Sea
B
=
{
v
1
,
v
2
,
.
.
.
,
v
q
}
{\displaystyle B=\{v_{1},v_{2},...,v_{q}\}}
V
K
{\displaystyle \mathrm {V_{\mathbb {K} }} }
K
{\displaystyle \mathbb {K} }
(
K
=
R
{\displaystyle {\bigl (}\mathbb {K} =\mathbb {R} }
K
=
C
)
{\displaystyle \mathbb {K} =\mathbb {C} {\bigr )}}
Se demuestra que
∀
x
∈
V
K
{\displaystyle \forall x\in \mathrm {V_{\mathbb {K} }} }
x
=
∑
i
=
1
q
α
i
v
i
{\displaystyle x=\sum _{i=1}^{q}\alpha _{i}v_{i}}
entonces
α
i
=
⟨
x
,
v
i
⟩
|
|
v
i
|
|
2
{\displaystyle \alpha _{i}={\frac {\langle x,v_{i}\rangle }{||v_{i}||^{2}}}}
i
=
1
,
2
,
.
.
.
,
q
{\displaystyle i=1,2,...,q}
donde
α
i
{\displaystyle \alpha _{i}}
B
{\displaystyle B}
x
{\displaystyle x}
x
=
∑
i
=
1
q
⟨
x
,
v
i
⟩
|
|
v
i
|
|
2
v
i
{\displaystyle x=\sum _{i=1}^{q}{\frac {\langle x,v_{i}\rangle }{||v_{i}||^{2}}}v_{i}}
Si la base
B
{\displaystyle B}
|
|
v
i
|
|
=
1
{\displaystyle ||v_{i}||=1}
x
≐
∑
i
=
1
q
⟨
x
,
v
i
⟩
v
i
{\displaystyle x\doteq \sum _{i=1}^{q}{\langle x,v_{i}\rangle }{}v_{i}}
Para este caso, puede calcularse:
|
|
x
|
|
2
=
⟨
x
,
x
⟩
=
⟨
∑
i
=
1
q
⟨
x
,
v
i
⟩
v
i
,
∑
i
=
1
q
⟨
x
,
v
i
⟩
v
i
⟩
=
⟨
⟨
v
1
,
x
⟩
v
1
,
⟨
v
1
,
x
⟩
v
1
⟩
+
⟨
⟨
v
2
,
x
⟩
v
2
,
⟨
v
2
,
x
⟩
v
2
⟩
+
⋯
+
⟨
⟨
v
q
,
x
⟩
v
q
,
⟨
v
q
,
x
⟩
v
q
⟩
{\displaystyle {\begin{aligned}||x||^{2}&=\langle x,x\rangle \\&=\langle \sum _{i=1}^{q}\langle x,v_{i}\rangle v_{i},\sum _{i=1}^{q}\langle x,v_{i}\rangle v_{i}\rangle \\&=\langle \langle v_{1},x\rangle v_{1},\langle v_{1},x\rangle v_{1}\rangle +\langle \langle v_{2},x\rangle v_{2},\langle v_{2},x\rangle v_{2}\rangle +\cdots +\langle \langle v_{q},x\rangle v_{q},\langle v_{q},x\rangle v_{q}\rangle \end{aligned}}}
Por dos de los axiomas del producto interno ,
(
x
,
z
y
)
=
z
¯
(
x
,
y
)
{\displaystyle (x,zy)={\bar {z}}(x,y)}
z
∈
K
{\displaystyle z\in \mathbb {K} }
(
x
,
y
)
=
(
y
,
x
)
¯
{\displaystyle (x,y)={\overline {(y,x)}}}
resulta
(
z
′
x
,
z
y
)
=
z
′
z
¯
(
y
,
x
)
¯
=
z
z
′
¯
(
y
,
x
)
¯
{\displaystyle (z'x,zy)={\overline {z'{\bar {z}}(y,x)}}=z{\bar {z'}}{\overline {(y,x)}}}
z
{\displaystyle z}
z
′
∈
K
{\displaystyle z'\in \mathbb {K} }
|
|
x
|
|
2
=
(
(
v
1
,
x
)
¯
(
v
1
,
x
)
(
v
1
,
v
1
)
)
+
(
(
v
2
,
x
)
¯
(
v
2
,
x
)
(
v
2
,
v
2
)
)
+
.
.
.
+
(
(
v
q
,
x
)
¯
(
v
q
,
x
)
(
v
q
,
v
q
)
)
{\displaystyle ||x||^{2}={\Bigl (}{\overline {(v_{1},x)}}(v_{1},x)(v_{1},v_{1}){\Bigr )}+{\Bigl (}{\overline {(v_{2},x)}}(v_{2},x)(v_{2},v_{2}){\Bigr )}+...+{\Bigl (}{\overline {(v_{q},x)}}(v_{q},x)(v_{q},v_{q}){\Bigr )}}
Como
(
v
i
,
v
i
)
=
|
|
v
i
|
|
2
{\displaystyle (v_{i},v_{i})=||v_{i}||^{2}}
B
{\displaystyle B}
⇒
|
|
v
i
|
|
=
1
{\displaystyle \Rightarrow ||v_{i}||=1}
Además, usando la propiedad de los número complejos ,
z
⋅
z
¯
=
|
z
|
2
{\displaystyle z\cdot {\bar {z}}=|z|^{2}}
z
∈
K
{\displaystyle z\in \mathbb {K} }
|
|
x
|
|
2
=
|
(
v
1
,
x
)
|
2
+
|
(
v
2
,
x
)
|
2
+
⋯
+
|
(
v
q
,
x
)
|
2
{\displaystyle ||x||^{2}=|(v_{1},x)|^{2}+|(v_{2},x)|^{2}+\cdots +|(v_{q},x)|^{2}}
quedando entonces la expresión
‖
x
‖
2
=
∑
i
=
1
q
|
(
v
i
,
x
)
|
2
{\displaystyle \|x\|^{2}=\sum _{i=1}^{q}\left|(v_{i},x)\right|^{2}}
editar
Datos: Q944238