Algèbre linéaire by Joseph Grifone

By Joseph Grifone

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9 Somme et somme directe de plusieurs sous-espaces Nota – La th´ eorie et les r´ esultats de ce paragraphe ne seront utilis´ es qu’au chapitre 6. On peut donc en diff´ erer l’´ etude jusqu’au commencement du chapitre 6. 35 – Soient E1 , . . , Ep des sous-espaces vectoriels d’un mˆeme espace vectoriel E. On note : E1 + · · · + Ep = {x ∈ E | ∃ x1 ∈ E1 , . . , ∃ xp ∈ Ep : x = x1 + · · · + xp } . E1 + · · · + Ep est un sous-espace vectoriel de E dit somme des sous-espaces Ei . 36 – Si G1 , . . , Gp sont des familles g´en´eratrices respectivement de E1 , .

Alors E = E1 ⊕ E2 ⇐⇒ 1. 2. E1 ∩ E2 = {0} dim E = dim E1 + dim E2 . 32). Montrons qu’elles sont suffisantes. Soient {v1 , . . , vp } une base de E1 et {wp+1 , . . , wn } une base de E2 , n ´etant la dimension de E. , wq } est une base. Soit λ1 v1 + · · · + λp vp + µp+1 wp+1 + · · · + µn wn = 0. On a : λ1 v1 + · · · + λp vp = − (µp+1 wp+1 + · · · + µn wn ) ∈E1 ∈E2 Soit y ce vecteur ; y ∈ E1 ∩ E2 donc y = 0. D’o` u : λ1 v1 + · · · + λp vp = 0 et, comme la famille {v1 , . . , vp } est libre : λ1 = 0, .

Jusqu’`a xp = 0 . APPLICATIONS I Extraire une base d’une famille g´en´eratrice et d´eterminer les relations liant les vecteurs. Exemple : D´eterminer une base du sous-espace de R4 engendr´e par les vecteurs v1 = (1, 1, 0, −1) , v2 = (−1, 1, 1, 0) , v3 = (0, 2, 1, −1) et les ´eventuelles relations lin´eaires. 2 3 1 1 0 −1 6 7 6 −1 1 1 07 4 5 0 2 1 −1 2 1 6 6 −→ 4 0 0 1 0 2 1 0 0 v2′ −1 2 v1 v2 v3 3 7 −1 7 5 0 1 6 −→ 6 40 0 1 0 2 1 2 1 −1 3 7 −1 7 5 −1 v1 ′ v2 =v1 +v2 v3 v1 ′ v2 ′ ′ v3 =v2 −v3 Ainsi v1 = (1, 1, 0, −1) et = (0, 2, 1, −1) forment une base de Vect{v1 , v2 , v3 }.

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