corrigé - Normalesup.org

•AR2
(2,6) inA −3(2,6) = (−6,−18) /∈A
•BR2
(0; 2) ∈B(−4; 0) ∈B(0,2) + (−4,0) = (−4,2) /∈B
•CR2(x, y)∈C(x0, y0)∈C x =y x0=y0x+x0=y+y0
(x, y) + (x0, y0)∈C x =y λx =λy C
•DR2(0; 0)
•ER2
•
u0r λu0
λr u0v0r s
u0+v0r+s
un+1 =un+r un=u0+nr
•
un= 2n+ 3nu1
u0
=5
2
u2
u1
=13
5
•
2un= 2n+3nvn= 4n+5n
2un+2 = 5un+1 −6un
vn+2 = 9vn+1 −20vn
•(un) (vn)
λ
•(un) (λun)
(un)n0(vn)n1(un+vn)
max(n0, n1)
•
un=1
nvn=1 n2−1 n
un∼1
nvn∼−1 nun+vn=1 n2
λ
nλn
1

F F ={(x, y, x+y)|z∈R}F=V ect((1,0,1); (0,1,1))
F
R3G=V ect((1,1,1); (−1,1,−3))
G F F ∩G=
{(a−b, a +b, a −3b)|a−b+a+b−a+ 3b= 0}={(a−b, a +b, a −3b)|a=−3b}=
{(−4b, −2b, −6b)|b∈R}=V ect((−4,−2,−6))
y1=x1+x2+x3
y2=x1+x2
y3= 2x1+x2−x3
⇔
y1−y2=x3
y2=x1+x2
y1+y3= 3x1+ 2x2
⇔
y1−y2=x3
3y2−y1−y3=x2
y1+y3−2y2=x1
x1x2x3y1y2y3
V ect(y1, y2, y3)
V ect(y1, y2, y3)⊂V ect(x1, x2, x3)yi
xi
y1
y2
y3
=
1 1 1
1 1 0
2 1 −1
x1
x2
x3
x1
x2
x3
=
1−2 1
−1 3 −1
1−1 0
y1
y2
y3
F t
z= 4x
y−7x= 6x−3y= 0
x=y= 0 z= 0 F={(0,0,0, t)|t∈R}=V ect((0,0,0,1))
F((0,0,0,1))
G G =V ect((1,2,4,2); (−2,−3,0,1); (1,0,2,−1)
G
λ(1,2,4,2) + µ(−2,−3,0,1) + ν(1,0,2,−1) = 0
2λ−3µ= 0 µ=3
2λ
4λ+ 2ν= 0 ν=−2λ
λ−2µ+ν= 0 −4λ= 0 λ= 0 µ=ν= 0
G3

H z =x+y
t=y−2x−2y−2x= 0
y=−x z = 0 t=−3x H ={(x, −x, 0,−3x)|x∈R}=
V ect((1,−1,0,−3)) H
R3(x, y, z)∈R3
λ, µ, ν (x, y, z) = λ(0,1,1)+µ(2,0,−1)+ν(2,1,1) 2µ+2ν=
x µ =x
2−ν λ +ν=y λ =y−ν
λ−µ+ν=z ν =−z+y+x
2λ=z−x 2
µ=z−yR3
(4,−1,1) λ µ ν
x= 4 y=−1z= 1 λ=−3µ= 2 ν= 0
(4,−1,1) ((0,1,1); (2,0,−1); (2,1,1)) (−3,2,0)
(1,0,0) −1
2,0,1 2
(x, y, z)∈R3(x, y, z) = λ(3,−1,1) +
µ(2,0,0) + ν(1,−2,4) −λ−2ν=y λ + 4ν=z
ν=y+z
2λ=−2y−z
3λ+2µ+ν=x2µ=x−3λ−ν=x+11
2y+5 2z(4,−1,1)
1,1
2,0(1,0,0) 0,1 2,0
3
(e1+e2, e2+e3, e3+e1)a(e1+e2) + b(e2+e3) +
c(e3+e1) = 0 ⇔(a+c)e1+ (a+b)e2+ (b+c)e3= 0 (e1, e2, e3)
a+c=a+b=b+c= 0 a=b=c= 0
(e1+e2, e2+e3, e3+e1)E
(e1, e1+e2, e1+e2+e3)ae1+b(e1+e2) + c(e1+e2+e3) = 0 ⇔
(a+b+c)e1+ (b+c)e2+ce3= 0 ⇔a+b+c=b+c=c= 0
E
X1X2∈F AX1=AX2= 0 A(X1+X2)=0 X1+X2∈F
AX = 0 A(λX) = 0 F
0R3
A
x−y+ 2z= 0
−2x+y+ 3z= 0
−x+ 5z= 0

x= 5z
−y+ 7z= 0 y−7z= 0
y= 7z F ={(5z, 7z, z)|z∈R}=V ect((5,7,1)) F
Y1Y2∈G2Y1=AX1Y2=AX2Y1+Y2=A(X1+X2)∈
G λY1=λAX1=A(λX1)λY1∈G0 = A×0∈G G
R3
Y=
a
b
c
∈F⇔
x−y+ 2z=a
−2x+y+ 3z=b
−x+ 5z=c
2y x z
a b c a −b+c= 0
G={(a, b, b −a)|a, b ∈R2}=V ect((1,0,−1); (0,1,1))
G
x
1a+b= 0
x > 1x2ln x
a
xln x+b ln x+c x+d= 0
x+∞d0
d= 0
x2d= 0
b b = 0
a+b= 0 b= 0 a= 0 c
cx ln x= 0 R+∗c= 0
x= 1
E4
(I, J, K, L)
aI +bJ +cK +dL = 0 
a c b d
d a c b
b d a c
c b d a
= 0
a=b=c=d= 0
(I, J, K, L)
4
J2=L K2=L L2=L J3=K K3=J L3=L
JK =JJ3=J4= (J2)2=L2=I KJ =I KL =LK =K3=J
JL =LJ =J3=K
E aI +bJ +cK +dL eI +fJ +hK +iL

(ae +bh +cf +di)I+ (af +be +ci +dh)J+ (ag +bi +ce +df)K+ (ai +bf +cg +de)L
E E