorem to prove that there exist a real number δ > 0 and two unique functions Λ ∈ C k ((−δ, δ), C) and Φ ∈ C k ((−δ, δ), Y ) such that Λ(0) = λ0 , Φ(0) = ϕ0 , and for all s ∈ (−δ, δ), Φ(s) − ϕ0 ∈ Y, A(s)Φ(s) = Λ(s)Φ(s). • Calculate the jth derivative of Λ at 0 for each 1 ≤ j ≤ k, and prove that Λ (0) = 0 if and only if A (0)ϕ0 ∈ R[A0 − λ0 I]. 5. Suppose Ω ⊂ R is an open interval containing zero, aij ∈ C(Ω, C) for each i, j ∈ {1, .

P } and pick 1 ≤ j ≤ p. 3 all have real components if λj ∈ R, whereas if λj = αj + iβj , with αj , βj ∈ R, βj = 0, then, their components are complex. Suppose this is the case, and reorder the eigenvalues, if necessary, so that ¯j . 30) where e¯ is the complex conjugate vector of e, that is, the vector whose components are the complex conjugate components of e. Therefore, for each k ≥ 0, dim N [(A − λj I)k ] = dim N [(A − λj+1 I)k ] and, in particular, ν(λj ) = ν(λj+1 ) and ma (λj ) = ma (λj+1 ).

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