Révisions d`algèbre linéaire

K=R C
E+·
(E, +,·)K
(E, +)
• ∀u, v ∈E, u +v∈E+
• ∀u, v, w ∈E, (u+v) + w=u+ (v+w) +
• ∃0E∈E∀u∈E, u + 0E= 0E+u=u
• ∀u∈E∃v∈E / u +v=v+u= 0Ev=−u
• ∀u, v ∈E u +v=v+u+
·
• ∀λ∈K∀u∈E, λ ·u∈E+
• ∀λ, µ ∈K,∀u∈E, λ ·(µ·u) = (λµ)·u
• ∃u∈E1·u=u
• ∀λ, µ ∈K,∀u∈E(λ+µ)·u=λ·u+µ·u
• ∀λ∈K,∀u, v ∈E, λ ·(u+v) = λ·u+λ·v
EK
λ·u λu

K K
KnK
C R
Mn,p(K)K
K[X]Kn[X]K
KNK
RIR
(v1, v2, . . . , vn)n E
(vi)16i6n
n
i=1
λivi=λ1v1+λ2v2+· · · +λnvn,
i∈J1, nKλiK
(v1, . . . , vn)
V ect(v1, v2, . . . , vn)
u= (1,−1,2), v = (5,3,1), w = (−2,2,−4)
x= (3,−1,−3) V ect(u, v, w)
a b c
x=au +bv +cw
x=au +bv +cw ⇐⇒ (3,−1,−3) = a(1,−1,2) + b(5,3,1) + c(−2,2,−4)
⇐⇒ (3,−1,−3) = (a+ 5b−2c, −a+ 3b+ 2c, 2a+b−4c)
⇐⇒
a+ 5b−2c= 3
−a+ 3b+ 2c=−1
2a+b−4c=−3
⇐⇒ b= 1
a= 2c−2
x= 4u+v+ 3w∈V ect(u, v, w)
(a, b, c)

(v1, . . . , vn)E
(v1, . . . , vn)E
E=V ect(v1, . . . , vn)
E
(v1, . . . , vn)
(1, X, X2, . . . , Xn−1, Xn)Rn[X]
((1,0,0),(0,1,0),(0,0,1)) C3
EKF E
FE
•F F
•F
∀u, v ∈F, ∀λ∈K, λu +v∈F
E
a∈CE a E
E⊂C[X]E={P∈C[X]/ P (a) = 0}EC[X]
•E a
•P, Q ∈E λ, µ ∈CλP +µQ ∈E
(λP +µQ)(a) = λP (a) + µQ(a)=0
E
EC[X]

V ect
E
E=x2x−y
y x +y, x, y ∈RE
E=x1 2
0 1 +y0−1 1 1 , x, y ∈R=V ect 1 2 0 1 ,0−1 1 1
EM2(R)
EK
•(v1, . . . , vn)E n (λ1, . . . , λn)∈Kn
p
i=1
λivi= 0 =⇒λ1=λ2=· · · =λn= 0
•
v1, . . . , vn
λ1v1+λ2v2+· · · +λnvn= 0
λ1, . . . , λn
(1 + X+X2,3 + X+ 5X2,2 + X+ 3X2)C[X]
a, b, c ∈C
a(1 + X+X2) + b(3 + X+ 5X2) + c(2 + X+ 3X2)=0
(a+ 3b+ 2c)+(a+b+c)X+ (a+ 5b+ 3c)X2= 0 ⇐⇒
a+ 3b+ 2c= 0
a+b+c= 0
a+ 5b+ 3c= 0
⇐⇒ a=b
c=−2a
(a, b, c)
((1,3,−3),(4,2,−3),(−1,7,6)) R3
a, b, c ∈R
a(1,3,−3) + b(4,2,−3) + c(−1,7,6) = (0,0,0)

(1 + 4b−c, 3a+ 2b+ 7c, −3a−3b+ 6c) = (0,0,0)
a+ 4b−c= 0
3a+ 2b+ 7c= 0
−3a−3b+ 6c= 0
⇐⇒
a+ 4b−c= 0
b=c
3b+c= 0
⇐⇒
a+ 4b−c= 0
b=c
c= 0
⇐⇒ a=b=c= 0
ER+∗x∈R+∗
f1(x) = 1, f2(x) = x, f3(x) = ln(x), f4(x) = ex
(f1, f2, f3, f4)E
a, b, c, d ∈R
af1+bf2+cf3+df4= 0
∀x∈R+∗, a +bx +cln(x) + dex= 0
0+cln(x)c6= 0
±∞ 0
c= 0
+∞dexd6= 0
±∞ 0
d= 0
∀x∈R+∗, a +bx = 0
a+bX R+∗
a=b= 0
(u, v)λ∈R
u=λv v =λu