Devoir Corrigé Algèbre Linéaire L2 MASS Lyon 1

ϕ(R2[X])2
∀P, Q ∈R2[X], ϕ(P, Q) = P(1)Q(−1) + P(−1)Q(1)
ϕ
P, Q R R2[X]λ∈R
ϕ(Q, P ) = Q(1)P(−1) + Q(−1)P(1) = ϕ(P, Q)
ϕ(λP +Q, R) = (λP +Q)(1)R(−1) + (λP +Q)(−1)R(1) = λϕ(P, R) + ϕ(Q, R)
ϕ
R2[X]
ϕ(P, Q)P, Q
ϕ(1,1) = 1.1+1.1 = 2 ϕ(1, X) = ϕ(X, 1) = 1.1+(−1).1 = 0
ϕ(X, X) = 1.(−1) + (−1).1 = −2ϕ(1, X2) = ϕ(X2,1) = 1.(−1)2+ 1.(1)2= 2
ϕ(X2, X2) = 12.(−1)2+ (−1)2.1 = 2 ϕ(X, X2) = ϕ(X2, X) = 1.(−1)2+ (−1).12= 0
ϕR2[X]
A=
2 0 2
0−2 0
2 0 2
B0= (1 −X2, X, X2)
B0R2[X]
PB0B
P=
1 0 0
0 1 0
−1 0 1
det(P)=16= 0 P P
B0R2[X] 3 B0R2[X]
ϕ
ϕ
A0=tP AP =
1 0 −1
0 1 0
0 0 1
2 0 2
0−2 0
2 0 2
1 0 0
0 1 0
−101
=
1 0 −1
0 1 0
0 0 1
002
0−2 0
002
=
000
0−2 0
002

ϕ q
P, Q R2[X]B0
P(X) = a0(1 −X2) + a1X+a2X2
Q(X) = b0(1 −X2) + b1X+b2X2
ϕ(P, Q) = −2a1b1+ 2a2b2
q(P) = −2a2
1+ 2a2 2
Jqq
Jq={P∈R2[X]/ q(P) = 0}
P∈R2[X]P=a0(1 −X2) + a1X+a2X2
P∈Jp⇐⇒ −2a2
1+ 2a2 2= 0 ⇐⇒ a2 1=a2 2⇐⇒ a1=±a2
a1=a2P(X) = a0(1 −X2) + a1(X+X2)P∈V ect(1 −X2, X +X2)
a1=−a2P(X) = a0(1 −X2) + a1(X−X2)P∈V ect(1 −X2, X −X2)
Jq=V ect(1 −X2, X +X2)∪V ect(1 −X2, X −X2)
F={P∈R2[X]/ P (0) = 0}
FR2[X]F
(1, X, X2)R2[X]
P∈R2[X]P=a0+a1X+a2X2(a0, a1, a2)∈R3
P∈F⇐⇒ P(0) = 0
⇐⇒ a0+a1.0 + a2.0 = 0
⇐⇒ a0= 0
⇐⇒ P=a1X+a2X2∈V ect(X, X2)
F=V ect(X, X2)
FR2[X] (X, X2)R2[X]
F
F ϕ
F⊥={P∈R2[X]/∀Q∈F, ϕ(P, Q) = 0}
P∈F⊥ϕ(P, X) = ϕ(P, X2) = 0
ϕ
½P(1).(−1) + P(−1).1 = 0
P(1).(−1)2+P(−1).12= 0
½P(−1) −P(1) = 0
P(−1) + P(1) = 0
P(−1) = P(1) = 0
P−1 1
P∈R2[X]P P (X) = a(X−1)(X+ 1) = a(X2−1)
F⊥=V ect(X2−1)

E=R4B(e1, e2, e3, e4)E
a, b, c ∈Rq
q(x) = x2
1+ 2x2 2+x2 3+a x2 4+ 2x1x2+ 4b x1x3+ 4b x2x3+ 2c x2x4
(x1, x2, x3, x4)xB
q q B
ϕ q x y E ϕ q
q(x) = ϕ(x, x)ϕ(x, y) = q(x+y)−q(x)−q(y)
2
x= (x1, x2, x3, x4)y= (y1, y2, y3, y4)R4
ϕ(x, y) = x1y1+ 2x2y2+x3y3+a x4y4+x1y2+y1x2+2b x1y3+2b y1x3+2b x2y3+2b y2x3+c x2y4+c y2x4
q ϕ
A=
1 1 2b0
1 2 2b c
2b2b1 0
0c0a
q
q(x) = x2
1+ 2x2 2+x2 3+a x2 4+ 2x1x2+ 4b x1x3+ 4b x2x3+ 2c x2x4
=£x2
1+ 2x1(x2+ 2bx3)¤+ 2x2 2+x2 3+a x2 4+ 4b x2x3+ 2c x2x4
=£(x1+x2+ 2bx3)2−x2
2−4b2x2 3−4bx2x3¤+ 2x2 2+x2 3+a x2 4+ 4b x2x3+ 2c x2x4
= (x1+x2+ 2bx3)2+x2
2−4b2x2 3+x2 3+ax2 4+ 2cx2x4
= (x1+x2+ 2bx3)2+£x2
2+ 2x2(cx4)¤+x2 3(1 −4b2) + ax2 4
= (x1+x2+ 2bx3)2+£(x2+cx4)2−c2x2
4¤+ +x2 3(1 −4b2) + ax2 4
= (x1+x2+ 2bx3)2+ (x2+cx4)2+ (1 −4b2)x2
3+ (a−c2)x2 4
q a b c
a−c2>0,
1−4b2>0, rg(q) = 4 sgn(q) = (4,0)
1−4b2= 0, rg(q) = 3 sgn(q) = (3,0)
1−4b2<0, rg(q) = 4 sgn(q) = (3,1)
a−c2= 0,
1−4b2>0, rg(q) = 3 sgn(q) = (3,0)
1−4b2= 0, rg(q) = 2 sgn(q) = (2,0)
1−4b2<0, rg(q) = 3 sgn(q) = (2,1)
a−c2<0,
1−4b2>0, rg(q) = 4 sgn(q) = (3,1)
1−4b2= 0, rg(q) = 3 sgn(q) = (2,1)
1−4b2<0, rg(q) = 4 sgn(q) = (2,2)

E q
(a, b, c) = (2,0,1)
1−4b2= 1 >0a−c2= 2 −1>0
rg(q) = 1 sgn(q) = (4,0)
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
X=x1+x2+ 2b x3=x1+x2
Y=x2+cx4=x2+x4
Z=√1−4b2x3=x3
T=√a−c2x4=x4
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
x1=X−Y+T
x2=Y−T
x3=Z
x4=T
P
x1
x2
x3
x4
=P
X
Y
Z
T
P=
1−1 0 1
0 1 0 −1
0 0 1 0
0 0 0 1
(v1, v2, v3, v4)q
v1=
1
0
0
0
, v2=
−1
1
0
0
, v3=
0
0
1
0
, v4=
1
−1
0
1