Espaces vectoriels - Institut de Mathématiques de Toulouse
1
◦
∼
R
R
R{0}[−1,1].
R2
(x
y)∈R2, x +y= 0,(x
y)∈R2, x +y= 1,
(x
y)∈R2,3x+y=x−y= 0,(x
y)∈R2, x >0,
(x
y)∈R2,|x|=|y|.
R3
nx
y
z∈R3, x2+y2+z2= 1o,nx
y
z∈R3, x +y+z>0o,
nx
y
z∈R3,3x+y+z=x−y+ 4z= 0o,nx
y
z∈R3, x −y+z= 0o.
F(R,R)R R
2π
2
R[X]
{P∈R[X],deg(P) = d},{P∈R[X],deg(P)6d},
{P∈R[X], P (0) = 0},{P∈R[X], P (1) = 1},
d
Vect
ERx1, . . . , xnn>1EVect(x1, . . . , xn)
x1, . . . , xn
2
3×3
63
(( 1
2),(1
3),(1
4)) R2
(v1, . . . , vr)R3R3
vi
v1=2
1
1v2=1
3
1v3=−2
1
3
v1=1
0
3v2=0
1
2v3=2
−3
0
v1=2
1
0v2=0
0
1v3=−4
−2
0
v1=1
−2
3v2=0
3
−4v3=7
3
0v4=−6
0
−1
R3
v1=
−1
−2
−5
v2=
−2
−1
−4
.
(v1, v2)R3
v1=
1
1
−4
v2=
1
2
−3
.
v1=
1
−1
4
v2=
1
2
3
.
{e1, e2, e3}R3
v1=e1+e2−e3, v2=−e1+ 3e2−e3, v3= 4e2−2e3
R4
v1=
1
2
−1
0
, v2=
2
1
−1
0
, v3=
−1
1
1
0
,
(v1, v2, v3)R4
(v1, . . . , vr)
Vect(v1, . . . , vr)R4
R4
v1=1
1
1
1v2=0
1
2
−1v3=1
0
−2
3v4=2
1
0
−1v5=4
3
2
1
v1=1
2
3
4v2=0
1
2
−1v3=3
4
5
16
v1=1
2
3
4v2=0
1
2
−1v3=2
1
0
11 v4=3
4
5
14
R4
u1=
1
0
2
0
, u2=
2
0
−3
0
, v1=
0
2
0
−2
, v2=
0
3
0
5
U= Vect(u1, u2)V= Vect(v1, v2)R4=
U⊕V
R2B= (e1, e2)
e0
1=e1+ 2e2e0
2=e1+e2v=e1+ 2e2w=−e1+ 3e2
e00
1=e1−e2e00
2= 2e1−e2
B0= (e0
1, e0
2)B00 = (e00
1, e00
2)R2
v w
R3C= (f1, f2, f3)
v=f1+ 2f2+ 3f3w=−f1+ 3f2+f3
C0= (f1+ 2f2+ 3f3, f1−f2−f3, f2+f3)C00 = (−f1+f2,−f1+ 3f3,4f1−3f2−2f3)
R3
v w
u=
−1
1
1
, v =
1
−1
1
, w =
1
1
−1
,
R3
(u, v, w)R3
2
−3
−1
ERd>1Rd
d+ 1
d+ 1
ERd>1Rd
E
E
E
E
E3
3
3
P D R3D6⊂ P
R3
R3
R
−1
12
4
,8x−y+ 5z= 0,12x+y= 0
4x+z= 0 ,R
1
3
−1
⊕R
−1
2
2
D P
D P
D P
3x−2y+z= 0 2x−y+ 3z= 0
−5x+ 3y+z= 0
R
1
2
−1
,R
2
1
−3
⊕R
1
5
−2
R3
R3{e1, e2, e3}
P4x+ 5y+ 2z D 1
1
1
P D
R3=D⊕P
P D
P=R
1
2
−1
⊕R
−1
−3
2
D=(x−y−z= 0
x+ 3y+ 2z= 0 .
D⊂P
P−3x−4y+ 2z= 0 D
1
2
−1P D
R3=D⊕P
P D
P=R
−1
2
−1
⊕R
1
−3
−1
D=(x−y= 0
7x−z= 0 .
D⊂P
P−2x−3y+z= 0 D
1
2
2P D
R3=D⊕P
P D
P=R
1
2
1
⊕R
1
3
1
D=(x+y+z= 0
x−y+z= 0 .
D⊂P
R3{e1, e2, e3}
P2x−3y+ 5z= 0 D
(1,3,2)
P D
R3=D⊕P
P D
P=R
1
−2
1
⊕R
1
3
−2
D=(y= 0
x+ 5z= 0 .
D⊂P
R3{e1, e2, e3}
P−x+ 3y+ 4z= 0 D
−1
1
−2P D
R3=D⊕P
P D
P=R
1
2
1
⊕R
1
3
1
D=(2x+y= 0
y+ 2z= 0 .
D⊂P
5
10
−15 ∈R1
2
−3 −4
12
−16 ∈R−1
3
4 −1
12
0∈R1
3
3+R2
1
5
1
1
3∈R1
1
1+R1
1
2 1
2
3∈Vect n1
0
0,0
1
0o 5
5
5∈ P1∩ P2,
P1=R1
1
0+R0
0
1P2=R0
1
1+R1
0
0
R1
−2
1=R−3
6
−3,R1
−1
2=R2
−2
1,R1
−2
3=R2
−4
6+R−3
6
−9,
R1
2
−3+R−1
−3
4=R0
−1
1+R1
1
−2,R0
1
−1+R0
2
1=R0
−2
2+R1
1
1.
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