∀
f
∈
L
(
R
n
;
R
)
=
(
R
n
)
∗
,
∃
!
y
∈
R
n
{\displaystyle \forall \;f\in L(\mathbb {R} ^{n};\mathbb {R} )=(\mathbb {R} ^{n})^{*},\exists !\;y\in \mathbb {R} ^{n}}
f
(
x
)
=
⟨
y
,
x
⟩
,
∀
x
∈
R
n
{\displaystyle f(x)=\langle y,x\rangle ,\forall x\in \mathbb {R} ^{n}}
Tome
f
∈
(
R
n
)
∗
{\displaystyle f\in (\mathbb {R} ^{n})^{*}}
f
:
R
n
→
R
{\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }
∃
A
n
×
1
∈
M
(
n
×
1
)
{\displaystyle \exists A_{n\times 1}\in M(n\times 1)}
f
(
x
)
=
A
x
=
(
A
11
A
12
.
.
.
A
1
n
)
⋅
(
x
1
x
2
.
.
.
x
n
)
t
=
A
11
x
1
+
A
12
x
2
+
.
.
.
+
A
1
n
x
n
=
⟨
A
,
x
⟩
{\displaystyle f(x)=Ax=(A_{11}A_{12}...A_{1n})\cdot (x_{1}x_{2}...x_{n})^{t}=A_{11}x_{1}+A_{12}x_{2}+...+A_{1n}x_{n}=\langle A,x\rangle }
A
1
i
=
y
i
{\displaystyle A_{1i}=y_{i}}
∀
f
∈
(
R
n
)
∗
,
f
(
x
)
=
⟨
A
,
x
⟩
=
⟨
y
,
x
⟩
,
∀
x
∈
R
n
{\displaystyle \forall f\in (\mathbb {R} ^{n})^{*},f(x)=\langle A,x\rangle =\langle y,x\rangle ,\forall x\in \mathbb {R} ^{n}}
Um conjunto
{
u
1
,
u
2
,
.
.
.
,
u
r
}
⊂
R
n
{\displaystyle \{u_{1},u_{2},...,u_{r}\}\subset \mathbb {R} ^{n}}
⟨
u
j
,
u
j
⟩
=
1
e
⟨
u
i
,
u
j
⟩
=
0
,
p
a
r
a
i
≠
j
{\displaystyle \langle u_{j},u_{j}\rangle =1\;e\;\langle u_{i},u_{j}\rangle =0,\;para\;i\not =j}
{
u
1
,
.
.
.
,
u
n
}
{\displaystyle \{u_{1},...,u_{n}\}\;}
x
=
∑
i
=
1
n
⟨
x
,
u
i
⟩
u
i
,
∀
x
∈
R
n
{\displaystyle x=\sum _{i=1}^{n}\langle x,u_{i}\rangle u_{i},\forall x\in \mathbb {R} ^{n}}
Seja
S
k
⊂
W
{\displaystyle S_{k}\subset W}
S
k
⊂
β
W
{\displaystyle S_{k}\subset \beta _{W}}
β
W
{\displaystyle \beta _{W}\;}
Seja
S
k
=
{
u
1
,
u
2
,
.
.
.
,
u
k
}
,
β
W
=
{
u
1
,
u
2
,
.
.
.
,
u
k
−
1
,
u
k
,
u
k
+
1
,
.
.
.
,
u
n
}
{\displaystyle S_{k}=\{u_{1},u_{2},...,u_{k}\},\beta _{W}=\{u_{1},u_{2},...,u_{k-1},u_{k},u_{k+1},...,u_{n}\}\;}
S
k
{\displaystyle S_{k}\;}
Tomando K = {o conjunto das combinações lineares dos elementos de
S
k
{\displaystyle S_{k}\;}
{
u
/
u
=
⟨
S
k
,
x
⟩
,
∀
x
∈
R
k
}
{\displaystyle \{u/u=\langle S_{k},x\rangle ,\forall x\in \mathbb {R} ^{k}\}}
S
k
⊂
W
{\displaystyle S_{k}\subset W}
Seja
v
∈
K
⇒
v
=
⟨
S
k
,
y
⟩
⇒
⟨
v
,
u
l
⟩
,
o
n
d
e
1
≤
l
≤
k
⇒
∑
i
=
1
k
⟨
y
i
u
i
,
u
l
⟩
⇒
∑
i
=
1
k
y
i
⟨
u
i
,
u
l
⟩
=
c
j
{\displaystyle v\in K\Rightarrow v=\langle S_{k},y\rangle \Rightarrow \langle v,u_{l}\rangle ,onde\;1\leq l\leq k\Rightarrow \sum _{i=1}^{k}\langle y_{i}u_{i},u_{l}\rangle \Rightarrow \sum _{i=1}^{k}y_{i}\langle u_{i},u_{l}\rangle =c_{j}}
Quando v = 0,
c
j
=
0
{\displaystyle c_{j}=0\;}
S
k
{\displaystyle S_{k}\;}
Tome
w
∈
W
⇒
w
=
w
1
u
1
+
.
.
.
+
w
n
u
n
,
{\displaystyle w\in W\Rightarrow w=w_{1}u_{1}+...+w_{n}u_{n},}
w
1
,
.
.
.
,
w
n
{\displaystyle w_{1},...,w_{n}\;}
β
W
{\displaystyle \beta _{W}\;}
⟨
w
,
u
i
⟩
=
⟨
w
1
u
1
+
.
.
.
+
w
n
u
n
,
u
i
⟩
=
⟨
w
1
u
1
,
u
i
⟩
+
.
.
.
+
⟨
w
n
u
n
,
u
i
⟩
=
w
1
⟨
u
1
,
u
i
⟩
+
.
.
.
+
w
n
⟨
u
n
,
u
i
⟩
=
w
i
{\displaystyle \langle w,u_{i}\rangle =\langle w_{1}u_{1}+...+w_{n}u_{n},u_{i}\rangle =\langle w_{1}u_{1},u_{i}\rangle +...+\langle w_{n}u_{n},u_{i}\rangle =w_{1}\langle u_{1},u_{i}\rangle +...+w_{n}\langle u_{n},u_{i}\rangle =w_{i}}
w
=
w
1
u
1
+
.
.
.
+
w
n
u
n
=
⟨
w
,
u
1
⟩
u
1
+
.
.
.
+
⟨
w
,
u
n
⟩
u
n
,
∀
w
∈
R
n
{\displaystyle w=w_{1}u_{1}+...+w_{n}u_{n}=\langle w,u_{1}\rangle u_{1}+...+\langle w,u_{n}\rangle u_{n},\forall w\in \mathbb {R} ^{n}}
Considere em
R
m
,
R
n
{\displaystyle \mathbb {R} ^{m},\mathbb {R} ^{n}}
A
:
R
m
→
R
n
{\displaystyle A:\mathbb {R} ^{m}\to \mathbb {R} ^{n}}
A
∗
:
R
n
→
R
m
{\displaystyle A^{*}:\mathbb {R} ^{n}\to \mathbb {R} ^{m}}
⟨
A
x
,
y
⟩
=
⟨
x
,
A
∗
y
⟩
,
∀
x
∈
R
m
,
y
∈
R
n
{\displaystyle \langle Ax,y\rangle =\langle x,A^{*}y\rangle ,\forall x\in \mathbb {R} ^{m},y\in \mathbb {R} ^{n}}
b
∈
R
n
{\displaystyle b\in \mathbb {R} ^{n}}
A
x
=
b
{\displaystyle Ax=b\;}
x
∈
R
m
{\displaystyle x\in \mathbb {R} ^{m}}
A
∗
{\displaystyle A^{*}\;}
A
∗
{\displaystyle A^{*}\;}
Vamos verificar que
⟨
A
x
,
y
⟩
=
⟨
x
,
A
∗
y
⟩
{\displaystyle \langle Ax,y\rangle =\langle x,A^{*}y\rangle }
Seja
y
∈
R
n
{\displaystyle y\in \mathbb {R} ^{n}}
f
:
R
m
→
R
{\displaystyle f:\mathbb {R} ^{m}\to \mathbb {R} }
f
(
x
)
=
⟨
A
x
,
y
⟩
,
∀
x
∈
R
m
,
y
∈
R
n
{\displaystyle f(x)=\langle Ax,y\rangle ,\forall x\in \mathbb {R} ^{m},y\in \mathbb {R} ^{n}}
R
m
{\displaystyle \mathbb {R} ^{m}}
∃
!
z
∈
R
m
{\displaystyle \exists !\;z\in \mathbb {R} ^{m}}
f
(
x
)
=
⟨
x
,
z
⟩
⇒
⟨
A
x
,
y
⟩
=
⟨
x
,
z
⟩
,
∀
x
∈
R
m
{\displaystyle f(x)=\langle x,z\rangle \Rightarrow \langle Ax,y\rangle =\langle x,z\rangle ,\forall x\in \mathbb {R} ^{m}}
A
∗
(
y
)
=
z
{\displaystyle A^{*}(y)=z\;}
⟨
A
x
,
y
⟩
=
⟨
x
,
A
∗
y
⟩
{\displaystyle \langle Ax,y\rangle =\langle x,A^{*}y\rangle }
Vamos verificar que
A
∗
{\displaystyle A^{*}\;}
⟨
x
,
A
∗
(
c
a
+
b
)
⟩
=
⟨
A
x
,
(
c
a
+
b
)
⟩
=
⟨
A
x
,
c
a
⟩
+
⟨
A
x
,
b
⟩
=
c
⟨
A
x
,
a
⟩
+
⟨
A
x
,
b
⟩
=
c
⟨
x
,
A
∗
a
⟩
+
⟨
x
,
A
∗
b
⟩
=
⟨
x
,
c
A
∗
a
⟩
+
⟨
x
,
A
∗
b
⟩
=
{\displaystyle \langle x,A^{*}(ca+b)\rangle =\langle Ax,(ca+b)\rangle =\langle Ax,ca\rangle +\langle Ax,b\rangle =c\langle Ax,a\rangle +\langle Ax,b\rangle =c\langle x,A^{*}a\rangle +\langle x,A^{*}b\rangle =\langle x,cA^{*}a\rangle +\langle x,A^{*}b\rangle =}
=
⟨
x
,
c
A
∗
a
+
A
∗
b
⟩
{\displaystyle =\langle x,cA^{*}a+A^{*}b\rangle }
unicidade de
A
∗
{\displaystyle A^{*}\;}
Para
y
∈
R
n
{\displaystyle y\in \mathbb {R} ^{n}}
z
=
A
∗
y
∈
R
m
{\displaystyle z=A^{*}y\in \mathbb {R} ^{m}}
⟨
A
x
,
y
⟩
=
⟨
x
,
A
∗
y
⟩
,
∀
x
∈
R
n
{\displaystyle \langle Ax,y\rangle =\langle x,A^{*}y\rangle ,\forall x\in \mathbb {R} ^{n}}
Dado
b
∈
R
n
{\displaystyle b\in \mathbb {R} ^{n}}
A
x
=
b
{\displaystyle Ax=b\;}
x
∈
R
m
⇒
{\displaystyle x\in \mathbb {R} ^{m}\Rightarrow }
A
∗
{\displaystyle A^{*}\;}
Tome
A
x
=
b
∈
R
n
{\displaystyle Ax=b\in \mathbb {R} ^{n}}
⟨
b
,
y
⟩
=
⟨
x
,
A
∗
y
⟩
{\displaystyle \langle b,y\rangle =\langle x,A^{*}y\rangle }
y
∈
R
n
{\displaystyle y\in \mathbb {R} ^{n}}
A
∗
y
=
0
⇒
⟨
b
,
y
⟩
=
⟨
x
,
0
⟩
=
0
⇒
b
⊥
y
,
y
∈
K
e
r
(
A
∗
)
{\displaystyle A^{*}y=0\Rightarrow \langle b,y\rangle =\langle x,0\rangle =0\Rightarrow b\perp y,y\in Ker(A^{*})}
Dado b é ortogonal a todo elemento do núcleo de
A
∗
⇒
b
∈
R
n
{\displaystyle A^{*}\Rightarrow b\in \mathbb {R} ^{n}}
A
x
=
b
{\displaystyle Ax=b\;}
x
∈
R
m
{\displaystyle x\in \mathbb {R} ^{m}}
Tome
b
,
y
∈
R
n
;
b
⊥
y
,
y
∈
K
e
r
(
A
∗
)
⇒
⟨
b
,
y
⟩
=
0
=
⟨
x
,
0
⟩
=
⟨
x
,
A
∗
y
⟩
=
⟨
A
x
,
y
⟩
⇒
A
x
=
b
{\displaystyle b,y\in \mathbb {R} ^{n};b\perp y,y\in Ker(A^{*})\Rightarrow \langle b,y\rangle =0=\langle x,0\rangle =\langle x,A^{*}y\rangle =\langle Ax,y\rangle \Rightarrow Ax=b}
A imagem de
A
∗
{\displaystyle A^{*}\;}
Como b é ortogonal aos vetores
y
∈
K
e
r
(
A
∗
)
{\displaystyle y\in Ker(A^{*})}
b
∈
I
m
(
A
∗
)
{\displaystyle b\in Im(A^{*})}
b
=
A
x
,
b
∈
I
m
(
A
)
{\displaystyle b=Ax,b\in Im(A)}
d
i
m
(
I
m
(
A
∗
)
)
=
d
i
m
(
I
m
(
A
)
)
{\displaystyle dim(Im(A^{*}))=dim(Im(A))\;}
