Determine uma base ortonormal para o subespaço U de \mathbb{R}^{4} gerado pelos vetores u_{1}= (1,1,1,1), u_{2}=(1,2,4,5) e u_{3}=(1,-3,-4,-2), considerando o produto interno usual.

Passo 1

O enunciado pede uma base ortonormal de
\displaystyle U=\operatorname{span}\{u_{1},u_{2},u_{3}\}\subset\mathbb{R}^{4}
com o produto interno usual.
O processo que vamos usar é o de Gram–Schmidt para gerar vetores ortogonais a partir de u_{1},u_{2},u_{3} e depois normalizar cada um.

Passo 2

Começamos com
u_{1}=(1,1,1,1).
Calculamos a norma:
\|u_{1}\|=\sqrt{1^{2}+1^{2}+1^{2}+1^{2}}=\sqrt{4}=2.
Definimos então
e_{1}=\frac{u_{1}}{\|u_{1}\|}=\bigl(\tfrac12,\tfrac12,\tfrac12,\tfrac12\bigr).

Passo 3

Para obter o segundo vetor, projetamos u_{2}=(1,2,4,5) sobre e_{1}:
\langle u_{2},e_{1}\rangle =1\cdot\tfrac12+2\cdot\tfrac12+4\cdot\tfrac12+5\cdot\tfrac12=6.
Então
v_{2}=u_{2}-\langle u_{2},e_{1}\rangle\,e_{1} =(1,2,4,5)-6\bigl(\tfrac12,\tfrac12,\tfrac12,\tfrac12\bigr) =(-2,-1,1,2).
Calculamos sua norma:
\|v_{2}\|=\sqrt{(-2)^{2}+(-1)^{2}+1^{2}+2^{2}}=\sqrt{10},
e normalizamos:
e_{2}=\frac{v_{2}}{\|v_{2}\|}=\bigl(-\tfrac{2}{\sqrt{10}},-\tfrac{1}{\sqrt{10}},\tfrac{1}{\sqrt{10}},\tfrac{2}{\sqrt{10}}\bigr).

Passo 4

Para o terceiro vetor, primeiro removemos a componente em e_{1} de
u_{3}=(1,-3,-4,-2):
\langle u_{3},e_{1}\rangle =1\cdot\tfrac12+(-3)\cdot\tfrac12+(-4)\cdot\tfrac12+(-2)\cdot\tfrac12=-4,
u_{3}-\langle u_{3},e_{1}\rangle\,e_{1} =(1,-3,-4,-2)-(-4)\bigl(\tfrac12,\tfrac12,\tfrac12,\tfrac12\bigr) =(3,-1,-2,0).
Em seguida projetamos esse resultado sobre e_{2}. Chamando-o de w=(3,-1,-2,0), temos
\langle w,v_{2}\rangle =3\cdot(-2)+(-1)\cdot(-1)+(-2)\cdot1+0\cdot2=-7, \quad \|v_{2}\|^{2}=10,
\text{proj}_{v_{2}}w=\frac{\langle w,v_{2}\rangle}{\|v_{2}\|^{2}}\,v_{2} =\frac{-7}{10}(-2,-1,1,2) =(\tfrac{14}{10},\tfrac{7}{10},-\tfrac{7}{10},-\tfrac{14}{10}).
Então
v_{3} =w-\text{proj}_{v_{2}}w =(3,-1,-2,0)-\bigl(\tfrac{14}{10},\tfrac{7}{10},-\tfrac{7}{10},-\tfrac{14}{10}\bigr) =\bigl(\tfrac{16}{10},-\tfrac{17}{10},-\tfrac{13}{10},\tfrac{14}{10}\bigr).
Podemos reescrever
v_{3}=\tfrac{1}{10}(16,\,-17,\,-13,\,14), \quad \|v_{3}\| =\frac{1}{10}\sqrt{16^{2}+(-17)^{2}+(-13)^{2}+14^{2}} =\frac{\sqrt{910}}{10}.
Logo
e_{3}=\frac{v_{3}}{\|v_{3}\|} =\frac{\tfrac1{10}(16,-17,-13,14)}{\tfrac{\sqrt{910}}{10}} =\frac{(16,-17,-13,14)}{\sqrt{910}}.

Passo 5

Os três vetores e_{1},e_{2},e_{3} são mutuamente ortogonais e de norma 1, portanto formam uma base ortonormal de U.

Resposta

e_{1}=\bigl(\tfrac12,\tfrac12,\tfrac12,\tfrac12\bigr),\quad e_{2}=\bigl(-\tfrac{2}{\sqrt{10}},-\tfrac{1}{\sqrt{10}},\tfrac{1}{\sqrt{10}},\tfrac{2}{\sqrt{10}}\bigr),\quad e_{3}=\frac{1}{\sqrt{910}}\,(16,\,-17,\,-13,\,14).

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