Lemme d'échange de Steinitz

Lemme d'échange de Steinitz (Redirected from Exchange theorem) Jump to navigation Jump to search The Steinitz exchange lemma is a basic theorem in linear algebra used, par exemple, to show that any two bases for a finite-dimensional vector space have the same number of elements. The result is named after the German mathematician Ernst Steinitz. The result is often called the Steinitz–Mac Lane exchange lemma, also recognizing the generalization[1] by Saunders Mac Lane of Steinitz's lemma to matroids.[2] Contenu 1 Déclaration 2 Preuve 3 Applications 4 Références 5 External links Statement Let {style d'affichage U} et {style d'affichage W.} be finite subsets of a vector space {style d'affichage V} . Si {style d'affichage U} is a set of linearly independent vectors, et {style d'affichage W.} spans {style d'affichage V} , alors: 1. {style d'affichage |tu|leq |O|} ; 2. There is a set {displaystyle W'subseteq W} avec {style d'affichage |W'|=|O|-|tu|} tel que {displaystyle Ucup W'} spans {style d'affichage V} .

Proof Suppose {style d'affichage U={tu_{1},des points ,tu_{m}}} et {style d'affichage W ={w_{1},des points ,w_{n}}} . We wish to show that for each {parent de style d'affichage {0,des points ,m}} , on a ça {displaystyle kleq n} , and that the set {style d'affichage {tu_{1},pointsc ,tu_{k},w_{k+1},pointsc ,w_{n}}} spans {style d'affichage V} (où le {displaystyle w_{j}} have possibly been reordered, and the reordering depends on {style d'affichage k} ). We proceed by induction on {style d'affichage k} .

For the base case, supposer {style d'affichage k} est zéro. Dans ce cas, the claim holds because there are no vectors {displaystyle u_{je}} , and the set {style d'affichage {w_{1},pointsc ,w_{n}}} spans {style d'affichage V} by hypothesis.

For the inductive step, assume the proposition is true for some {style d'affichage k

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