Fiche 1 : Algèbre - PCSI

m
(m−1)x+y+z= 0
x+ (m−1)y+z= 0
x+y+ (m−1)z= 0
f:R3→R3
(x, y, z)7→ (x−y−z, −x+y−z, −x−y+z)
fR3
A f
m1∆ = Ker(f−m1Id)m2
P=Ker(f−m2Id)
∆PR3
1∆23P
D f B= (1, 2, 3)R3
n DnAn
ER Rn[X]n∈Nn≥2
(P0, P1,· · · , Pn)E
P0= 1 ; P1=X;Pk=1
k!X(X−k)k−1(2 ≤k≤n).
f E E
∀P∈E f(P) = Q Q(X) = P(X)−P0(X+ 1) .
(P0, P1,· · · , Pn)E
∀k∈ {1, ..., n}Pk
0(X+ 1) = Pk−1(X)
k∈ {0, ..., n}Qk=f(Pk)
Qk0≤k≤n Pj0≤j≤n
f E f∈GL(E)
Pk0≤k≤n Qj
n= 3
X3B= (P0, P1, P2, P3)
PR3[X]
P(X)−P0(X+ 1) = X3.
P
0≤k≤n Pkk
(P0, P1,· · · , Pn)n+ 1
Rn[X]n+ 1

1≤k≤n
P0
k(X) = 1
k!((X−k)k−1+X(k−1)(X−k)k−2) = 1 k!(X−k)k−2k(X−1) = (X−1)(X−k)k−2 (k−1)!
Pk
0(X+ 1) = Pk−1(X)
Q0=P0k≥1Qk(X) = Pk(X)−P0
k(X+ 1) = Pk(X)−Pk−1(X)
Qk=Pk−Pk−1
(Q0, Q1,· · · , Qn)
E f E
E E E
Q0=P0
Q1=P1−P0
Q2=P2−P1
. . .
Qn=Pn−Pn−1
P0=Q0
P1=Q1+Q0
Q2=Q2+Q1+Q0
. . .
Qn=Qn+... +Q1+Q0
n= 3 P0= 1 P1=X P2=X2
2−X P3=X3 6−X2+3X 2
X3= 6P3+ 12P2+ 3P1
X3= 6(Q0+Q1+Q2+Q3) + 12(Q0+Q1+Q2) + 3(Q0+Q1)
= 6Q3+ 18Q2+ 21Q1+ 21Q0
k∈ {0, ..., n}Qk=f(Pk)
X3=f(P)(X) = f(6Q3+ 18Q2+ 21Q1+ 21Q0)(X)
f P = 6Q3+ 18Q2+ 21Q1+ 21Q0=X3+ 3X2+ 12X+ 2