Algèbre linéaire pour l`économie Feuille de TD 6 : Matrices d

fR3
M=1
3
1 2 −2
212
−2 2 1
.
v1= (1,−1,1), v2= (1,1,0), v3= (−1,1,2)
B= (v1, v2, v3)R3
fB
f
gR3R2(e1, e2, e3)
(f1, f2)
A=2−1 1
3 2 −3.
R3e0
1=e2+e3, e0 2=e1+e3, e0 3=e1+e2B g
R2f0
1=1
2(f1+f2), f0
2=1
2(f1−f2)C g
(e0
1, e0 2, e0 3) (f0 1, f0 2)
(e1, e2, e3, e4)R4fR4R3
f(x, y, z, t)=(X, Y, Z)
X=x−y+z+t, Y =x+ 2z−t, Z =x+y+ 3z−3t.
f
(f(e1), f(e2)) imfBR3
ker f2 (u1, u2) ker f
(e1, e2, u1, u2)R4f(e1, e2, u1, u2)
B
(e1, e2, e3)R3ε1= (1,1,1) ε2= (1,2,1)
ε1, ε2e1R3
fR3R3
f(ε1) = ε2, f(ε2) = ε2et f(e1) = e1+e2.
fR3
f(ε1, ε2, e1)
f:R3→R3
f(x, y, z)=(x+y, 2x−y+z, x +z).
fR3

v1= (2,1,−1) v2= (2,−1,2) v3= (3,0,1) R3
f(v1, v2, v3)
Rn(e1, . . . , en)v
v=e1+. . . +enf:Rn→Rni= 1, . . . , n f(ei) = ei+v
ker f v 0
f
fRn
f:R3→R3g=f−Id g◦g6= 0
g◦g◦g= 0
vR3g◦g(v)6= 0 v g(v)g◦g(v)
R3
g f {v, g(v), g ◦g(v)}
fR3
A=
2 1 0
0 1 1
−1−1 0
.
v= (1,0,0) f v
f(v, g(v), g ◦g(v))
(e1, e2, e3)R3f:R3→R3
f(x, y, z)=(−x+y+z, −2x+y+z, −6x+ 2y+ 4z).
fR3
v1= (1,0,2) v2= (1,1,2) v3= (2,1,5) R3(v1, v2, v3)
R3f
ER3 (e1, e2, e3)E f :E→E
(e1, e2, e3)
1 1 −1
4 1 −2
6 3 −4
.
ker f
v1=e1+ 2e2+ 3e3v2=e2+e3v3=e1+ 2e3(v1, v2, v3)E
f
f◦f=−f
B= (e1, e2, e3)R3f:R3→R3
f(e1) = −e1+e2+e3, f(e2) = −2e1+ 2e3etf(e3) = −4e1+e2+ 4e3.
fB
imfimf
t f −tId

v1ker(f−3Id) v2ker f v1v2
v3R3(v1, v2, v3)R3
f(v1, v2, v3)
E3B= (e1, e2, e3)E f
E E B
103
021
2−1 2
.
f(e3, e2, e1)
f(e1+e3, e3, e2−e3)
R3(e1, e2, e3)
u
u(e1) = 2e1−3e2+e3, u(e2) = e1−e2et u(e3) = e2−e3.
M u M2M3= 0
ker uker u2imuimu2
e0
1=u2(e1), e0 2=−u(e3), e0 3=e3(e0 1, e0 2, e0 3)R3
N u
S(e1, e2, e3) (e0
1, e0 2, e0 3)S2S−1
S−1MS =N
φ:R≤4[X]→R≤4[X]
P7→ P0
A φ (1, X, X2
2! ,X3 3! ,X4 4! )
A5
ϕR≤3[X]
ϕ(P)(X) = P(X+ 1).
A ϕ (1, X, X2, X3)
A
ϕR≤3[X]R[X]
ϕ(P)(X) = P(X)−XP 0(X).
ϕR≤3[X]
ϕ
M ϕ (1, X, X2, X3)
M
uR≤4[X]
u(P) = X2P00 (X)−3XP 0(X).

uR≤4[X]R≤4[X]u
R≤4[X]
ker uImu
u(P)=1 R≤4[X]u(P) = X3
P1P2P1−P2
u(P) = X3
A=0 1
−1 0 
ψ:M2(R)→M2(R)
M7→ AM −MA
ψ ψ  a b
c d  ker ψimψ
K ψ M2(R)
M2(R)ψ
L=
0 0 0 0
0 0 0 0
000−2
0 0 2 0
.