Corrigé DM4. - Lycée Jean Perrin

TSI1
no4
ERn
f E p
16p6nKer fp⊕Im fp=E(1)
fp=f◦f◦ · · · ◦ f
p
f E f Ker f={0}
Im f=E
Ker f⊕Im f=E, p = 1
n= 3 B= (e1, e2, e3)E f
EB
A=
4−1 5
−2−1−1
−4 1 −5
u=xe1+ye2+ze3X=
x
y
z
u
Bu∈Ker f f(x) = 0 AX = 0
AX = 0 ⇐⇒ 
4x−y+ 5z= 0
−2x−y−z= 0
−4x+y−5z= 0
⇐⇒ 4x−y+ 5z= 0
6x+ 6z= 0 L2←L2−L1
AX = 0 ⇐⇒ z=−x
y=−x⇐⇒ 
x
y
z
=
x
−x
−x
=x
1
−1
−1
(e′
1) = (e1−e2−e3) Ker f
Im f
f(e1), f(e2), f(e3)=4e1−2e2−4e3,−e1−e2+e3,5e1−e2−5e3
dim Im f= 2
(e′
2, e′ 3) = (4e1−2e2−4e3,−e1−e2+e3) Im f

detB(e′
1, e′ 2, e′ 3) = 0
e′
1=1
3(e′
2+e′ 3)
(e′
1, e′ 2, e′ 3)EKer fIm f
E
p= 1
Bf2
A2=
−2 2 −4
−2 2 −4
2−2 4
u=xe1+ye2+ze3Ker f
−2x+ 2y−4z= 0 
x
y
z
=
y−2z
y
z
=y
1
1
0
+z
−2
0
1
(e′′
1, e′′ 2) = (e1+e2,−2e1+e3) Ker f2
Im f2= Vect f2(e1), f2(e2), f2(e3)= Vect(−2e1−2e2+ 2e3)
(e′′
3) = (e1+e2−e3) Im f2
det
B(e′′
1, e′′ 2, e′′ 3) =
1−2 1
1 0 1
0 1 −1
=−2̸= 0
(e′′
1, e′′ 2, e′′ 3)E
E= Ker f2+ Im f2dim E= dim Ker f2+ dim Im f2
E= Ker f2⊕Im f2
p= 2
m n = 4 B= (e1, e2, e3, e4)E f
EB
Am=
0−1 0 0
0m0 0
1 0 −m−1
0 1 0 0

TSI1
u=xe1+ye2+ze3+te4Ker f
Am
x
y
z
t
= 0 
−y= 0
my = 0
x−mz −t= 0
y= 0
y= 0 x=mz +t
x
y
z
t
=
mz +t
0
z
t
=z
m
0
1
0
+t
1
0
0
1
(me1+e3, e1+e4) Ker f
(e′
1, e′ 2) = (me1+e3, e1+e4) Ker f
f(e1), f(e2), f(e3), f(e4)Im f
(e′
3, e′ 4) = (e3,−e1+me2+e4) Im f
p= 1 (e′
1, e′ 2, e′ 3, e′ 4)E
detB(e′
1, e′ 2, e′ 3, e′ 4)̸= 0
det
B(e′
1, e′ 2, e′ 3, e′ 4) = 
m1 0 −1
0 0 0 m
1 0 1 0
0 1 0 1
=
m1−1
0 0 m
0 1 1
=−mm1
0 1 =−m2
detB(e′
1, e′ 2, e′ 3, e′ 4)̸= 0 m̸= 0
p= 1 m̸= 0
p(1) m= 0
A0=
0−1 0 0
0000
100−1
0100
A2
0=
0 0 0 0
0 0 0 0
0−200
0 0 0 0
(e1, e3, e4) Ker f2(e3) Im f2
Ker f2Im f2p̸= 2 A3=
0 (e1, e2, e3, e4) Ker f3Ker f3⊕Im f3=E
m= 0 p= 3
f E
k∈N
⋆Ker fk⊂Ker fk+1
x∈Ker fkfk(x) = 0 fk+1(x) = ffk(x)=f(0) = 0
x∈Ker fk+1

⋆Im fk+1 ⊂Im fk
y∈Im fk+1 x∈E y =fk+1(x)
y=fkf(x)=fk(z)z=f(x)
y∈Im fk
∀k∈NKer fk⊂Ker fk+1 Im fk+1 ⊂Im fk
∀k∈Nak= dim Ker fk
∀k∈NKer fk⊂Ker fk+1 dim Ker fk6dim Ker fk+1
∀k∈Nak6ak+1
(ak)k∈NN
(ak)n= dim E E
E
k ak+1 > akak
ak+1 >ak+ 1
k ak>k+a0
an+1 >n+ 1 + a0>n+ 1
FNak=ak+1
F
N
F
p∈[[1, n]] F F ap=ap+1
k < p ak̸=ak+1 p∈[[1, n]]
• ∀k∈[[0, p −1]] dim Ker fk̸= dim Ker fk+1
•dim Ker fp= dim Ker fp+1
p∈[[1, n]]
• ∀k∈[[0, p −1]] Ker fk̸= Ker fk+1
•Ker fp= Ker fp+1
k>p∀k∈[|p, +∞[|Ker fk= Ker fp
⋆ k =p
⋆ k p Ker fk= Ker fp
Ker fk⊂Ker fk+1 Ker fp⊂Ker fk+1
x∈Ker fk+1 f(x)∈Ker fkf(x)∈Ker fp
fpf(x)= 0 = fp+1(x)x∈Ker fp+1
Ker fp= Ker fp+1 x∈Ker fp
Ker fk+1 ⊂Ker fp

TSI1
⋆
∀k∈[|p, +∞[|Ker fk= Ker fp
E= Ker fp⊕Im fp
⋆ x ∈Ker fp∩Im fpfp(x) = 0 y∈E x =fp(y)
f2p(y) = fpfp(y)=fp(x) = 0
y∈Ker f2pKer f2p= Ker fp
y∈Ker fp
x=fp(y) = 0
Ker fp∩Im fp={0}
⋆dim Ker fp+ dim Im fp= dim E
E= Ker fp⊕Im fp