Prove que
‖
x
‖
≤
∑
i
=
1
n
|
x
i
|
{\displaystyle \left\|x\right\|\leq \sum _{i=1}^{n}|x_{i}|}
Devemos mostrar primeiro que
∑
i
=
1
n
x
i
2
≤
∑
i
=
1
n
x
i
2
{\displaystyle {\sqrt {\sum _{i=1}^{n}x_{i}^{2}}}\leq \sum _{i=1}^{n}{\sqrt {x_{i}^{2}}}}
Seja
(
a
1
+
a
2
)
2
=
a
1
+
a
2
+
2
a
1
a
2
⇒
a
1
+
a
2
=
a
1
+
a
2
+
2
a
1
a
2
≥
a
1
+
a
2
,
∀
a
1
,
a
2
∈
R
+
{\displaystyle ({\sqrt {a_{1}}}+{\sqrt {a_{2}}})^{2}=a_{1}+a_{2}+2{\sqrt {a_{1}a_{2}}}\Rightarrow {\sqrt {a_{1}}}+{\sqrt {a_{2}}}={\sqrt {a_{1}+a_{2}+2{\sqrt {a_{1}a_{2}}}}}\geq {\sqrt {a_{1}+a_{2}}},\forall a_{1},a_{2}\in \mathbb {R} ^{+}}
Suponha ser verdade que
a
1
+
a
2
+
.
.
.
+
a
n
≥
a
1
+
a
2
+
.
.
.
+
a
n
,
∀
a
1
,
a
2
,
.
.
.
,
a
n
∈
R
+
{\displaystyle {\sqrt {a_{1}}}+{\sqrt {a_{2}}}+...+{\sqrt {a_{n}}}\geq {\sqrt {a_{1}+a_{2}+...+a_{n}}},\forall a_{1},a_{2},...,a_{n}\in \mathbb {R} ^{+}}
Tome
a
1
+
a
2
+
.
.
.
+
a
n
+
a
n
+
1
≥
a
1
+
a
2
+
.
.
.
+
a
n
+
a
n
+
1
≤
a
1
+
a
2
+
.
.
.
+
a
n
+
a
n
+
1
,
∀
a
1
,
a
2
,
.
.
.
,
a
n
,
a
n
+
1
∈
R
+
{\displaystyle {\sqrt {a_{1}}}+{\sqrt {a_{2}}}+...+{\sqrt {a_{n}}}+{\sqrt {a_{n+1}}}\geq {\sqrt {a_{1}+a_{2}+...+a_{n}}}+{\sqrt {a_{n+1}}}\leq {\sqrt {a_{1}+a_{2}+...+a_{n}+a_{n+1}}},\forall a_{1},a_{2},...,a_{n},a_{n+1}\in \mathbb {R} ^{+}}
‖
x
‖
=
∑
i
=
1
n
x
i
2
≤
∑
i
=
1
n
x
i
2
≤
∑
i
=
1
n
|
x
i
|
{\displaystyle \left\|x\right\|={\sqrt {\sum _{i=1}^{n}x_{i}^{2}}}\leq \sum _{i=1}^{n}{\sqrt {x_{i}^{2}}}\leq \sum _{i=1}^{n}|x_{i}|}
Prove que
‖
x
−
y
‖
≤
‖
x
‖
+
‖
y
‖
{\displaystyle \left\|x-y\right\|\leq \left\|x\right\|+\left\|y\right\|}
Verificar que
‖
x
−
y
‖
≤
‖
x
‖
+
‖
y
‖
,
∀
x
,
y
∈
R
n
{\displaystyle \left\|x-y\right\|\leq \left\|x\right\|+\left\|y\right\|,\forall x,y\in \mathbb {R} ^{n}}
Tome
‖
x
−
y
‖
2
=
⟨
x
−
y
,
x
−
y
⟩
=
⟨
x
,
x
⟩
+
⟨
x
,
−
y
⟩
+
⟨
−
y
,
x
⟩
+
⟨
−
y
,
−
y
⟩
=
‖
x
‖
2
−
2
⟨
x
,
y
⟩
+
‖
y
‖
2
≤
‖
x
‖
2
+
2
‖
x
‖
‖
y
‖
+
‖
y
‖
2
=
{\displaystyle \left\|x-y\right\|^{2}=\langle x-y,x-y\rangle =\langle x,x\rangle +\langle x,-y\rangle +\langle -y,x\rangle +\langle -y,-y\rangle =\left\|x\right\|^{2}-2\langle x,y\rangle +\left\|y\right\|^{2}\leq \left\|x\right\|^{2}+2\left\|x\right\|\left\|y\right\|+\left\|y\right\|^{2}=}
=
(
‖
x
‖
+
‖
y
‖
)
2
⇒
‖
x
−
y
‖
≤
‖
x
‖
+
‖
y
‖
{\displaystyle =(\left\|x\right\|+\left\|y\right\|)^{2}\Rightarrow \left\|x-y\right\|\leq \left\|x\right\|+\left\|y\right\|}
Verificar quando é válido que
‖
x
−
y
‖
=
‖
x
‖
+
‖
y
‖
{\displaystyle \left\|x-y\right\|=\left\|x\right\|+\left\|y\right\|}
Tome
y
=
−
c
x
,
c
∈
R
+
,
‖
x
−
y
‖
=
‖
x
−
(
−
c
x
)
‖
=
‖
x
+
c
x
‖
=
‖
(
1
+
c
)
x
‖
=
|
c
+
1
|
‖
x
‖
=
|
c
|
‖
x
‖
+
‖
x
‖
=
‖
c
x
‖
+
‖
x
‖
=
{\displaystyle y=-cx,c\in \mathbb {R} ^{+},\left\|x-y\right\|=\left\|x-(-cx)\right\|=\left\|x+cx\right\|=\left\|(1+c)x\right\|=|c+1|\left\|x\right\|=|c|\left\|x\right\|+\left\|x\right\|=\left\|cx\right\|+\left\|x\right\|=}
=
‖
−
y
‖
+
‖
x
‖
=
‖
y
‖
+
‖
x
‖
{\displaystyle =\left\|-y\right\|+\left\|x\right\|=\left\|y\right\|+\left\|x\right\|}
|
‖
x
‖
−
‖
y
‖
|
≤
‖
x
−
y
‖
,
∀
x
,
y
∈
R
n
{\displaystyle |\left\|x\right\|-\left\|y\right\||\leq \left\|x-y\right\|,\forall x,y\in \mathbb {R} ^{n}}
‖
x
−
y
+
y
‖
≤
‖
x
−
y
‖
+
‖
y
‖
⇒
‖
x
‖
−
‖
y
‖
≤
‖
x
−
y
‖
,
∀
x
,
y
∈
R
n
{\displaystyle \left\|x-y+y\right\|\leq \left\|x-y\right\|+\left\|y\right\|\Rightarrow \left\|x\right\|-\left\|y\right\|\leq \left\|x-y\right\|,\forall x,y\in \mathbb {R} ^{n}}
‖
y
‖
−
‖
x
−
y
‖
≤
‖
y
−
(
y
−
x
)
‖
⇒
−
‖
x
−
y
‖
≤
‖
x
‖
−
‖
y
‖
,
∀
x
,
y
∈
R
n
{\displaystyle \left\|y\right\|-\left\|x-y\right\|\leq \left\|y-(y-x)\right\|\Rightarrow -\left\|x-y\right\|\leq \left\|x\right\|-\left\|y\right\|,\forall x,y\in \mathbb {R} ^{n}}
Uma transformação linear
T
:
R
n
→
R
n
{\displaystyle T:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}
‖
T
(
x
)
‖
=
‖
x
‖
{\displaystyle \left\|T(x)\right\|=\left\|x\right\|}
⟨
T
(
x
)
,
T
(
y
)
⟩
=
⟨
x
,
y
⟩
{\displaystyle \langle T(x),T(y)\rangle =\langle x,y\rangle }
Prove que T preserva a norma se, e somente, se T preserva o produto interno.
Prove que uma tal transformação linear T é injetiva e a
T
−
1
{\displaystyle T^{-1}\;}
Se T preserva a norma, então T preserva o produto interno.
‖
T
(
x
+
y
)
‖
2
=
‖
x
+
y
‖
2
⇒
{\displaystyle \left\|T(x+y)\right\|^{2}=\left\|x+y\right\|^{2}\Rightarrow }
⇒
‖
T
(
x
)
+
T
(
y
)
‖
2
=
‖
x
+
y
‖
2
⇒
{\displaystyle \Rightarrow \left\|T(x)+T(y)\right\|^{2}=\left\|x+y\right\|^{2}\Rightarrow }
⇒
⟨
T
(
x
)
+
T
(
y
)
,
T
(
x
)
+
T
(
y
)
⟩
=
⟨
x
+
y
,
x
+
y
⟩
⇒
{\displaystyle \Rightarrow \langle T(x)+T(y),T(x)+T(y)\rangle =\langle x+y,x+y\rangle \Rightarrow }
⇒
⟨
T
(
x
)
,
T
(
x
)
⟩
+
2
⟨
T
(
x
)
,
T
(
y
)
⟩
+
⟨
T
(
y
)
,
T
(
y
)
⟩
=
⟨
x
,
x
⟩
+
2
⟨
x
,
y
⟩
+
⟨
y
,
y
⟩
⇒
{\displaystyle \Rightarrow \langle T(x),T(x)\rangle +2\langle T(x),T(y)\rangle +\langle T(y),T(y)\rangle =\langle x,x\rangle +2\langle x,y\rangle +\langle y,y\rangle \Rightarrow }
⇒
‖
T
(
x
)
‖
2
+
2
⟨
T
(
x
)
,
T
(
y
)
⟩
+
‖
T
(
y
)
‖
2
=
‖
x
‖
2
+
2
⟨
x
,
y
⟩
+
‖
y
‖
2
{\displaystyle \Rightarrow \left\|T(x)\right\|^{2}+2\langle T(x),T(y)\rangle +\left\|T(y)\right\|^{2}=\left\|x\right\|^{2}+2\langle x,y\rangle +\left\|y\right\|^{2}}
Como T preserva a norma,
‖
T
(
x
)
‖
2
+
‖
T
(
y
)
‖
2
=
‖
x
‖
2
+
‖
y
‖
2
{\displaystyle \left\|T(x)\right\|^{2}+\left\|T(y)\right\|^{2}=\left\|x\right\|^{2}+\left\|y\right\|^{2}}
2
⟨
T
(
x
)
,
T
(
y
)
⟩
=
2
⟨
x
,
y
⟩
⇒
{\displaystyle 2\langle T(x),T(y)\rangle =2\langle x,y\rangle \Rightarrow }
⇒
⟨
T
(
x
)
,
T
(
y
)
⟩
=
⟨
x
,
y
⟩
{\displaystyle \Rightarrow \langle T(x),T(y)\rangle =\langle x,y\rangle }
Se T preserva o produto interno, então T preserva a norma.
⟨
T
(
x
)
,
T
(
x
)
⟩
=
⟨
x
,
x
⟩
⇒
{\displaystyle \langle T(x),T(x)\rangle =\langle x,x\rangle \Rightarrow }
⇒
‖
T
(
x
)
‖
2
=
‖
x
‖
2
⇒
{\displaystyle \Rightarrow \left\|T(x)\right\|^{2}=\left\|x\right\|^{2}\Rightarrow }
⇒
‖
T
(
x
)
‖
=
‖
x
‖
{\displaystyle \Rightarrow \left\|T(x)\right\|=\left\|x\right\|}
T é injetiva
Seja
a
,
b
∈
R
n
{\displaystyle a,b\in \mathbb {R} ^{n}}
T
(
a
)
=
T
(
b
)
⇒
T
(
a
)
−
T
(
b
)
=
0
⇒
T
(
a
−
b
)
=
0
{\displaystyle T(a)=T(b)\Rightarrow T(a)-T(b)=0\Rightarrow T(a-b)=0}
Mas
0
=
‖
0
‖
=
‖
T
(
a
−
b
)
‖
=
‖
a
−
b
‖
⇒
a
−
b
=
0
⇒
a
=
b
{\displaystyle 0=\left\|0\right\|=\left\|T(a-b)\right\|=\left\|a-b\right\|\Rightarrow a-b=0\Rightarrow a=b}
T
−
1
{\displaystyle T^{-1}}
Seja
c
∈
R
,
x
=
T
(
a
)
,
y
=
T
(
b
)
∈
R
n
,
T
(
c
a
+
b
)
=
c
T
(
a
)
+
T
(
b
)
=
c
x
+
y
⇒
T
−
1
(
c
x
+
y
)
=
c
a
+
b
=
c
T
−
1
(
x
)
+
T
−
1
(
y
)
{\displaystyle c\in \mathbb {R} ,x=T(a),y=T(b)\in \mathbb {R} ^{n},T(ca+b)=cT(a)+T(b)=cx+y\Rightarrow T^{-1}(cx+y)=ca+b=cT^{-1}(x)+T^{-1}(y)}
T
−
1
{\displaystyle T^{-1}}
Seja
a
=
T
(
x
)
,
‖
T
(
x
)
‖
=
‖
T
−
1
(
T
(
x
)
)
‖
⇒
‖
T
−
1
(
a
)
‖
=
‖
a
‖
,
∀
a
∈
R
n
{\displaystyle a=T(x),\left\|T(x)\right\|=\left\|T^{-1}(T(x))\right\|\Rightarrow \left\|T^{-1}(a)\right\|=\left\|a\right\|,\forall a\in \mathbb {R} ^{n}}
T
−
1
{\displaystyle T^{-1}}
Seja
c
=
T
(
a
)
,
e
=
T
(
b
)
∈
R
n
{\displaystyle c=T(a),e=T(b)\in \mathbb {R} ^{n}}
T
−
1
(
c
)
=
T
−
1
(
e
)
⇒
T
(
c
)
−
1
−
T
(
e
)
−
1
=
0
⇒
T
−
1
(
c
−
e
)
=
0
{\displaystyle T^{-1}(c)=T^{-1}(e)\Rightarrow T(c)^{-1}-T(e)^{-1}=0\Rightarrow T^{-1}(c-e)=0}
T
−
1
{\displaystyle T^{-1}\;}
Mas
0
=
‖
0
‖
=
‖
T
−
1
(
c
−
e
)
‖
=
‖
c
−
e
‖
⇒
c
−
e
=
0
⇒
c
=
e
{\displaystyle 0=\left\|0\right\|=\left\|T^{-1}(c-e)\right\|=\left\|c-e\right\|\Rightarrow c-e=0\Rightarrow c=e}
Se
x
,
y
∈
R
n
{\displaystyle x,y\in \mathbb {R} ^{n}}
∠
(
x
,
y
)
{\displaystyle \angle (x,y)}
c
o
s
−
1
⟨
x
,
y
⟩
‖
x
‖
⋅
‖
y
‖
{\displaystyle cos^{-1}{\langle x,y\rangle \over \left\|x\right\|\cdot \left\|y\right\|}}
⟨
x
,
y
⟩
≤
‖
x
‖
⋅
‖
y
‖
{\displaystyle \langle x,y\rangle \leq \left\|x\right\|\cdot \left\|y\right\|}
x
,
y
≠
0
{\displaystyle x,y\not =0}
∠
(
T
x
,
T
y
)
=
∠
(
x
,
y
)
{\displaystyle \angle (Tx,Ty)=\angle (x,y)}
Prove que se T preserva a norma, então T preserva o ângulo.
Se existe uma base
{
x
1
,
.
.
.
,
x
n
}
{\displaystyle \{x_{1},...,x_{n}\}\;}
R
n
{\displaystyle \mathbb {R} ^{n}}
{
k
1
,
.
.
.
,
k
n
}
{\displaystyle \{k_{1},...,k_{n}\}\;}
T
x
i
=
k
i
x
i
{\displaystyle Tx_{i}=k_{i}x_{i}\;}
|
k
i
|
{\displaystyle |k_{i}|\;}
Quais todos os
T
:
R
n
→
R
n
{\displaystyle T:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}
Mostrar que T preserva o ângulo: T preserva a norma, logo preserva o produto interno, isto é,
‖
T
(
x
)
‖
=
‖
x
‖
e
⟨
T
(
x
)
,
T
(
y
)
⟩
=
⟨
x
,
y
⟩
,
∀
x
,
y
∈
R
n
{\displaystyle \left\|T(x)\right\|=\left\|x\right\|\;e\;\langle T(x),T(y)\rangle =\langle x,y\rangle ,\forall x,y\in \mathbb {R} ^{n}}
Assim
∠
(
T
x
,
T
y
)
=
c
o
s
−
1
⟨
T
(
x
)
,
T
(
y
)
⟩
‖
T
(
x
)
‖
⋅
‖
T
(
y
)
‖
=
c
o
s
−
1
⟨
x
,
y
⟩
‖
x
‖
⋅
‖
y
‖
=
∠
(
x
,
y
)
{\displaystyle \angle (Tx,Ty)=cos^{-1}{\langle T(x),T(y)\rangle \over \left\|T(x)\right\|\cdot \left\|T(y)\right\|}=cos^{-1}{\langle x,y\rangle \over \left\|x\right\|\cdot \left\|y\right\|}=\angle (x,y)}
No exercício anterior provamos que toda T linear é injetiva.
T preserva o ângulo. Mostrar que
|
k
i
|
{\displaystyle |k_{i}|\;}
i
=
1
,
.
.
.
,
n
{\displaystyle i=1,...,n\;}
Seja
y
=
y
1
x
1
+
.
.
.
+
y
n
x
n
⇒
T
(
y
)
=
T
(
y
1
x
1
+
.
.
.
+
y
n
x
n
)
=
y
1
T
(
x
1
)
+
.
.
.
+
y
n
T
(
x
n
)
=
y
1
k
1
x
1
+
.
.
.
+
y
n
k
n
x
n
=
(
y
1
k
1
,
.
.
.
,
y
n
k
n
)
{\displaystyle y=y_{1}x_{1}+...+y_{n}x_{n}\Rightarrow T(y)=T(y_{1}x_{1}+...+y_{n}x_{n})=y_{1}T(x_{1})+...+y_{n}T(x_{n})=y_{1}k_{1}x_{1}+...+y_{n}k_{n}x_{n}=(y_{1}k_{1},...,y_{n}k_{n})}
Mas T preserva o ângulo, então
∠
(
T
a
,
T
b
)
=
∠
(
a
,
b
)
⇒
⟨
T
(
a
)
,
T
(
b
)
⟩
‖
T
(
a
)
‖
⋅
‖
T
(
b
)
‖
=
⟨
a
,
b
⟩
‖
a
‖
⋅
‖
b
‖
⇒
{\displaystyle \angle (Ta,Tb)=\angle (a,b)\Rightarrow {\langle T(a),T(b)\rangle \over \left\|T(a)\right\|\cdot \left\|T(b)\right\|}={\langle a,b\rangle \over \left\|a\right\|\cdot \left\|b\right\|}\Rightarrow }
⇒
a
1
b
1
k
1
2
+
.
.
.
+
a
n
b
n
k
n
2
(
a
1
2
k
1
2
+
.
.
.
+
a
n
2
k
n
2
)
(
b
1
2
k
1
2
+
.
.
.
b
n
2
k
n
2
)
=
a
1
b
1
+
.
.
.
+
a
n
b
n
(
a
1
2
+
.
.
.
+
a
n
2
)
(
b
1
2
+
.
.
.
+
b
n
2
)
⇒
{\displaystyle \Rightarrow {a_{1}b_{1}k_{1}^{2}+...+a_{n}b_{n}k_{n}^{2} \over {\sqrt {(a_{1}^{2}k_{1}^{2}+...+a_{n}^{2}k_{n}^{2})(b_{1}^{2}k_{1}^{2}+...b_{n}^{2}k_{n}^{2})}}}={a_{1}b_{1}+...+a_{n}b_{n} \over {\sqrt {(a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})}}}\Rightarrow }
⇒
(
a
1
b
1
k
1
2
+
.
.
.
+
a
n
b
n
k
n
2
)
2
(
a
1
2
k
1
2
+
.
.
.
+
a
n
2
k
n
2
)
(
b
1
2
k
1
2
+
.
.
.
b
n
2
k
n
2
)
=
(
a
1
b
1
+
.
.
.
+
a
n
b
n
)
2
(
a
1
2
+
.
.
.
+
a
n
2
)
(
b
1
2
+
.
.
.
+
b
n
2
)
{\displaystyle \Rightarrow {(a_{1}b_{1}k_{1}^{2}+...+a_{n}b_{n}k_{n}^{2})^{2} \over (a_{1}^{2}k_{1}^{2}+...+a_{n}^{2}k_{n}^{2})(b_{1}^{2}k_{1}^{2}+...b_{n}^{2}k_{n}^{2})}={(a_{1}b_{1}+...+a_{n}b_{n})^{2} \over (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})}}
k
l
=
m
i
n
{
k
i
}
,
k
p
=
m
i
n
{
k
i
}
,
1
≤
i
≤
n
{\displaystyle k_{l}=min\{k_{i}\},k_{p}=min\{k_{i}\},1\leq i\leq n\;}
⇒
(
a
1
b
1
k
1
2
+
.
.
.
+
a
n
b
n
k
n
2
)
2
(
a
1
2
k
1
2
+
.
.
.
+
a
n
2
k
n
2
)
(
b
1
2
k
1
2
+
.
.
.
b
n
2
k
n
2
)
=
(
a
1
b
1
+
.
.
.
+
a
n
b
n
)
2
(
a
1
2
+
.
.
.
+
a
n
2
)
(
b
1
2
+
.
.
.
+
b
n
2
)
{\displaystyle \Rightarrow {(a_{1}b_{1}k_{1}^{2}+...+a_{n}b_{n}k_{n}^{2})^{2} \over (a_{1}^{2}k_{1}^{2}+...+a_{n}^{2}k_{n}^{2})(b_{1}^{2}k_{1}^{2}+...b_{n}^{2}k_{n}^{2})}={(a_{1}b_{1}+...+a_{n}b_{n})^{2} \over (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})}}
