Algèbre linéaire 1. Fiche n◦3 Applications linéaires

◦3
K
K=Rf: (x, y, z)∈R37→ (x+y, y +z, z +x)∈R3
K=Cf: (x, y)∈C27→ (xy, y)∈C2
K=Rf: (x, y)∈R27→ (2x+ 1, x −y)∈R2
K=Rf: (x, y)∈R27→ (|x|, y, 0) ∈R3
K=RC1C0R
C1→ C0
f7→ f0
K=R
C0([−1,1]) →R
f7→ Z1
−1
f(t)dt C0([−1,1]) R
[−1,1] R
f:R2→R3f(x, y)=(x−y, y −x, 0)
f:R3→R3f(x, y, z) = (x−y, y −z, z −x)
f:R3[X]→R3[X]f(P(X)) = P(X)−(X+ 1)P0(X)
f:R3→R2
f(1,0,0) = (1,0), f(0,1,0) = (1,1), f(0,0,1) = (0,1).
Im(f)
Im(f)Ker(f)
f:
R3[X]→R2
P7→ (P(0), P (1))
f
Ker(f)f
f f
Ker(f)⊕R1[X] = R3[X]
m fm:R3→R3
fm(x, y, z) = x+y+z, mx +y+ (m−1)z, x +my +z.
m fm
m fm
m fm
(e1, e2, e3)R3m
fm(e1)fm(e2)fm(e3)R3

uR3Im(u)v1=
(1,2,0) v2= (1,1,−1) Ker(u)v3= (1,−1,0)
ER(e1, e2, e3, e4)E
u E
u(e1) = −e2+e3−e4
u(e2) = e1−e2+e3
u(e3) = e1+e4
u(e4) = e2−e3+e4
u2:= u◦u Im(u)⊂Ker(u)
Ker(u)Im(u) = Ker(u)
ERp E
p◦p=p
p(p(x) = x⇔x∈Im(p))
E=Ker(p)⊕Im(p)
p Id −p
p Im(p) = Ker(Id−p)Ker(p) = Im(Id−p)
M N E
p Im(p) = M Ker(p) = N p
E M N
E=R3p E F ={(x, y, z)∈E|x+y+z= 0}
G={(x, y, z)|x=y=z}p(x, y, z) (x, y, z)
ERn u E
k uk=u◦u◦... ◦u k u
p p up= 0
x∈E{x, u(x), u2(x), ..., up−1(x)}
un= 0
ER{e1, e2, e3}E
f E f(e1) = e2f(e2) = e3f(e3) = e1
x E f(x)
f3=IdE
V=Ker(f−IdE)W=Ker(f2+f+IdE)
V W
E=V⊕W
f(V) = V f(W) = W
E F f, g ∈L(E, F )
Ker(f)⊂Ker(g)⇐⇒ ∃h∈L(F)g=h◦f.
E, F G f ∈L(E, F )g∈L(F, G)
g◦f= 0 ⇐⇒ Imf ⊂Kerg
E=F=G

f◦g=g◦f fKer(g)⊂Ker(g)fIm(g)⊂Im(g)
g
E f ∈L(E)Ker(f), Ker(f−Id)Ker(f+Id)
E f ∈L(E)
Ker(f) = Ker(f2)
Im(f)∩Ker(f) = {0E}
Im(f)⊕Ker(f) = E
Im(f) = Im(f2)
1×x x /∈Q
n C1, . . . , Cnc1, . . . , cn
R1×x E =R K =Q
V x, c1, . . . , cn
{1, x}
f:V→Rf(1) = 1 f(x) =
−1
v a ×b f(a)f(b)
v(R) = Pn
i=1 v(Ci)