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4 Dans R3 muni des lois habituelles, lequel vectoriel ?  F = (x, y, z) ∈ R3 de ces sous-ensembles est-il un sous-espace ˛  ff ˛ x−y+z =0 ˛ ˛ 2x −y = 0 G = {(x, y, z) ∈ R3 | (x − y)2 = 2x + y} 5 Dans M2 (R) muni les lois de l’exemple 3, page 5, lequel de ces sous-ensembles est-il un sous-espace vectoriel ?  „ «ff a+b c E = A ∈ M2 (R) | A = 2c −b  „ «ff 1 a F = A ∈ M2 (R) | A = 0 b 6 Soit R[a,b] = {f : [a, b] −→ R} muni des lois de l’exemple 4, page 5. Lequel de ces sousensembles est-il un sous-espace vectoriel ?

6. 3 Application aux familles libres et aux familles g´ en´ eratrices I La famille {v1 , . . , vp } est-elle libre ? Exemple : V´erifier que les vecteurs de R3 v1 = (1, −2, −3) , v2 = (2, 3, −1) , v3 = (3, 2, 1) forment une famille libre. Il s’agit de savoir si x1 v1 + x2 v2 + x3 v3 = 0 =⇒ x1 le syst`eme : 8 x1 + 2 x2 + 3 x3 = > > < −2 x1 + 3 x2 + 2 x3 = > > : −3 x1 − x2 + x3 = admet comme seule solution, la solution nulle. On a : 8 L1 x1 + 2 x2 + 3 x3 = 0 L1 > > < 7 x2 + 8 x3 = 0 L′2 L′2 =L2 +2L1 > > : 7L3 −5L′2 L′3 =L3 +3L1 5 x2 + 10 x3 = 0 = 0, x2 = 0, x3 = 0 c’est-` a-dire si 0 0 0 8 x1 + 2 x2 + 3 x3 = 0 > > < 7 x2 + 8 x3 = 0 > > : 30 x3 = 0 d’o` u : x1 = 0, x2 = 0, x3 = 0.

Bp sont libres, et, par cons´equent, λ1 = 0, . . , λr = 0, . . , µ1 = 0, . . , µs = 0. Notons que E = E1 ⊕ · · · ⊕ Ep est un sous-espace vectoriel de E et peut, bien entendu, ˆetre diff´erent de E. 9 Somme et somme directe de plusieurs sous-espaces 27 E2 v O O E2 E = E1 ⊕ E2 E1 E1 E1 = axe Ox, E2 = axe Oy. E = E1 ⊕ E2 = plan xOy. On a : E ⊂ R3 . E1 = plan Oxy, E2 = Vect{v}, (v ∈ / E1 ). E = E1 ⊕ E2 . On a : E = R3 . Figure 12 Figure 13 = Il faut donc bien distinguer les notions «les Ei sont en somme directe» (cf.

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