corrigé - IMJ-PRG

A1
x1+√3x2= 0 √3e1−e2
√3x1−x2= 0 e1+√3e2
A1
A1X2−5X+ 6
λ A1A1−λId
A1A1−λId
A1X=λX
P=1
2√3 1
−1√3D=1 0
0 5 
tP.P =I2det(P) = 1 P
tP.A1.P =1 0
0 5 =D .
P D
X P −1X(y1, y2)
tX.A1.X =tX.tP−1.D.P −1.X =y2
1+ 5y2 2
xR2QA1(x)P−1X
X P A1S+
2(R)
P
A P
D
P
(O, (√3e1−e2)/2,(e1+√3e2)/2) ΣA1
y2
1+ 5y2 2= 1
e=q1−1
5
1=2
√5.

x2
1/a2+x2 2/b2= 1 0 < b < a c =√a2−b2
c e =c/a e
a= 1 b= 1/√5
xR2
QA2(x) = 2x2
1+ 4√2x1x2+ 4x2 2= 2(x1+√2x2)2≥0
QA2√2e1−e2
A2S+
2(R)
ΣA2
2(x1+√2x2)2= 1
ΣA2√2x1+ 2x2= 1
√2x1+ 2x2=−1
A
PtP AP (λ1, . . . , λn)
A X Rn(y1, . . . , yn)P−1X
QA(x) =
n
X
i=1
λiy2
i.
P x Rn(y1, . . . , yn)x
yii n
A QA
A S+
n(R) (ii)⇒(i)
ΣAQA
λ A x
λ QA(x) = λ||x||2=λ x/√λΣAΣA
A
ΣAλ x A
x/√λ+Ker(A) ΣAΣAA
A µ y
A t xt
cosh(t)x
√λ+ sinh(t)y
√−µ
ΣA||xt|| q1
λ−1 µett+∞ΣA
AΣA
A(iii)⇒(ii)
A S+
n(R)QARnR
a b x 0< a ≤QA(x)≤b QA
∀x∈Rn\ {0}a≤QAx
||x||=1 ||x||2QA(x)≤b
∀x∈Rna||x||2≤QA(x)≤b||x||2.

ΣAO1/√a
QAR
ΣAxRnQA(x)x/pQA(x) ΣA
(i)⇒(iii)
(i) (ii) (iii) ΣA
A
ΠRnQAΠ
A S+
n(R)QARn\ {0}Π\ {0}
QAΠ
Π
ΣA∩Π = {x∈Π|QA|Π(x) = 1}.
A S+
n(R) Π RnQAΠ
Π ΣA∩Π
Π ΣA∩ΠQA
QA
Rn\ {0}QAA S+
n(R)
AΣAΣA
O1/√λ1O
1/√λ1
λ µ µ ∆⊥
∆µ
r∆xΣAx=u+v u ∆v∆⊥
r(x) = u+r(v)r(v) ∆⊥v
QA(r(x)) = λ||u||2+µ||r(v)||2=λ||u||2+µ||v||2=QA(x) = 1
ΣA∆
Π ∆ ΣAΓ ΣA
Γ ∆ Γ ∆
Π ΣAOΓ Γ O
Γd O
Γ ∆ O d
(f1, f2, f3)A(λ1, λ2, λ3)x=
x1f1+x2f2+x3f3ΣAO d
QA(x) = 1 ||x||2=d2QA(x)−λ(x2
1+x2 2+x2 3) = 1 −λd2
λ A (µ−λ)x2
3= 1−λd2(µ−λ)x2 1= 1−λd2
λ λ1λ3λd2= 1 λ−µ
ΣAO d
∆ Π1Π2

Π ∆ Π1Π2Π
Γ Γ ΣA∩Π Γ
∆ ΣAxR3
u+v u ∆v∆xΣAΠv QA(u+v) = 1
v= 0 λ||u||2= 1 λ
ΣA
ΣAΠ
QAΠQA
Π Π ∩Π0
ΣAΠ∩Π0ΣA∩Π0
QAΠ0AΠ0A
AΠ0
λ2λ1λ2>0
λ3−λ2λ3−λ1
√λ3−λ2x3−√λ3−λ1x1= 0
R3
(x1, x2, x3)∈ΣA∩Π⇔QA(x) = 1 pλ3−λ2x3−pλ3−λ1x1= 0
⇔λ2||x||2+ (pλ3−λ2x3−pλ3−λ1x1)(pλ3−λ2x3−pλ3+λ1x1) = 1
pλ3−λ2x3−pλ3−λ1x1= 0
⇔λ2||x||2= 1 pλ3−λ2x3−pλ3−λ1x1= 0
λ2ΣA∩ΠO1/√λ2Π
Π0√λ3−λ2x3+√λ3−λ1x1= 1 ΣA∩Π0
O1/√λ2Π0√λ3−λ1Π Π0
Π Π0∆
f2(f1, f3)
D D0∆ Π Π0Π Π0
∆f1f3Π Π0f1f3D
D0
u D0D0∆u(f1, f3)
u=αf1+βf3α β D0f1
f3u(f1, f3)α2+β2= 1
(Π,Π0) (f1, f2)
Π0Π Π0Π

f1f2−Id
f3(f1, f2, f3)f3
u αf1−βf3(αf1−βf3, f2) Π
s t
QA(s(αf1±βf3) + tf2) = α2u11 +β2u33 ±2αβu13s2+u22t2+ 2(αu12 ±βu23)st
Π∩ΣAΠ0∩ΣA(αf1−βf3, f2) (αf1+βf3, f2)
α2u11 +β2u33 −2αβu13s2+u22t2+ 2(αu12 −βu23)st = 1
α2u11 +β2u33 + 2αβu13s2+u22t2+ 2(αu12 +βu23)st = 1 .
s2+t2
α2u11 +β2u33 −2αβu13=u22 =α2u11 +β2u33 + 2αβu13
2(αu12 −βu23) = 0 = 2(αu12 +βu23).
α β u12 =u23 = 0
u13 = 0 α2u11 +β2u33 =u22
(f1, f2, f3)QAu11 u22 u33
A u22 α2β2u11 u33 f2
u22 f1f3
Π Π0ΣA
ΣAΠ
ΣA
xR3
QA3(x) = 4x2
1+ 5x2 2+ 4x2 3−2x1x2−2x2x3= 3(x2 1+x2 2+x2 3)+(x1−x2)2+ (x2−x3)2≥3||x||2.
A3ΣA3
xR3
QA3(x) = 4(x2
1+x2 2+x2 3) + x2 2−2x1x2−2x2x3= 4(x2 1+x2 2+x2 3) + x2(x2−2x1−2x3)
ΣA3x2= 0 x2−2x1−2x3= 0 O
1/2
ΣA3
A3ΣA3
(f1, f2, f3)R3(f1, f2) ΣA3
(s, t)
QA3(sf1+tf2) = QA3(f1)s2+ 2ΦA3(f1, f2)st +QA3(f2)t2