Exercices avec corrigés pour le soutien - Espaces

E=K3F={(x, y, z)∈K3|x+y+ 2z= 0}G={(x, y, z)∈K3|y−z= 0}
F G F +G F ∩G
~u = (1,1,1) ~v = (1,2,3) ~w = (a, a2, a3)R3
a(~u, ~v, ~w)
R4a= (3,2,1,4) b= (1,1,1,3) c= (4,2,0,2) d= (−1,0,1,2) e= (0,3,2,1)
U= (a, b, c, d, e)
(a, b, c)
(a, c, d)
U
UR4
F= (a, b, c)G= (d, e)F G
UF+G F +G
F∩G F ∩G
B= (−1,3 + X, 5+4X2,2X2, X +X2+X3)R3[X]
fR4R3f(x, y, z, t)=(x+y, z −t, x +t)
f(f)f(f)
fR4R3
f∈ L(K3)f(1,0,0) = (0,1,2), f(0,1,0) = (−1,1,1) f(0,0,1) = (2,0,0) f
K3
F a = (−2,0,1) b= (−1,1,0)
G c = (1,0,0)
d= (0,1,1)
F+G a, b, c d
(a, b, c)F+G
(F∩G) = 1
(x, y, z)R3F∩G(x+y+ 2z+ 0
y−z= 0 ⇐⇒ (y=z x=−3z
F∩G= ((−3,1,1))
~u, ~v ~w p, q, r
p~u+q~v+r ~w =~ 0⇐⇒ 
p+q+ra = 0
p+ 2q+ra2= 0
p+ 3q+ra3= 0
⇐⇒ 
p+q+ra = 0
q+r(a2−a)=0
2q+r(a3−a)=0
⇐⇒
p+q+ra = 0
q+ra(a−1) = 0
ra(a2−1) −2ra(a−1) = 0
q⇐⇒ 
p+q+ra = 0
q+ra(a−1) = 0
ra(a−1)2= 0
a= 0 a= 1

−2a+ 2b+c= 0
d=a+c
(a, c)b d (a, b, c, d) (a, c)
(a, c, e)U
UR4UR4
F(a, b)F= (a, b)
(d, e)G G
UR4
U(F+G) = 3 F+G= (a, b, e)
(F∩G)=1
F∩G xa +yb =zd +te
x, y, z, t F ∩G a c d e
3x+y=−z
2x+y= 3t
x+y=z+ 2t
4x+ 3y= 2z+t
x=−1, y = 2 z= 1 t= 0
xa +yb = (−1,0,1,2) F∩G
R3[X]B
(−1,3 + X, 2X2, X +X2+X3)
R3[X]
BR3[X]
f f((x, y, z, t)+λ(x0, y0, z0, t0)) = f(x+λx0, y +λy0, z +λz0, t +λt0)=(x+λx0+y+λy0, z +λz0−(t+
λt0), x+λx0+t+λt0) = ((x+y)+λ(x0+y0),(z−t)+λ(z0−t0), x+t+λ(x0+t0)) = (x+y, z−t, x+t)+λ(x0+y0, z0−t0, x0+t0) =
f(x, y, z) + λf (x0, y0, z0)
(R4) = ( (f)) + ( (f)) f
( (f)) = 0 ( (f)) = 4 (f)R3
(f) (x, y, z, t)∈(f)⇐⇒ 
x+y= 0
z−t= 0
x+t= 0
⇐⇒ 
x=−t
y=t
z=−t
(f) = ((−1,1,1,1))
( (f)) = 3 (f) = R3f
f A =
1 1 0 0
001−1
1 0 0 1 
f
(0,1,2),(−1,1,1),(2,0,0) f
f(x, y, z)=(−y+ 2z, x +y, 2x)f(x, y, z)
(f)
−y+ 2z= 0
x+y= 0
2x= 0
x=y=z= 0 f
R3
(a, b, c)∈R3f
−y+ 2z=a
x+y=b
2x=c
⇐⇒ 
x=a
2
y=b−c
2
z=a
2+b 2−c 4
f−1(a, b, c)=(a
2, b −c 2,a 2+b 2−c 4)