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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : RangApplication linéaire/Exercices/Rang », n'a pu être restituée correctement ci-dessus.
Soient
A
∈
M
3
,
2
(
R
)
{\displaystyle A\in {\mathcal {M}}_{3,2}(\mathbb {R} )}
B
∈
M
2
,
3
(
R
)
{\displaystyle B\in {\mathcal {M}}_{2,3}(\mathbb {R} )}
A
B
=
(
0
−
1
−
1
−
1
0
−
1
1
1
2
)
{\displaystyle AB=\left({\begin{array}{ccc}0&-1&-1\\-1&0&-1\\1&1&2\\\end{array}}\right)}
B
A
=
I
2
{\displaystyle BA=I_{2}}
Soient
u
∈
L
(
E
,
F
)
{\displaystyle u\in \operatorname {L} (E,F)}
w
∈
L
(
E
,
G
)
{\displaystyle w\in \operatorname {L} (E,G)}
ker
u
⊂
ker
w
{\displaystyle \ker u\subset \ker w}
v
∈
L
(
F
,
G
)
{\displaystyle v\in \operatorname {L} (F,G)}
w
=
v
∘
u
{\displaystyle w=v\circ u}
Soient
ψ
,
φ
1
,
…
,
φ
n
{\displaystyle \psi ,\varphi _{1},\dots ,\varphi _{n}}
E
{\displaystyle E}
ker
ψ
{\displaystyle \ker \psi }
ker
φ
i
{\displaystyle \ker \varphi _{i}}
ψ
{\displaystyle \psi }
φ
i
{\displaystyle \varphi _{i}}
Solution
Soit
E
0
{\displaystyle E_{0}}
ker
u
{\displaystyle \ker u}
E
{\displaystyle E}
u
′
{\displaystyle u'}
E
0
→
im
u
,
x
↦
u
(
x
)
{\displaystyle E_{0}\to \operatorname {im} u,\;x\mapsto u(x)}
F
0
{\displaystyle F_{0}}
im
u
{\displaystyle \operatorname {im} u}
F
{\displaystyle F}
v
∈
L
(
F
,
G
)
{\displaystyle v\in \operatorname {L} (F,G)}
im
u
{\displaystyle \operatorname {im} u}
F
0
{\displaystyle F_{0}}
v
{\displaystyle v}
F
0
{\displaystyle F_{0}}
y
∈
im
u
{\displaystyle y\in \operatorname {im} u}
v
(
y
)
=
w
(
u
′
−
1
(
y
)
)
{\displaystyle v(y)=w\left({u'}^{-1}(y)\right)}
x
∈
E
{\displaystyle x\in E}
x
=
x
0
+
z
{\displaystyle x=x_{0}+z}
x
0
∈
E
0
{\displaystyle x_{0}\in E_{0}}
z
∈
ker
u
⊂
ker
w
{\displaystyle z\in \ker u\subset \ker w}
y
=
u
(
x
0
)
{\displaystyle y=u\left(x_{0}\right)}
w
(
x
)
=
w
(
x
0
)
=
v
(
y
)
=
v
(
u
(
x
)
)
{\displaystyle w\left(x\right)=w\left(x_{0}\right)=v\left(y\right)=v\left(u\left(x\right)\right)}
D'après la question précédente appliquée à
G
=
K
{\displaystyle G=K}
w
=
ψ
{\displaystyle w=\psi }
F
=
K
n
{\displaystyle F=K^{n}}
u
:
E
→
F
,
x
↦
(
φ
1
(
x
)
,
…
,
φ
n
(
x
)
)
{\displaystyle u:E\to F,\;x\mapsto \left(\varphi _{1}(x),\dots ,\varphi _{n}(x)\right)}
v
{\displaystyle v}
K
n
{\displaystyle K^{n}}
v
(
k
1
,
…
,
k
n
)
=
∑
λ
i
k
i
{\displaystyle v\left(k_{1},\dots ,k_{n}\right)=\sum \lambda _{i}k_{i}}
∀
x
∈
E
ψ
(
x
)
=
v
(
φ
1
(
x
)
,
…
,
φ
n
(
x
)
)
=
∑
λ
i
φ
i
(
x
)
{\displaystyle \forall x\in E\quad \psi \left(x\right)=v\left(\varphi _{1}(x),\dots ,\varphi _{n}(x)\right)=\sum \lambda _{i}\varphi _{i}\left(x\right)}
ψ
=
∑
λ
i
φ
i
{\displaystyle \psi =\sum \lambda _{i}\varphi _{i}}
Soit
f
{\displaystyle f}
R
3
{\displaystyle \mathbb {R} ^{3}}
f
∘
f
=
0
{\displaystyle f\circ f=0}
Montrer que
rang
(
f
)
≤
1
{\displaystyle \operatorname {rang} (f)\leq 1}
En déduire qu'il existe un vecteur
v
∈
R
3
{\displaystyle v\in \mathbb {R} ^{3}}
g
{\displaystyle g}
∀
u
∈
R
3
f
(
u
)
=
g
(
u
)
v
{\displaystyle \forall u\in \mathbb {R} ^{3}\quad f(u)=g(u)\,v}
