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Espaces Vectoriels
Représentation matricielle des applications linéaires - Exercice 1
10 min
15
Question 1
Soit
f
f
f
l’application linéaire définie par
f
:
{
R
3
⟶
R
2
(
x
,
y
,
z
)
⟼
(
x
+
y
,
x
−
z
)
f:\left\{ \begin{array}{ccc}{\mathbb{R}}^3 & \longrightarrow & {\mathbb{R}}^2 \\ \left(x,y,z\right) & \longmapsto & \left(x+y,x-z\right) \end{array}\right.
f
:
{
R
3
(
x
,
y
,
z
)
⟶
⟼
R
2
(
x
+
y
,
x
−
z
)
Écrire la matrice
A
A
A
de
f
f
f
relativement à la base canonique de
R
3
{\mathbb{R}}^3
R
3
.
Correction
Soient
E
E
E
et
F
F
F
deux
K
\mathbb{K}
K
-espaces vectoriels de dimension respective
n
n
n
et
p
p
p
. On note
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
B=\left(e_1,e_2,\cdots ,e_n\right)
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
une base de
E
E
E
et
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
B'=\left(e'_1,e'_2,\cdots ,e'_p\right)
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
une base de
F
F
F
.
On appelle matrice de
f
f
f
dans les bases
B
B
B
et
B
′
B'
B
′
et on note
M
(
f
,
B
,
B
′
)
M\left(f,B,B'\right)
M
(
f
,
B
,
B
′
)
le tableau dont les colonnes sont les composantes des vecteurs
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
\left(f\left(e_1\right),f\left(e_2\right),\cdots ,f\left(e_n\right)\right)
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
dans la base
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
\left(e'_1,e'_2,\cdots ,e'_p\right)
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
.
Autrement dit :
{
f
(
e
1
)
=
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
f
(
e
2
)
=
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
⋮
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
(
a
11
⋯
a
1
n
a
21
⋯
a
2
n
⋮
⋯
⋮
a
p
1
⋯
a
p
n
)
\left\{ \begin{array}{c} \begin{array}{ccc}f\left(e_1\right) & = & a_{11}{e'}_1+a_{21}{e'}_2+\cdots +a_{p1}{e'}_p \\ f\left(e_2\right) & = & a_{12}{e'}_1+a_{22}{e'}_2+\cdots +a_{p2}{e'}_p \\ & \vdots & \end{array} \\ \begin{array}{ccc}f\left(e_n\right) & = & a_{1n}{e'}_1+a_{2n}{e'}_2+\cdots +a_{pn}{e'}_p \end{array} \end{array}\right. \Longleftrightarrow M\left(f,B,B'\right)=\left( \begin{array}{ccc}a_{11} &\cdots & a_{1n} \\a_{21} &\cdots & a_{2n} \\ \vdots & \cdots & \vdots \\ a_{p1} & \cdots & a_{pn} \end{array}\right)
⎩
⎨
⎧
f
(
e
1
)
f
(
e
2
)
=
=
⋮
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
⎝
⎛
a
11
a
21
⋮
a
p
1
⋯
⋯
⋯
⋯
a
1
n
a
2
n
⋮
a
p
n
⎠
⎞
.
Soient
E
=
R
3
E={\mathbb{R}}^3
E
=
R
3
et
F
=
R
2
F={\mathbb{R}}^2
F
=
R
2
On note
B
=
(
e
1
,
e
2
,
e
3
)
B=\left(e_1,e_2,e_3\right)
B
=
(
e
1
,
e
2
,
e
3
)
une base de
R
3
{\mathbb{R}}^3
R
3
avec
e
1
=
(
1
,
0
,
0
)
e_1=\left(1,0,0\right)~
e
1
=
(
1
,
0
,
0
)
;
e
2
=
(
0
,
1
,
0
)
;
e
3
=
(
0
,
0
,
1
)
e_2=\left(0,1,0\right);\ e_3=\left(0,0,1\right)
e
2
=
(
0
,
1
,
0
)
;
e
3
=
(
0
,
0
,
1
)
On note
B
′
=
(
e
′
1
,
e
′
2
)
B'=\left({e'}_1,{e'}_2\right)
B
′
=
(
e
′
1
,
e
′
2
)
une base de
R
2
{\mathbb{R}}^2
R
2
avec
e
′
1
=
(
1
,
0
)
{e'}_1=\left(1,0\right)~
e
′
1
=
(
1
,
0
)
;
e
′
2
=
(
0
,
1
)
{e'}_2=\left(0,1\right)
e
′
2
=
(
0
,
1
)
On a :
{
f
(
e
1
)
=
f
(
1
,
0
,
0
)
f
(
e
2
)
=
f
(
0
,
1
,
0
)
f
(
e
3
)
=
f
(
0
,
0
,
1
)
=
=
=
(
1
,
1
)
(
1
,
0
)
(
0
,
−
1
)
=
=
=
1
e
′
1
+
1
e
′
2
1
e
′
1
+
0
e
′
2
0
e
′
1
−
1
e
′
2
\left\{ \begin{array}{ccc}f\left(e_1\right) & = & f\left(1,0,0\right) \\ f\left(e_2\right) & = & f\left(0,1,0\right) \\ f\left(e_3\right) & = & f\left(0,0,1\right) \end{array}\right. \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}\left(1,1\right) \\ \left(1,0\right) \\ \left(0,-1\right) \end{array} \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}\red{1}{e'}_1+\red{1}{e'}_2 \\ \purple{1}{e'}_1+\purple{0}{e'}_2 \\ \pink{0}{e'}_1\pink{-1}{e'}_2 \end{array}
⎩
⎨
⎧
f
(
e
1
)
f
(
e
2
)
f
(
e
3
)
=
=
=
f
(
1
,
0
,
0
)
f
(
0
,
1
,
0
)
f
(
0
,
0
,
1
)
=
=
=
(
1
,
1
)
(
1
,
0
)
(
0
,
−
1
)
=
=
=
1
e
′
1
+
1
e
′
2
1
e
′
1
+
0
e
′
2
0
e
′
1
−
1
e
′
2
Ainsi :
A
=
M
(
f
,
B
,
B
′
)
=
(
1
1
0
1
0
−
1
)
A=M\left(f,B,B'\right)=\left( \begin{array}{ccc}\red{1} & \purple{1} & \pink{0} \\ \red{1} & \purple{0} & \pink{-1} \end{array}\right)
A
=
M
(
f
,
B
,
B
′
)
=
(
1
1
1
0
0
−
1
)
Question 2
Soit
f
f
f
l’application linéaire définie par
f
:
{
R
2
[
X
]
⟶
R
2
[
X
]
P
⟼
P
+
P
′
f:\left\{ \begin{array}{ccc}{\mathbb{R}}_2\left[X\right] & \longrightarrow & {\mathbb{R}}_2\left[X\right] \\ P & \longmapsto & P+P' \end{array}\right.
f
:
{
R
2
[
X
]
P
⟶
⟼
R
2
[
X
]
P
+
P
′
Écrire la matrice
A
A
A
de
f
f
f
relativement à la base canonique de
R
2
[
X
]
{\mathbb{R}}_2\left[X\right]
R
2
[
X
]
notée
(
1
,
X
,
X
2
)
\left(1,X,X^2\right)
(
1
,
X
,
X
2
)
.
Correction
Soient
E
E
E
et
F
F
F
deux
K
\mathbb{K}
K
-espaces vectoriels de dimension respective
n
n
n
et
p
p
p
. On note
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
B=\left(e_1,e_2,\cdots ,e_n\right)
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
une base de
E
E
E
et
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
B'=\left(e'_1,e'_2,\cdots ,e'_p\right)
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
une base de
F
F
F
.
On appelle matrice de
f
f
f
dans les bases
B
B
B
et
B
′
B'
B
′
et on note
M
(
f
,
B
,
B
′
)
M\left(f,B,B'\right)
M
(
f
,
B
,
B
′
)
le tableau dont les colonnes sont les composantes des vecteurs
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
\left(f\left(e_1\right),f\left(e_2\right),\cdots ,f\left(e_n\right)\right)
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
dans la base
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
\left(e'_1,e'_2,\cdots ,e'_p\right)
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
.
Autrement dit :
{
f
(
e
1
)
=
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
f
(
e
2
)
=
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
⋮
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
(
a
11
⋯
a
1
n
a
21
⋯
a
2
n
⋮
⋯
⋮
a
p
1
⋯
a
p
n
)
\left\{ \begin{array}{c} \begin{array}{ccc}f\left(e_1\right) & = & a_{11}{e'}_1+a_{21}{e'}_2+\cdots +a_{p1}{e'}_p \\ f\left(e_2\right) & = & a_{12}{e'}_1+a_{22}{e'}_2+\cdots +a_{p2}{e'}_p \\ & \vdots & \end{array} \\ \begin{array}{ccc}f\left(e_n\right) & = & a_{1n}{e'}_1+a_{2n}{e'}_2+\cdots +a_{pn}{e'}_p \end{array} \end{array}\right. \Longleftrightarrow M\left(f,B,B'\right)=\left( \begin{array}{ccc}a_{11} &\cdots & a_{1n} \\a_{21} &\cdots & a_{2n} \\ \vdots & \cdots & \vdots \\ a_{p1} & \cdots & a_{pn} \end{array}\right)
⎩
⎨
⎧
f
(
e
1
)
f
(
e
2
)
=
=
⋮
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
⎝
⎛
a
11
a
21
⋮
a
p
1
⋯
⋯
⋯
⋯
a
1
n
a
2
n
⋮
a
p
n
⎠
⎞
.
Soient
E
=
R
2
[
X
]
E={\mathbb{R}}_2\left[X\right]
E
=
R
2
[
X
]
et
F
=
R
2
[
X
]
F={\mathbb{R}}_2\left[X\right]
F
=
R
2
[
X
]
On note
B
=
(
1
,
X
,
X
2
)
B=\left(1,X,X^2\right)
B
=
(
1
,
X
,
X
2
)
une base de
R
2
[
X
]
{\mathbb{R}}_2\left[X\right]
R
2
[
X
]
On note
B
′
=
(
1
,
X
,
X
2
)
B'=\left(1,X,X^2\right)
B
′
=
(
1
,
X
,
X
2
)
une base de
R
2
[
X
]
{\mathbb{R}}_2\left[X\right]
R
2
[
X
]
On a :
{
f
(
1
)
=
1
+
(
1
)
′
f
(
X
)
=
X
+
(
X
)
′
f
(
X
2
)
=
X
2
+
(
X
2
)
′
=
=
=
1
1
+
X
2
X
+
X
2
=
=
=
1
⋅
1
+
0
X
+
0
X
2
1
⋅
1
+
1
X
+
0
X
2
0
⋅
1
+
2
X
+
1
X
2
\left\{ \begin{array}{ccc}f\left(1\right) & = & 1+\left(1\right)' \\ f\left(X\right) & = & X+\left(X\right)' \\ f\left(X^2\right) & = & X^2+\left(X^2\right)' \end{array}\right. \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}1\\ 1+X \\ 2X+X^2 \end{array} \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}\red{1}\cdot1+\red{0}X+\red{0}X^2\\\purple{1}\cdot1+\purple{1}X+ \purple{0}X^2\\ \pink{0}\cdot1+\pink{2}X+\pink{1}X^2 \end{array}
⎩
⎨
⎧
f
(
1
)
f
(
X
)
f
(
X
2
)
=
=
=
1
+
(
1
)
′
X
+
(
X
)
′
X
2
+
(
X
2
)
′
=
=
=
1
1
+
X
2
X
+
X
2
=
=
=
1
⋅
1
+
0
X
+
0
X
2
1
⋅
1
+
1
X
+
0
X
2
0
⋅
1
+
2
X
+
1
X
2
Ainsi :
A
=
M
(
f
,
B
,
B
′
)
=
(
1
1
0
0
1
2
0
1
1
)
A=M\left(f,B,B'\right)=\left( \begin{array}{ccc}\red{1} & \purple{1} & \pink{0} \\ \red{0} & \purple{1} & \pink{2}\\ \red{0} & \purple{\purple{1}} & \pink{1} \end{array}\right)
A
=
M
(
f
,
B
,
B
′
)
=
⎝
⎛
1
0
0
1
1
1
0
2
1
⎠
⎞
Question 3
Soit
f
f
f
l’application linéaire définie par
f
:
{
R
2
[
X
]
⟶
R
4
[
X
]
P
⟼
X
2
P
−
X
P
′
+
X
2
P
′
′
f:\left\{ \begin{array}{ccc}{\mathbb{R}}_2\left[X\right] & \longrightarrow & {\mathbb{R}}_4\left[X\right] \\ P & \longmapsto & X^2P-XP'+X^2P'' \end{array}\right.
f
:
{
R
2
[
X
]
P
⟶
⟼
R
4
[
X
]
X
2
P
−
X
P
′
+
X
2
P
′′
Écrire la matrice
A
A
A
de
f
f
f
relativement à la base canonique de
R
2
[
X
]
{\mathbb{R}}_2\left[X\right]
R
2
[
X
]
notée
(
1
,
X
,
X
2
)
\left(1,X,X^2\right)
(
1
,
X
,
X
2
)
et à la base de
R
4
[
X
]
{\mathbb{R}}_4\left[X\right]
R
4
[
X
]
notée
(
X
4
,
X
3
,
X
2
,
X
,
1
)
\left(X^4,X^3,X^2,X,1\right)
(
X
4
,
X
3
,
X
2
,
X
,
1
)
.
Correction
Soient
E
E
E
et
F
F
F
deux
K
\mathbb{K}
K
-espaces vectoriels de dimension respective
n
n
n
et
p
p
p
. On note
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
B=\left(e_1,e_2,\cdots ,e_n\right)
B
=
(
e
1
,
e
2
,
⋯
,
e
n
)
une base de
E
E
E
et
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
B'=\left(e'_1,e'_2,\cdots ,e'_p\right)
B
′
=
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
une base de
F
F
F
.
On appelle matrice de
f
f
f
dans les bases
B
B
B
et
B
′
B'
B
′
et on note
M
(
f
,
B
,
B
′
)
M\left(f,B,B'\right)
M
(
f
,
B
,
B
′
)
le tableau dont les colonnes sont les composantes des vecteurs
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
\left(f\left(e_1\right),f\left(e_2\right),\cdots ,f\left(e_n\right)\right)
(
f
(
e
1
)
,
f
(
e
2
)
,
⋯
,
f
(
e
n
)
)
dans la base
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
\left(e'_1,e'_2,\cdots ,e'_p\right)
(
e
1
′
,
e
2
′
,
⋯
,
e
p
′
)
.
Autrement dit :
{
f
(
e
1
)
=
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
f
(
e
2
)
=
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
⋮
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
(
a
11
⋯
a
1
n
a
21
⋯
a
2
n
⋮
⋯
⋮
a
p
1
⋯
a
p
n
)
\left\{ \begin{array}{c} \begin{array}{ccc}f\left(e_1\right) & = & a_{11}{e'}_1+a_{21}{e'}_2+\cdots +a_{p1}{e'}_p \\ f\left(e_2\right) & = & a_{12}{e'}_1+a_{22}{e'}_2+\cdots +a_{p2}{e'}_p \\ & \vdots & \end{array} \\ \begin{array}{ccc}f\left(e_n\right) & = & a_{1n}{e'}_1+a_{2n}{e'}_2+\cdots +a_{pn}{e'}_p \end{array} \end{array}\right. \Longleftrightarrow M\left(f,B,B'\right)=\left( \begin{array}{ccc}a_{11} &\cdots & a_{1n} \\a_{21} &\cdots & a_{2n} \\ \vdots & \cdots & \vdots \\ a_{p1} & \cdots & a_{pn} \end{array}\right)
⎩
⎨
⎧
f
(
e
1
)
f
(
e
2
)
=
=
⋮
a
11
e
′
1
+
a
21
e
′
2
+
⋯
+
a
p
1
e
′
p
a
12
e
′
1
+
a
22
e
′
2
+
⋯
+
a
p
2
e
′
p
f
(
e
n
)
=
a
1
n
e
′
1
+
a
2
n
e
′
2
+
⋯
+
a
p
n
e
′
p
⟺
M
(
f
,
B
,
B
′
)
=
⎝
⎛
a
11
a
21
⋮
a
p
1
⋯
⋯
⋯
⋯
a
1
n
a
2
n
⋮
a
p
n
⎠
⎞
.
Soient
E
=
R
2
[
X
]
E={\mathbb{R}}_2\left[X\right]
E
=
R
2
[
X
]
et
F
=
R
4
[
X
]
F={\mathbb{R}}_4\left[X\right]
F
=
R
4
[
X
]
On note
B
=
(
1
,
X
,
X
2
)
B=\left(1,X,X^2\right)
B
=
(
1
,
X
,
X
2
)
une base de
R
2
[
X
]
{\mathbb{R}}_2\left[X\right]
R
2
[
X
]
On note
B
′
=
(
X
4
,
X
3
,
X
2
,
X
,
1
)
B'=\left(X^4,X^3,X^2,X,1\right)
B
′
=
(
X
4
,
X
3
,
X
2
,
X
,
1
)
une base de
R
4
[
X
]
{\mathbb{R}}_4\left[X\right]
R
4
[
X
]
On a :
{
f
(
1
)
=
X
2
×
1
−
X
×
(
1
)
′
+
X
2
×
(
1
)
′
′
f
(
X
)
=
X
2
×
X
−
X
×
(
X
)
′
+
X
2
×
(
X
)
′
′
f
(
X
2
)
=
X
2
×
X
2
−
X
×
(
X
2
)
′
+
X
2
×
(
X
2
)
′
′
=
=
=
X
2
X
3
−
X
X
4
−
2
X
2
+
2
X
2
=
=
=
0
X
4
+
0
X
3
+
1
X
2
+
0
X
+
0
⋅
1
0
X
4
+
1
X
3
+
0
X
2
−
1
X
+
0
⋅
1
1
X
4
+
0
X
3
+
0
X
2
+
0
X
+
0
⋅
1
\left\{ \begin{array}{ccc}f\left(1\right) & = & X^2 \times1-X\times\left(1\right)'+X^2\times\left(1\right)'' \\ f\left(X\right) & = & X^2 \times X-X\times\left(X\right)'+X^2\times\left(X\right)''\\ f\left(X^2\right) & = & X^2 \times X^2-X\times\left(X^2\right)'+X^2\times\left(X^2\right)'' \end{array}\right. \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}X^2\\ X^3-X \\ X^4-2X^2+2X^2 \end{array} \begin{array}{c}= \\ = \\ = \end{array} \begin{array}{c}\red{0}X^4+\red{0}X^3+\red{1}X^2+\red{0}X+\red{0}\cdot1\\ \purple{0}X^4+\purple{1}X^3+\purple{0}X^2\purple{-1}X+\purple{0}\cdot1\\ \pink{1}X^4+\pink{0}X^3+\pink{0}X^2+\pink{0}X+\pink{0}\cdot1 \end{array}
⎩
⎨
⎧
f
(
1
)
f
(
X
)
f
(
X
2
)
=
=
=
X
2
×
1
−
X
×
(
1
)
′
+
X
2
×
(
1
)
′′
X
2
×
X
−
X
×
(
X
)
′
+
X
2
×
(
X
)
′′
X
2
×
X
2
−
X
×
(
X
2
)
′
+
X
2
×
(
X
2
)
′′
=
=
=
X
2
X
3
−
X
X
4
−
2
X
2
+
2
X
2
=
=
=
0
X
4
+
0
X
3
+
1
X
2
+
0
X
+
0
⋅
1
0
X
4
+
1
X
3
+
0
X
2
−
1
X
+
0
⋅
1
1
X
4
+
0
X
3
+
0
X
2
+
0
X
+
0
⋅
1
Ainsi :
A
=
M
(
f
,
B
,
B
′
)
=
(
0
0
1
0
1
0
1
0
0
0
−
1
0
0
0
0
)
A=M\left(f,B,B'\right)=\left( \begin{array}{ccc}\red{0} & \purple{0} & \pink{1} \\ \red{0} & \purple{1} & \pink{0} \\ \red{1} & \purple{0} & \pink{0} \\ \red{0} & \purple{-1} & \pink{0} \\ \red{0} & \purple{0} & \pink{0} \end{array}\right)
A
=
M
(
f
,
B
,
B
′
)
=
⎝
⎛
0
0
1
0
0
0
1
0
−
1
0
1
0
0
0
0
⎠
⎞