Origem: Wikilivros, livros abertos por um mundo aberto.
Mostre que as operações usuais de soma de aplicações e produto de uma aplicação por um número real fazem do conjunto
L
(
R
m
;
R
n
)
{\displaystyle L(\mathbb {R} ^{m};\mathbb {R} ^{n})}
M
(
n
×
m
)
{\displaystyle M(n\times m)}
R
n
m
{\displaystyle R^{nm}}
L
(
R
m
;
R
n
)
e
M
(
n
×
m
)
{\displaystyle L(\mathbb {R} ^{m};\mathbb {R} ^{n})\;e\;M(n\times m)}
Mostrar que
L
(
R
m
;
R
n
)
{\displaystyle L(\mathbb {R} ^{m};\mathbb {R} ^{n})}
Para isso tome
A
,
B
,
C
∈
L
(
R
m
;
R
n
)
{\displaystyle A,B,C\in L(\mathbb {R} ^{m};\mathbb {R} ^{n})}
A
:
R
m
→
R
n
,
B
:
R
m
→
R
n
,
C
:
R
m
→
R
n
{\displaystyle A:\mathbb {R} ^{m}\to \mathbb {R} ^{n},B:\mathbb {R} ^{m}\to \mathbb {R} ^{n},C:\mathbb {R} ^{m}\to \mathbb {R} ^{n}}
A
(
c
x
+
y
)
=
c
A
(
x
)
+
A
(
y
)
,
(
A
+
B
)
(
x
)
=
A
(
x
)
+
B
(
x
)
,
(
c
A
)
(
x
)
=
c
A
(
x
)
,
∀
c
∈
R
,
∀
x
,
y
∈
R
m
{\displaystyle A(cx+y)=cA(x)+A(y),(A+B)(x)=A(x)+B(x),(cA)(x)=cA(x),\forall c\in \mathbb {R} ,\forall x,y\in \mathbb {R} ^{m}}
(
A
+
B
)
(
c
x
+
y
)
=
A
(
c
x
+
y
)
+
B
(
c
x
+
y
)
=
c
A
(
x
)
+
A
(
y
)
+
c
B
(
x
)
+
B
(
y
)
=
c
(
A
(
x
)
+
B
(
x
)
)
+
A
(
y
)
+
B
(
y
)
=
c
(
A
+
B
)
(
x
)
+
(
A
+
B
)
(
y
)
,
∀
x
,
y
∈
R
m
,
c
∈
R
{\displaystyle (A+B)(cx+y)=A(cx+y)+B(cx+y)=cA(x)+A(y)+cB(x)+B(y)=c(A(x)+B(x))+A(y)+B(y)=c(A+B)(x)+(A+B)(y),\forall x,y\in \mathbb {R} ^{m},c\in \mathbb {R} }
(
A
+
B
)
(
x
)
=
A
(
x
)
+
B
(
x
)
=
B
(
x
)
+
A
(
x
)
=
(
B
+
A
)
(
x
)
,
∀
x
∈
R
m
{\displaystyle (A+B)(x)=A(x)+B(x)=B(x)+A(x)=(B+A)(x),\forall x\in \mathbb {R} ^{m}}
(
A
+
(
B
+
C
)
)
(
x
)
=
A
(
x
)
+
(
B
+
C
)
(
x
)
=
A
(
x
)
+
B
(
x
)
+
C
(
x
)
=
(
A
+
B
)
(
x
)
+
C
(
x
)
=
(
(
A
+
B
)
+
C
)
(
x
)
,
∀
x
∈
R
m
{\displaystyle (A+(B+C))(x)=A(x)+(B+C)(x)=A(x)+B(x)+C(x)=(A+B)(x)+C(x)=((A+B)+C)(x),\forall x\in \mathbb {R} ^{m}}
A
(
x
)
+
B
(
x
)
=
A
(
x
)
⇒
B
(
x
)
=
A
(
x
)
−
A
(
x
)
⇒
B
(
x
)
=
0
,
∀
x
∈
R
m
{\displaystyle A(x)+B(x)=A(x)\Rightarrow B(x)=A(x)-A(x)\Rightarrow B(x)=0,\forall x\in \mathbb {R} ^{m}}
A
(
x
)
+
B
(
x
)
=
0
(
x
)
⇒
B
(
x
)
=
−
A
(
x
)
,
∀
x
∈
R
m
{\displaystyle A(x)+B(x)=0(x)\Rightarrow B(x)=-A(x),\forall x\in \mathbb {R} ^{m}}
(
k
A
)
(
c
x
+
y
)
=
k
[
A
(
c
x
+
y
)
]
=
k
[
c
A
(
x
)
+
A
(
y
)
]
=
k
c
A
(
x
)
+
k
A
(
y
)
=
c
k
A
(
x
)
+
(
k
A
)
(
y
)
=
c
(
k
A
)
(
x
)
+
(
k
A
)
(
y
)
,
∀
x
∈
R
m
{\displaystyle (kA)(cx+y)=k[A(cx+y)]=k[cA(x)+A(y)]=kcA(x)+kA(y)=ckA(x)+(kA)(y)=c(kA)(x)+(kA)(y),\forall x\in \mathbb {R} ^{m}}
c
A
(
x
)
=
A
(
x
)
⇒
c
A
(
x
)
−
A
(
x
)
=
0
⇒
(
c
−
1
)
A
(
x
)
=
0
⇒
c
−
1
=
0
⇒
c
=
1
,
∀
x
∈
R
m
{\displaystyle cA(x)=A(x)\Rightarrow cA(x)-A(x)=0\Rightarrow (c-1)A(x)=0\Rightarrow c-1=0\Rightarrow c=1,\forall x\in \mathbb {R} ^{m}}
1
A
(
x
)
=
A
(
x
)
,
∀
x
∈
R
m
{\displaystyle 1A(x)=A(x),\forall x\in \mathbb {R} ^{m}}
(
b
c
)
A
(
x
)
=
A
(
b
c
x
)
=
b
A
(
c
x
)
=
b
(
c
A
(
x
)
)
,
∀
x
∈
R
m
,
∀
b
,
c
∈
R
{\displaystyle (bc)A(x)=A(bcx)=bA(cx)=b(cA(x)),\forall x\in \mathbb {R} ^{m},\forall b,c\in \mathbb {R} }
c
(
A
+
B
)
(
x
)
=
(
A
+
B
)
(
c
x
)
=
A
(
c
x
)
+
B
(
c
x
)
=
c
A
(
x
)
+
c
B
(
x
)
,
∀
x
∈
R
m
,
∀
c
∈
R
{\displaystyle c(A+B)(x)=(A+B)(cx)=A(cx)+B(cx)=cA(x)+cB(x),\forall x\in \mathbb {R} ^{m},\forall c\in \mathbb {R} }
(
b
+
c
)
A
(
x
)
=
A
(
(
b
+
c
)
x
)
=
A
(
b
x
+
c
x
)
=
b
A
(
x
)
+
c
A
(
x
)
,
∀
x
∈
R
m
,
∀
b
,
c
∈
R
{\displaystyle (b+c)A(x)=A((b+c)x)=A(bx+cx)=bA(x)+cA(x),\forall x\in \mathbb {R} ^{m},\forall b,c\in \mathbb {R} }
Mostrar que
M
(
n
×
m
)
{\displaystyle M(n\times m)}
Para isso tome
A
,
B
,
C
∈
M
(
n
×
m
)
{\displaystyle A,B,C\in M(n\times m)}
A
=
(
A
i
j
)
K
,
K
=
{
i
j
/
1
≤
i
≤
n
,
1
≤
j
≤
m
}
{\displaystyle A=(A_{ij})_{K},K=\{ij/1\leq i\leq n,1\leq j\leq m\}}
A
+
B
=
(
A
i
j
)
K
+
(
B
i
j
)
K
=
(
A
i
j
+
B
i
j
)
K
=
(
B
i
j
+
A
i
j
)
K
=
(
B
i
j
)
K
+
(
A
i
j
)
K
=
B
+
A
,
∀
A
i
j
,
B
i
j
∈
R
{\displaystyle A+B=(A_{ij})_{K}+(B_{ij})_{K}=(A_{ij}+B_{ij})_{K}=(B_{ij}+A_{ij})_{K}=(B_{ij})_{K}+(A_{ij})_{K}=B+A,\forall A_{ij},B_{ij}\in \mathbb {R} }
A
+
(
B
+
C
)
=
(
A
i
j
)
K
+
(
(
B
i
j
)
K
+
(
C
i
j
)
K
)
=
(
A
i
j
)
K
+
(
B
i
j
+
C
i
j
)
K
=
(
A
i
j
+
B
i
j
+
C
i
j
)
K
=
(
A
i
j
+
B
i
j
)
K
+
(
C
i
j
)
K
=
(
(
A
i
j
)
K
+
(
B
i
j
)
K
)
+
(
C
i
j
)
K
=
(
A
+
B
)
+
C
{\displaystyle A+(B+C)=(A_{ij})_{K}+((B_{ij})_{K}+(C_{ij})_{K})=(A_{ij})_{K}+(B_{ij}+C_{ij})_{K}=(A_{ij}+B_{ij}+C_{ij})_{K}=(A_{ij}+B_{ij})_{K}+(C_{ij})_{K}=((A_{ij})_{K}+(B_{ij})_{K})+(C_{ij})_{K}=(A+B)+C\;}
∀
A
i
j
,
B
i
j
,
C
i
j
∈
R
{\displaystyle \forall A_{ij},B_{ij},C_{ij}\in \mathbb {R} }
A
+
B
=
A
⇒
B
=
A
−
A
⇒
B
=
0
,
∀
A
i
j
,
B
i
j
∈
R
{\displaystyle A+B=A\Rightarrow B=A-A\Rightarrow B=0,\forall A_{ij},B_{ij}\in \mathbb {R} }
0
i
j
{\displaystyle 0_{ij}\;}
A
+
B
=
0
⇒
B
=
−
A
,
∀
A
i
j
,
B
i
j
∈
R
{\displaystyle A+B=0\Rightarrow B=-A,\forall A_{ij},B_{ij}\in \mathbb {R} }
c
A
=
A
⇒
c
A
−
A
=
0
⇒
(
c
−
1
)
A
=
0
⇒
c
−
1
=
0
⇒
c
=
1
,
∀
A
i
j
,
B
i
j
∈
R
{\displaystyle cA=A\Rightarrow cA-A=0\Rightarrow (c-1)A=0\Rightarrow c-1=0\Rightarrow c=1,\forall A_{ij},B_{ij}\in \mathbb {R} }
1
A
=
A
,
∀
A
i
j
∈
R
{\displaystyle 1A=A,\forall A_{ij}\in \mathbb {R} }
(
b
c
)
A
=
b
c
(
A
i
j
)
K
=
(
b
c
A
i
j
)
K
=
b
(
c
A
i
j
)
K
=
b
(
c
A
)
,
∀
A
i
j
,
b
,
c
∈
R
{\displaystyle (bc)A=bc(A_{ij})_{K}=(bcA_{ij})_{K}=b(cA_{ij})_{K}=b(cA),\forall A_{ij},b,c\in \mathbb {R} }
c
(
A
+
B
)
=
c
[
(
A
i
j
)
K
+
(
B
i
j
)
K
]
=
c
(
A
i
j
+
B
i
j
)
K
=
[
c
(
A
i
j
+
B
i
j
)
]
K
=
(
c
A
i
j
+
c
B
i
j
)
K
=
(
c
A
i
j
)
K
+
(
c
B
i
j
)
K
=
c
(
A
i
j
)
K
+
c
(
B
i
j
)
K
=
c
A
+
c
B
,
∀
A
i
j
,
B
i
j
,
c
∈
R
{\displaystyle c(A+B)=c[(A_{ij})_{K}+(B_{ij})_{K}]=c(A_{ij}+B_{ij})_{K}=[c(A_{ij}+B_{ij})]_{K}=(cA_{ij}+cB_{ij})_{K}=(cA_{ij})_{K}+(cB_{ij})_{K}=c(A_{ij})_{K}+c(B_{ij})_{K}=cA+cB,\forall A_{ij},B_{ij},c\in \mathbb {R} }
(
b
+
c
)
A
=
(
b
+
c
)
(
A
i
j
)
K
=
[
(
b
+
c
)
A
i
j
]
K
=
[
b
A
i
j
+
c
A
i
j
]
K
=
(
b
A
i
j
)
K
+
(
c
A
i
j
)
K
=
b
(
A
i
j
)
K
+
c
(
A
i
j
)
K
=
b
A
+
c
A
,
∀
A
i
j
,
b
,
c
∈
R
,
∀
b
,
c
∈
R
{\displaystyle (b+c)A=(b+c)(A_{ij})_{K}=[(b+c)A_{ij}]_{K}=[bA_{ij}+cA_{ij}]_{K}=(bA_{ij})_{K}+(cA_{ij})_{K}=b(A_{ij})_{K}+c(A_{ij})_{K}=bA+cA,\forall A_{ij},b,c\in \mathbb {R} ,\forall b,c\in \mathbb {R} }
Mostre que as bijeções estabelecidas no texto entre esses conjuntos e
R
n
m
{\displaystyle R^{nm}}
Tome
ϕ
M
(
n
×
m
)
→
R
n
m
,
M
↦
x
{\displaystyle \phi M(n\times m)\to \mathbb {R} ^{nm},M\mapsto x}
x
=
(
M
11
,
M
12
,
.
.
.
,
M
1
m
,
M
21
,
.
.
.
,
M
2
m
,
.
.
.
,
M
n
1
,
.
.
.
,
M
n
m
)
{\displaystyle x=(M_{11},M_{12},...,M_{1m},M_{21},...,M_{2m},...,M_{n1},...,M_{nm})\;}
ϕ
(
M
)
=
x
{\displaystyle \phi (M)=x}
injetiva
sobrejetiva
linear Exiba explicitamente bases para os espaços
L
(
R
m
;
R
n
)
e
M
(
n
×
m
)
{\displaystyle L(\mathbb {R} ^{m};\mathbb {R} ^{n})\;e\;M(n\times m)}
Base de
L
(
R
m
;
R
n
)
{\displaystyle L(\mathbb {R} ^{m};\mathbb {R} ^{n})\;}
Base de
M
(
n
×
m
)
{\displaystyle M(n\times m)\;}
{
e
i
j
,
1
≤
i
≤
n
,
1
≤
j
≤
m
,
e
i
j
=
A
,
o
n
d
e
A
i
j
=
1
,
A
l
=
0
,
l
≠
i
j
}
{\displaystyle \{e_{ij},1\leq i\leq n,1\leq j\leq m,e_{ij}=A,onde\;A_{ij}=1,A_{l}=0,l\not =ij\}}
Seja
E
=
L
(
R
m
,
R
n
;
R
p
)
{\displaystyle E=L(\mathbb {R} ^{m},\mathbb {R} ^{n};\mathbb {R} ^{p})}
ϕ
:
R
m
×
R
n
→
R
p
{\displaystyle \phi :\mathbb {R} ^{m}\times \mathbb {R} ^{n}\to \mathbb {R} ^{p}}
Mostrar que
L
(
R
m
,
R
n
;
R
p
)
{\displaystyle L(\mathbb {R} ^{m},\mathbb {R} ^{n};\mathbb {R} ^{p})}
Para isso tome
A
,
B
,
C
∈
L
(
R
m
,
R
n
;
R
p
)
{\displaystyle A,B,C\in L(\mathbb {R} ^{m},\mathbb {R} ^{n};\mathbb {R} ^{p})}
A
:
R
m
×
R
n
→
R
p
,
B
:
R
m
×
R
n
→
R
p
,
C
:
R
m
×
R
n
→
R
p
{\displaystyle A:\mathbb {R} ^{m}\times \mathbb {R} ^{n}\to \mathbb {R} ^{p},B:\mathbb {R} ^{m}\times \mathbb {R} ^{n}\to \mathbb {R} ^{p},C:\mathbb {R} ^{m}\times \mathbb {R} ^{n}\to \mathbb {R} ^{p}}
A
(
x
+
x
′
,
y
)
=
A
(
x
,
y
)
+
A
(
x
′
,
y
)
,
A
(
x
,
y
+
y
′
)
=
A
(
x
,
y
)
+
A
(
x
,
y
′
)
,
A
(
c
x
,
y
)
=
c
A
(
x
,
y
)
,
A
(
x
,
c
y
)
=
c
A
(
x
,
y
)
,
∀
c
∈
R
,
∀
x
∈
R
m
,
∀
y
∈
R
n
{\displaystyle A(x+x',y)=A(x,y)+A(x',y),A(x,y+y')=A(x,y)+A(x,y'),A(cx,y)=cA(x,y),A(x,cy)=cA(x,y),\forall c\in \mathbb {R} ,\forall x\in \mathbb {R} ^{m},\forall y\in \mathbb {R} ^{n}}
(
A
+
B
)
(
c
x
+
y
)
=
A
(
c
x
+
y
)
+
B
(
c
x
+
y
)
=
c
A
(
x
)
+
A
(
y
)
+
c
B
(
x
)
+
B
(
y
)
=
c
(
A
(
x
)
+
B
(
x
)
)
+
A
(
y
)
+
B
(
y
)
=
c
(
A
+
B
)
(
x
)
+
(
A
+
B
)
(
y
)
,
∀
x
,
y
∈
R
m
,
c
∈
R
{\displaystyle (A+B)(cx+y)=A(cx+y)+B(cx+y)=cA(x)+A(y)+cB(x)+B(y)=c(A(x)+B(x))+A(y)+B(y)=c(A+B)(x)+(A+B)(y),\forall x,y\in \mathbb {R} ^{m},c\in \mathbb {R} }
(
A
+
B
)
(
x
)
=
A
(
x
)
+
B
(
x
)
=
B
(
x
)
+
A
(
x
)
=
(
B
+
A
)
(
x
)
,
∀
x
∈
R
m
{\displaystyle (A+B)(x)=A(x)+B(x)=B(x)+A(x)=(B+A)(x),\forall x\in \mathbb {R} ^{m}}
(
A
+
(
B
+
C
)
)
(
x
)
=
A
(
x
)
+
(
B
+
C
)
(
x
)
=
A
(
x
)
+
B
(
x
)
+
C
(
x
)
=
(
A
+
B
)
(
x
)
+
C
(
x
)
=
(
(
A
+
B
)
+
C
)
(
x
)
,
∀
x
∈
R
m
{\displaystyle (A+(B+C))(x)=A(x)+(B+C)(x)=A(x)+B(x)+C(x)=(A+B)(x)+C(x)=((A+B)+C)(x),\forall x\in \mathbb {R} ^{m}}
A
(
x
)
+
B
(
x
)
=
A
(
x
)
⇒
B
(
x
)
=
A
(
x
)
−
A
(
x
)
⇒
B
(
x
)
=
0
,
∀
x
∈
R
m
{\displaystyle A(x)+B(x)=A(x)\Rightarrow B(x)=A(x)-A(x)\Rightarrow B(x)=0,\forall x\in \mathbb {R} ^{m}}
A
(
x
)
+
B
(
x
)
=
0
(
x
)
⇒
B
(
x
)
=
−
A
(
x
)
,
∀
x
∈
R
m
{\displaystyle A(x)+B(x)=0(x)\Rightarrow B(x)=-A(x),\forall x\in \mathbb {R} ^{m}}
(
k
A
)
(
c
x
+
y
)
=
k
[
A
(
c
x
+
y
)
]
=
k
[
c
A
(
x
)
+
A
(
y
)
]
=
k
c
A
(
x
)
+
k
A
(
y
)
=
c
k
A
(
x
)
+
(
k
A
)
(
y
)
=
c
(
k
A
)
(
x
)
+
(
k
A
)
(
y
)
,
∀
x
∈
R
m
{\displaystyle (kA)(cx+y)=k[A(cx+y)]=k[cA(x)+A(y)]=kcA(x)+kA(y)=ckA(x)+(kA)(y)=c(kA)(x)+(kA)(y),\forall x\in \mathbb {R} ^{m}}
c
A
(
x
)
=
A
(
x
)
⇒
c
A
(
x
)
−
A
(
x
)
=
0
⇒
(
c
−
1
)
A
(
x
)
=
0
⇒
c
−
1
=
0
⇒
c
=
1
,
∀
x
∈
R
m
{\displaystyle cA(x)=A(x)\Rightarrow cA(x)-A(x)=0\Rightarrow (c-1)A(x)=0\Rightarrow c-1=0\Rightarrow c=1,\forall x\in \mathbb {R} ^{m}}
1
A
(
x
)
=
A
(
x
)
,
∀
x
∈
R
m
{\displaystyle 1A(x)=A(x),\forall x\in \mathbb {R} ^{m}}
(
b
c
)
A
(
x
)
=
A
(
b
c
x
)
=
b
A
(
c
x
)
=
b
(
c
A
(
x
)
)
,
∀
x
∈
R
m
,
∀
b
,
c
∈
R
{\displaystyle (bc)A(x)=A(bcx)=bA(cx)=b(cA(x)),\forall x\in \mathbb {R} ^{m},\forall b,c\in \mathbb {R} }
c
(
A
+
B
)
(
x
)
=
(
A
+
B
)
(
c
x
)
=
A
(
c
x
)
+
B
(
c
x
)
=
c
A
(
x
)
+
c
B
(
x
)
,
∀
x
∈
R
m
,
∀
c
∈
R
{\displaystyle c(A+B)(x)=(A+B)(cx)=A(cx)+B(cx)=cA(x)+cB(x),\forall x\in \mathbb {R} ^{m},\forall c\in \mathbb {R} }
(
b
+
c
)
A
(
x
)
=
A
(
(
b
+
c
)
x
)
=
A
(
b
x
+
c
x
)
=
b
A
(
x
)
+
c
A
(
x
)
,
∀
x
∈
R
m
,
∀
b
,
c
∈
R
{\displaystyle (b+c)A(x)=A((b+c)x)=A(bx+cx)=bA(x)+cA(x),\forall x\in \mathbb {R} ^{m},\forall b,c\in \mathbb {R} }
![{\displaystyle (kA)(cx+y)=k[A(cx+y)]=k[cA(x)+A(y)]=kcA(x)+kA(y)=ckA(x)+(kA)(y)=c(kA)(x)+(kA)(y),\forall x\in \mathbb {R} ^{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a5c9598a917932052f54bec2fd11882c9d8eecf)
![{\displaystyle c(A+B)=c[(A_{ij})_{K}+(B_{ij})_{K}]=c(A_{ij}+B_{ij})_{K}=[c(A_{ij}+B_{ij})]_{K}=(cA_{ij}+cB_{ij})_{K}=(cA_{ij})_{K}+(cB_{ij})_{K}=c(A_{ij})_{K}+c(B_{ij})_{K}=cA+cB,\forall A_{ij},B_{ij},c\in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3958636095a2ffd71269bd7a523863a909ea681)
![{\displaystyle (b+c)A=(b+c)(A_{ij})_{K}=[(b+c)A_{ij}]_{K}=[bA_{ij}+cA_{ij}]_{K}=(bA_{ij})_{K}+(cA_{ij})_{K}=b(A_{ij})_{K}+c(A_{ij})_{K}=bA+cA,\forall A_{ij},b,c\in \mathbb {R} ,\forall b,c\in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f94a82f0ec5d247087442ad865377305b8b98d84)
