Diagonalisation (quelques exemples)
Michel Rigo
http://www.discmath.ulg.ac.be/
A=
1 −1
−12 32
.
On va regarder
1 0
, . . . ,An 1
0
, . . . 0
1
, . . . ,An 0
1
, . . . −2
−1
, . . . ,An −2
−1
, . . .
A0 1 0
A1 1 0
A2 1 0
A3 1 0
A4 1 0
1
0 , . . . ,A4 1 0
A0 0 1
A1 0 1
A2 0 1
A3 0 1
0
1 , . . . ,A3 0 1
A0 −2
−1
A1 −2
−1
A2 −2
−1
A3 −2
−1
A4 −2
−1
det(A−λI) = 1−λ −1
−12 32 −λ =λ2−5
2λ+1 = (λ−2) λ− 1 2 . Deux valeurs propres simples : 12 et 2
Espace propre associ´e `a 1/2
A−1
2I x1 x2
= 0⇔ 1
2x1−x2 = 0
−12x1+x2 = 0 x1 = 2x2
E1/2=i 2
1
h
(A−2I) x1
x2
= 0⇔
−x1−x2 = 0
−12x1− 12x2 = 0 x1=−x2
E2 =i −1
1
h
R2 =E1/2⊕E2 =i 2
1
h⊕i −1
1
h
On dispose d’une base de R2 form´ee de vecteurs propres de A.
Ainsi,
1 0
= 1 3
2 1
+(−1 3)
−1 1
2
E1/2
1 0
= 1 3
2 1
+(−1 3)
−1 1
1 0 = 1
3 2
1 +(−1 3) −1
1 A
1 0
= 1 3A
2 1
|{z}
∈E1/2
+(−1 3)A
−1 1
| {z }
∈E2
A
1
0
= 1 3
1 2
2
1
+(−1 3) 2
−1 1
1 0 = 1
3 2
1 +(−1 3) −1
1 A
1 0
= 1 3A
2 1
|{z}
∈E1/2
+(−1 3)A
−1 1
| {z }
∈E2
A 1
0
= 1 3
1 2
2 1
+(−1 3) 2
−1 1
E2
E1/2
1 0
= 1 3
2 1
+(−1 3)
−1 1
E2
E1/2
A 1
0
= 1 3
1 2
2 1
+(−1 3) 2
−1 1
8 2
1 + −1 1
8 2
1 + −1 1
A(8 2
1 + −1
1 ) = 8A 2
1 +A −1 1
A2(8 2
1 + −1
1 ) = 8A2 2
1 +A2 −1 1
A3(8 2
1 + −1
1 ) = 8A3 2
1 +A3 −1 1
A4(8 2
1 + −1
1 ) = 8A4 2
1 +A4 −1 1
propres non r´eels,
R= 9 10
cosπ4 −sinπ4 sinπ4 cosπ4
R= 9 10
cosπ4 −sinπ4 sinπ4 cosπ4
det(R−λI) =λ2−9√ 2 10 λ+ 81
100 =
λ−9√ 2 20 (1+i)
λ−9√
2 20 (1−i)
Deux valeurs propres simples (conjugu´ees car χA∈R[λ]) :
9√ 2
20 (1 +i) = 109eiπ/4 et 920√2(1−i) = 109 e−iπ/4 E9√
2
20 (1+i) =i i
1
h, E9√ 2
20 (1−i)=i −i
1
h
S =
i −i 1 1
et S−1RS =
9√ 2
20 (1 +i) 0 0 920√2(1−i)
!
C2=i i
1 h⊕i −i 1 h
∀a,b ∈C: a
b
= b−ai 2
i 1
+b+ai 2
−i 1
et R
a b
= b−ai 2
9 10eiπ/4
i 1
+b+ai 2
9 10e−iπ/4
−i 1
Rn a
b
= b−ai 2
9 10eiπ/4
n i 1
+b+ai 2
9 10e−iπ/4
n
−i 1
∈ Rn
a b
= xn,1
xn,2
= b−ai 2
9 10eiπ/4
n i 1
+b+ai 2
9 10e−iπ/4
n
−i 1
.
xn,1 =
b−ai 2
9 10eiπ/4
n
−b+ai 2
9 10e−iπ/4
n i
puisque z−z = 2iIm(z), xn,1=
9 10
n
(acos(nπ/4)−bsin(nπ/4))
xn,2 = b−ai 2
9 10eiπ/4
n
+b+ai 2
9 10e−iπ/4
n
puisque z+z = 2Re(z), xn,2=
9 10
n
(asin(nπ/4) +bcos(nπ/4))
∈
R a
b
=aR 1
0
+bR 0
1
Un dernier exemple dans ,
S =
1 1 0
1 1 −1
1 2 1
A=S
1.1 0 0
0 0.9 0 0 0 0.7
S−1 = 1 10
15 −2 −2 8 5 −2
4 0 7
7/2 1/2 9
=
1 1 1
+ 5 2
1 1 2
+ 3
0
−1 1
A (7/2,1/2,9)e =E1.1⊕E0.9⊕E0.7
A (7/2,1/2,9)e =E1.1⊕E0.9⊕E0.7
A (7/2,1/2,9)e =E1.1⊕E0.9⊕E0.7
A (7/2,1/2,9)e =E1.1⊕E0.9⊕E0.7
A (7/2,1/2,9)e =E1.1⊕E0.9⊕E0.7
A = (S∆S ) =S∆ S avec ∆ = diag(1.1,0.9,0.7) donc,
An =
1 1 0 1 1 −1 1 2 1
1.1n 0 0 0 0.9n 0 0 0 0.7n
3 −1 −1
−2 1 1 1 −1 0
31.1n−20.9n 0.9n−1.1n 0.9n−1.1n
−0.7n+ 31.1n−20.9n 0.7n+0.9n−1.1n 0.9n−1.1n 0.7n+ 31.1n−40.9n −0.7n−1.1n+ 20.9n −1.1n+ 20.9n
!
a b c d e f g h i
| {z }
S
λn1 0 0 0 λn2 0 0 0 λn3
| {z }
∆n
=
aλn1 bλn2 cλn3 dλn1 eλn2 f λn3 gλn1 hλn2 iλn3
S∆nS−1 =
aλn1 bλn2 cλn3 dλn1 eλn2 f λn3 gλn1 hλn2 iλn3
⋆ ⋆ ⋆
⋆ ⋆ ⋆
⋆ ⋆ ⋆
D’une mani`ere g´en´erale, si Aest une matrice diagonalisable r×r, A=S∆S−1 o`u ∆ = diag(λ1· · ·λr), alors
[An]i,j = Xr p=1
cp(i,j)λnp
