TSI1
no1
2
A2A∈ M2(R)
A∈ M2(R)A=a b
c d
χA(x) = x−a−c
−b x −d= (x−a)(x−d)−bc
=x2−(a+d)
| {z }
Tr(A)
x+ (ad −bc)
| {z }
det(A)
χA(x) = x2−Tr(A)x+ det(A)
∆ = Tr(A)2−4 det(A)
? A M2(C) ∆ 6=0 ∆=0
A λ0P∈GL2(C)
A=Pλ00
0λ0P−1=λ0I2
?∆6= 0 λ0∈CA=λ0I2
∆6= 0 A n = 2 A
M2(C)
A=λ0I2A
AM2(C)
Tr(A)2−4 det(A)6= 0 ∃λ0∈CA=λ0I2
? A M2(R)A
R∆>0 ∆ >0 ∆ = 0
∆=0 A λ0P∈GL2(R)
A=Pλ00
0λ0P−1=λ0I2
?∆>0λ0∈CA=λ0I2
∆>0A n = 2 A
M2(R)
A=λ0I2A
AM2(R)
Tr(A)2−4 det(A)>0∃λ0∈RA=λ0I2
2
(uk)k∈N(vk)k∈N
u0= 1
v0= 2 ∀k∈N,uk+1 = 4uk−2vk
vk+1 =uk+vk
k∈NXk=uk
vk
A=4−2
1 1
k Xk+1 =AXk
kNXk=AkX0
? k = 0 X0=I2X0=A0X0
? Xk=AkX0Xk+1 =AXk=A(AkX0)=(AAk)X0=Ak+1X0
?
∀k∈N, Xk=AkX0
χA(x) = x2−5x+ 6 Tr A2−4 det A= 1 >0
AM2(R)
Spec(A) = {2,3}E2
(A−2I)x
y=0
0⇐⇒ 2x−2y= 0
x−y= 0
⇐⇒ x=y
E3
(A−3I)x
y=0
0⇐⇒ x−2y= 0
x−2y= 0
⇐⇒ x= 2y
P−1AP =2 0
0 3 =D P =1 2
1 1
P−1=−1 2
1−1
kNAk=P DkP−1
? k = 0 P D0P−1=P I2P−1=P P −1=I2=A0
? Ak=P DkP−1
Ak+1 =AAk=P DP −1P DkP−1=P DDkP−1=P Dk+1P−1
TSI1
?
∀k∈N, Ak=P DkP−1
k∈N
Ak=P DkP−1=1 2
1 1 2k0
0 3k−1 2
1−1
∀k∈N, Ak=−2k+ 2 ×3k2k+1 −2×3k
−2k+ 3k2k+1 −3k
∀k∈N, Xk=AkX0uk
vk=−2k+ 2 ×3k2k+1 −2×3k
−2k+ 3k2k+1 −3k1
2
∀k∈N,uk= 3 ×2k−2×3k
vk= 3 ×2k−3k
ukvkk
3
(a, b, c)∈R3M(a, b, c)M3(R)
M(a, b, c) =
a+c b c
b a + 2c b
c b a +c
F F ={M(a, b, c) ; (a, b, c)∈R3}
I=M(1,0,0) J=M(0,1,0) K=M(0,0,1)
F= Vect(I, J, K)I, J K
aI +bJ +cK = 0 =⇒a=b=c= 0 (I, J, K)
(I, J, K)F
M(a, b, c)∈F
M(a, b, c)
J K
3
J=
010
101
010
χJ(x) =
x−1 0
−1x−1
0−1x
=x
1−1 0
0x−1
−1−1x
(C1←C1−C3)
=x
1−1 0
0x−1
0−2x
(L3←L3−L1)
=x(x2−2) = x(x−√2)(x+√2)
Spec(J) = {0,√2,−√2)}
E0
J
x
y
z
=
0
0
0
⇐⇒ y= 0
x+z= 0
⇐⇒ y= 0
z=−x
E0= Vect
1
0
−1
E√2
(J−√2I)
x
y
z
=
0
0
0
⇐⇒
−√2x+y= 0
x−√2y+z= 0
y−√2z= 0
⇐⇒ (y=√2x
z=1
√2y=x
E√2= Vect
1
√2
1
E−√2
(J+√2I)
x
y
z
=
0
0
0
⇐⇒
√2x+y= 0
x+√2y+z= 0
y+√2z= 0
⇐⇒ (y=−√2x
z=−1
√2y=x
E−√2= Vect
1
−√2
1
TSI1
K=
101
020
101
χK(x) =
x−1 0 −1
0x−2 0
−1 0 x−1
= (x−2)
1 0 −1
1x−2 0
1 0 x−1
(C1←C1+C2+C3)
= (x−2)
1 0 −1
0x−2 1
0 0 x
(L2←L2−L1) (L3←L3−L1)
=x(x−2)2
Spec(J) = {0,2}
E0
0
K
x
y
z
=
0
0
0
⇐⇒ x+z= 0
2y= 0
⇐⇒ y= 0
z=−x
E0
0= Vect
1
0
−1
=E0
E0
2
(K−2I)
x
y
z
=
0
0
0
⇐⇒ −x+z= 0
x−z= 0
⇐⇒ x=z
E0
2= Vect
1
0
1
,
0
1
0
J K
1
√2
1
=
1
0
1
+√2
0
1
0
1
−√2
1
=
1
0
1
−√2
0
1
0
Vect
1
√2
1
,
1
−√2
1
⊂Vect
1
0
1
,
0
1
0
Vect
1
0
1
,
0
1
0
= Vect
1
√2
1
,
1
−√2
1
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