Questão 5: Lembrando que:
Processo de Gram-Schmidt
Para converter uma base \left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \ldots, \mathbf{u}_{r}\right\} numa base ortogonal \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{r}\right\}, efetue as seguintes contas.
Passo 1. \mathbf{v}_{1}=\mathbf{u}_{1}
Passo 2. \quad \mathbf{v}_{2}=\mathbf{u}_{1}-\frac{\left\langle\mathbf{u}_{2}, \mathbf{v}_{1}\right\rangle}{\left\|\mathbf{v}_{1}\right\|^{2}} \mathbf{v}_{1}
Passo 3. \quad \mathbf{v}_{3}=\mathbf{u}_{3}-\frac{\left\langle\mathbf{u}_{3}, \mathbf{v}_{1}\right\rangle}{\left\|\mathbf{v}_{1}\right\|^{2}} \mathbf{v}_{1}-\frac{\left\langle\mathbf{u}_{3}, \mathbf{v}_{2}\right\rangle}{\left\|\mathbf{v}_{2}\right\|^{2}} \mathbf{v}_{2}
Passo 4. \quad \mathbf{v}_{4}=\mathbf{u}_{4}-\frac{\left\langle\mathbf{u}_{4}, \mathbf{v}_{1}\right\rangle}{\left\|\mathbf{v}_{1}\right\|^{2}} \mathbf{v}_{1}-\frac{\left\langle\mathbf{u}_{4}, \mathbf{v}_{2}\right\rangle}{\left\|\mathbf{v}_{2}\right\|^{2}} \mathbf{v}_{2}-\frac{\left\langle\mathbf{u}_{4}, \mathbf{v}_{3}\right\rangle}{\left\|\mathbf{v}_{3}\right\|^{2}} \mathbf{v}_{3}
(continue até r passos)
Passo opcional. Para converter a base ortogonal numa base ortonormal \left\{\mathbf{q}_{1}, \mathbf{q}_{2}, \ldots, \mathbf{q}_{r}\right\}, normalize os vetores da base ortogonal.
Usando o processo de Gram-Schmidt transforme os vetores de base u_{1}=(1,1,1), u_{2}=(0,1,1), u_{3}=(0,0,1) em uma base ortogonal \left\{v_{1}, v_{2}, v_{3}\right\} e, depois, normalize os vetores da base ortogonal para obter uma base ortonormal \left\{q_{1}, q_{2}, q_{3}\right\}.

Passo 1

O enunciado pede para aplicar o processo de Gram–Schmidt à base \mathbf{u}_1=(1,1,1), \mathbf{u}_2=(0,1,1), \mathbf{u}_3=(0,0,1)
para obter primeiro uma base ortogonal \{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\} e, em seguida, normalizar esses vetores para achar a base ortonormal \{\mathbf{q}_1,\mathbf{q}_2,\mathbf{q}_3\}. Vamos usar produto interno usual em \mathbb{R}^3 e a fórmula
\mathbf{v}_k=\mathbf{u}_k-\sum_{j=1}^{k-1}\frac{\langle\mathbf{u}_k,\mathbf{v}_j\rangle}{\|\mathbf{v}_j\|^2}\,\mathbf{v}_j.

Passo 2

Tomamos
\mathbf{v}_1=\mathbf{u}_1=(1,1,1).

Passo 3

Para \mathbf{v}_2 calculamos \langle\mathbf{u}_2,\mathbf{v}_1\rangle=0\cdot1+1\cdot1+1\cdot1=2
e \|\mathbf{v}_1\|^2=1^2+1^2+1^2=3.
Logo
\mathbf{v}_2 =\mathbf{u}_2-\frac{2}{3}\mathbf{v}_1 =(0,1,1)-\bigl(\tfrac{2}{3},\tfrac{2}{3},\tfrac{2}{3}\bigr) =\bigl(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{1}{3}\bigr).

Passo 4

Para \mathbf{v}_3 achamos \langle\mathbf{u}_3,\mathbf{v}_1\rangle=1, \|\mathbf{v}_1\|^2=3, \langle\mathbf{u}_3,\mathbf{v}_2\rangle=0\cdot(-\tfrac{2}{3})+0\cdot\tfrac{1}{3}+1\cdot\tfrac{1}{3}=\tfrac{1}{3}, \|\mathbf{v}_2\|^2=(-\tfrac{2}{3})^2+(\tfrac{1}{3})^2+(\tfrac{1}{3})^2=\tfrac{6}{9}=\tfrac{2}{3}.
Assim
\mathbf{v}_3 =\mathbf{u}_3 -\frac{1}{3}\mathbf{v}_1 -\frac{\tfrac{1}{3}}{\tfrac{2}{3}}\mathbf{v}_2 =(0,0,1)-\bigl(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}\bigr) -\tfrac12\bigl(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{1}{3}\bigr) =\bigl(0,-\tfrac12,\tfrac12\bigr).

Passo 5

A base ortogonal obtida é
\mathbf{v}_1=(1,1,1),\quad \mathbf{v}_2=\bigl(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{1}{3}\bigr),\quad \mathbf{v}_3=\bigl(0,-\tfrac12,\tfrac12\bigr).

Passo 6

Para normalizar cada \mathbf{v}_j calculamos \|\mathbf{v}_1\|=\sqrt{3}, \|\mathbf{v}_2\|=\sqrt{\tfrac{2}{3}}, \|\mathbf{v}_3\|=\tfrac{\sqrt2}{2}.
Logo a base ortonormal é
\mathbf{q}_1=\frac{\mathbf{v}_1}{\sqrt{3}} =\frac{1}{\sqrt{3}}(1,1,1),\quad \mathbf{q}_2=\frac{\mathbf{v}_2}{\sqrt{\tfrac{2}{3}}} =\frac{1}{\sqrt6}(-2,1,1),\quad \mathbf{q}_3=\frac{\mathbf{v}_3}{\tfrac{\sqrt2}{2}} =\frac{1}{\sqrt2}(0,-1,1).

Resposta

v_{1}=(1,1,1)
v_{2}=(−2/3,1/3,1/3)
v_{3}=(0,−1/2,1/2)
q_{1}=(1/√3,1/√3,1/√3)
q_{2}=(−2/√6,1/√6,1/√6)
q_{3}=(0,−1/√2,1/√2)

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