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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
1/7
f LK(E) E =n f= 0
f2 =f f=I dE
f
A=
3 1 10 2 0
1 1 1
A A
A1
An
n N
n 2
A=
0 1 . . . 1
1 0
0 11 1 1 0
In
(In+A)2
M(X) 2 A M(X)
A A
A1 An n N
A=
1 3 44 7 8
6 7 7
A
E=1 E=3
A
P
P1AP = 3 0 00 1 1
0 0 1
=T
Pour toutes remarque , merci de me contecter .
[email protected]
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
2/7
f E
f f
f2 =ff
A
A
() =det(A I3) =|A I3|=
3 1 1
0 2 01 1 1
=
(2 )[(3 )(1 ) + 1] = (2 )(3 4+2 + 1) = (2 )3
() = (2 )3
= X
A
P1(X) = 2 X , P2(X) = (2 X)2 ou P3(X) = (2 X)
3
A
A 2I3=
1 1 10 0 0
1 1 1
= 0M3(R)
(A 2I3)2 =
1 1 10 0 0
1 1 1
2
=
0 0 00 0 0
0 0 0
A
P2
A(X) = (X 2)2
A
A
A
A
(A 2I3)2 = 0 =A2 4A.I3+ 4I3 (car A et I 3 commutent)
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
3/7
A(4I3 A) = 4I3
A
A1 =1
4(4I3 A) =I3
1
4A=
14
14
14
0 12
014
14
34
An n N n= 0 1 A= I3 A= A n 2
P2
P2(X) = (X 2)2
A Xn P2
R(X) =aX+ b
(a, b) R R
X
n
= (X 2)2
Q(X) +R(X)(Xn)
= 2(X 2)Q(X) + (X 2)2Q
(X) +R
(X)
Xn = (X 2)2Q(X) +aX+ bnXn1 = 2(X 2)Q(X) + (X 2)2Q
(X) +a
X
2
a= n2n1
b= 2n(1 n) R(X) =n2n1X+ 2n(1 n)
An =n2n1A+ 2n(1 n)I3
An =
3n2
n1 + 2(1 n) n2n1 n2n1
0 n2n1 + 2n(n 1) 0n2n1 n2n1 n2n1 + 2n(1 n)
n
(In+A)
2
n N, n 2
(In+A)2 =
1 1 . . . 1
1 1
1 11 1 1 1
1 1 . . . 1
1 1
1 11 1 1 1
=
n n . . . n
n n
n n
n n n n
=n
1 1 . . . 1
1 1
1 11 1 1 1
(A+In)2 =n(A+In)
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
4/7
A
In
A2 + (2 n)A+ (1 n)In = 0
M(X)
M(X) =X2 + (2 n)X+ 1 n= (X+ 1)(X (n 1))
A
A A M(X)
A
A
n N
A1
A
n= 1 A n= 1
2
A2 + (2 n)A+ (1 n)In = 0
A[ 1
n 1(A+ (2 n)In)] =In
A
A1 = 1
n 1(A+ (2 n)In)
1
An A
X
M(X)
R(X) =aX+ b
(a, b) R R
Xn =M(X)Q(X) +R(X)
Xn = (X+ 1)(X+ (1 n))Q(X) +aX+ b
1
(n 1)
X
a= 1n
((n 1)n + (1)n+1) (1)n =1 (1)n = (1)n+1
b= 1n
((n 1)n + (1)n(n 1))
R(X) = [1
n((n 1)n + (1)n+1)]X+
1
n((n 1)n + (1)n(n 1))
An =M(A)Q(A) +R(A)
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
5/7
M(A)Q(A) = 0
M(A) = 0
M(X)
A
An = [1
n((n 1)n + (1)n+1)]A+ [
1
n((n 1)n + (1)n(n 1))]In
n= 0 An n
A
() =det(A I3) =|A I3|=
1 3 4
4 7 86 7 7
=
(1 )
7 87 7
4
3 47 7
+ 6
3 47 8
=
3 2 5 3
= 1
= 3
1
3
A
()
() =3 2 5 3 = ( 3)(+ 1)2
A
E=i={XMn,1(R), AX=iX}
= 1
X1 = (x,y ,z) M3,1(R)
AX=X ()
()
1 3 44 7 8
6 7 7
xy
z
=
xy
z
x 3y+ 4z=x
4x 7y+ 8z=y6x 7y+ 7z=z
x= x
y = 2xz=x
X1 = (x, 2x, x) E=1 = vect((1, 2, 1))
= 3
E=3 = vect(X2) = vect((12
, 1, 1))
AX= 3X ()
()
1 3 44 7 86 7 7
x
y
z
= 3
x
y
z
x 3y+ 4z= 3x4x 7y+ 8z= 3y6x 7y+ 7z= 3z
x= 12
y
y = yz=y
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
6/7
= 1
2
E=1) 1
1
A
A
T
P GL3(R)
P
X= (x,y ,z) R3 =M3,1(R)
AX=X1
X (
)
()
1 3 44 7 8
6 7 7
xy
z
=
12
1
xy
z
,
X1 = 1
x 3y+ 4z= 1 x4x 7y+ 8z= 2 y6x 7y+ 7z= 1 z
12
012
P =
12
1 12
1 2 01 1 1
2
P1AP =T
p A
D
A
D Dp(R) P GLp(R) A
A= P DP1
u
u
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7/23/2019 Correction d'Examen d' Algbre 4 Session d'Automne
2014-2015
7/7
An = (P DP1)n = P DnP1 = Pdiag(n1 , . . . , np)P
1 A P p
P GLn(R)
A
exp(A) =exp(P DP1) =P exp(D)P1 =Pdiag(exp(1), . . . , e x
p(p))P1
A
u
E A u
A
u
E
E spec(A)
u
E
A
A1, tA
exp(A)
A
exp(A)
A1
tA
tr(A)
A
spec(A)
A A A A
3
C
C
C
A k N Ak = 0 Ak1 = 0
R
A Mn(R) 0 spect(A) A GLn(R)
