Dada a base \alpha=\{(1,1,1),(-1,0,-1),(-1,2,3)\} de \mathbb{R}^{3}, utilize o processo de Gram-Schmidt para obter uma base ortonormal de \mathbb{R}^{3}.

Passo 1

O enunciado pede para aplicar o processo de Gram–Schmidt à base
\alpha=\{(1,1,1),\,(-1,0,-1),\,(-1,2,3)\}
de \mathbb{R}^3 e obter uma base ortonormal. Vamos usar os vetores
\mathbf{v}_1=(1,1,1), \mathbf{v}_2=(-1,0,-1), \mathbf{v}_3=(-1,2,3)
e gerar vetores \mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3 ortogonais, depois normalizar cada um.

Passo 2

Tomamos
\mathbf{u}_1=\mathbf{v}_1=(1,1,1).
O comprimento é
\|\mathbf{u}_1\|=\sqrt{1^2+1^2+1^2}=\sqrt{3},
logo o primeiro vetor unitário é
\mathbf{e}_1 = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} = \Bigl(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\Bigr).

Passo 3

Subtraímos de \mathbf{v}_2 sua projeção em \mathbf{e}_1:
\mathrm{proj}_{\mathbf{e}_1}(\mathbf{v}_2) = \bigl\langle \mathbf{v}_2,\mathbf{e}_1\bigr\rangle\,\mathbf{e}_1 = \frac{-1-0-1}{\sqrt{3}}\;\mathbf{e}_1 = -\frac{2}{\sqrt{3}}\;\mathbf{e}_1 = \Bigl(-\tfrac{2}{3},-\tfrac{2}{3},-\tfrac{2}{3}\Bigr).
Então
\mathbf{u}_2 = \mathbf{v}_2 - \mathrm{proj}_{\mathbf{e}_1}(\mathbf{v}_2) = (-1,0,-1) - \bigl(-\tfrac{2}{3},-\tfrac{2}{3},-\tfrac{2}{3}\bigr) = \bigl(-\tfrac{1}{3},\tfrac{2}{3},-\tfrac{1}{3}\bigr).
O comprimento é
\|\mathbf{u}_2\| = \sqrt{\bigl(-\tfrac{1}{3}\bigr)^2+\bigl(\tfrac{2}{3}\bigr)^2+\bigl(-\tfrac{1}{3}\bigr)^2} = \frac{\sqrt{6}}{3},
logo
\mathbf{e}_2 = \frac{\mathbf{u}_2}{\|\mathbf{u}_2\|} = \Bigl(-\tfrac{1}{\sqrt{6}},\,\tfrac{2}{\sqrt{6}},\,-\tfrac{1}{\sqrt{6}}\Bigr).

Passo 4

Para \mathbf{v}_3 subtraímos projeções em \mathbf{e}_1 e \mathbf{e}_2.
Proj. em \mathbf{e}_1:
\bigl\langle \mathbf{v}_3,\mathbf{e}_1\bigr\rangle\,\mathbf{e}_1 = \frac{-1+2+3}{\sqrt{3}}\;\mathbf{e}_1 = \frac{4}{\sqrt{3}}\;\mathbf{e}_1 = \Bigl(\tfrac{4}{3},\tfrac{4}{3},\tfrac{4}{3}\Bigr).
Proj. em \mathbf{e}_2:
\bigl\langle \mathbf{v}_3,\mathbf{e}_2\bigr\rangle\,\mathbf{e}_2 = \frac{(-1)(-1)+2\cdot2+3(-1)}{\sqrt{6}}\;\mathbf{e}_2 = \frac{2}{\sqrt{6}}\;\mathbf{e}_2 = \Bigl(-\tfrac{1}{3},\tfrac{2}{3},-\tfrac{1}{3}\Bigr).
Então
\mathbf{u}_3 = \mathbf{v}_3 - \Bigl(\tfrac{4}{3},\tfrac{4}{3},\tfrac{4}{3}\Bigr) - \Bigl(-\tfrac{1}{3},\tfrac{2}{3},-\tfrac{1}{3}\Bigr) = (-2,0,2).
O comprimento de \mathbf{u}_3 é \sqrt{(-2)^2+0^2+2^2}=\!2\sqrt{2}, portanto
\mathbf{e}_3 = \frac{\mathbf{u}_3}{\|\mathbf{u}_3\|} = \Bigl(-\tfrac{1}{\sqrt{2}},\,0,\,\tfrac{1}{\sqrt{2}}\Bigr).

Resposta

Uma base ortonormal obtida é
\mathbf{e}_1=\Bigl(\tfrac{1}{\sqrt{3}},\,\tfrac{1}{\sqrt{3}},\,\tfrac{1}{\sqrt{3}}\Bigr), \quad \mathbf{e}_2=\Bigl(-\tfrac{1}{\sqrt{6}},\,\tfrac{2}{\sqrt{6}},\,-\tfrac{1}{\sqrt{6}}\Bigr), \quad \mathbf{e}_3=\Bigl(-\tfrac{1}{\sqrt{2}},\,0,\,\tfrac{1}{\sqrt{2}}\Bigr).

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