Questão 3: Determinar uma base Ortonormal do subespaço vetorial W do \mathrm{R}^{4} utilizando o processo de Gram-Schmidt:
W=[(1,1,0,0),(0,1,2,0),(0,0,3,4)] .
\mathrm{R}:\left\{\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0,0\right),\left(-\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}, 0\right),\left(-\frac{2}{3 \sqrt{17}},-\frac{2}{3 \sqrt{17}}, \frac{1}{3 \sqrt{17}}, \frac{4}{\sqrt{17}}\right)\right\}

Passo 1

O enunciado pede para encontrarmos uma base ortonormal do subespaço \,W=\mathrm{span}\{(1,1,0,0),(0,1,2,0),(0,0,3,4)\}\subset\mathbb R^4
usando o processo de Gram–Schmidt. Precisaremos gerar vetores mutuamente ortogonais a partir dos vetores iniciais e depois normalizá-los.

Passo 2

Chamemos
v_{1}=(1,1,0,0).
Este será nosso primeiro vetor não-normalizado u_{1}=v_{1}. Agora normalizamos u_{1} para obter e_{1}:
\|u_{1}\|=\sqrt{1^{2}+1^{2}}=\sqrt{2}, \qquad e_{1}=\frac{u_{1}}{\|u_{1}\|} =\Bigl(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\Bigr).

Passo 3

Tomamos
v_{2}=(0,1,2,0).
Projetamos v_{2} sobre e_{1} e subtraímos para obter
\langle v_{2},e_{1}\rangle =\frac{1}{\sqrt{2}}, \qquad u_{2} =v_{2}-\langle v_{2},e_{1}\rangle e_{1} =(0,1,2,0)-\frac{1}{\sqrt{2}}\Bigl(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0,0\Bigr) =\bigl(-\tfrac12,\tfrac12,2,0\bigr).
Calculamos \|u_{2}\|:
\|u_{2}\| =\sqrt{\bigl(-\tfrac12\bigr)^{2}+\bigl(\tfrac12\bigr)^{2}+2^{2}} =\frac{3}{\sqrt{2}},
e normalizamos para obter
e_{2} =\frac{u_{2}}{\|u_{2}\|} =\Bigl(-\tfrac{\sqrt{2}}{6},\,\tfrac{\sqrt{2}}{6},\,\tfrac{2\sqrt{2}}{3},\,0\Bigr) =\Bigl(-\tfrac{1}{3\sqrt{2}},\,\tfrac{1}{3\sqrt{2}},\,\tfrac{4}{3\sqrt{2}},\,0\Bigr).

Passo 4

Finalmente, seja
v_{3}=(0,0,3,4).
Subtraímos as projeções sobre e_{1} e e_{2}:
\langle v_{3},e_{1}\rangle=0, \qquad \langle v_{3},e_{2}\rangle =3\cdot\frac{4}{3\sqrt{2}}=\frac{4}{\sqrt{2}}.
Logo
u_{3} =v_{3}-\frac{4}{\sqrt{2}}\,e_{2} =(0,0,3,4)-\frac{4}{\sqrt{2}}\Bigl(-\tfrac{1}{3\sqrt{2}},\tfrac{1}{3\sqrt{2}},\tfrac{4}{3\sqrt{2}},0\Bigr) =\Bigl(\tfrac{2}{3},-\tfrac{2}{3},\tfrac{1}{3},4\Bigr).
Calculamos \|u_{3}\|:
\|u_{3}\| =\sqrt{\bigl(\tfrac{2}{3}\bigr)^{2}+\bigl(-\tfrac{2}{3}\bigr)^{2}+\bigl(\tfrac{1}{3}\bigr)^{2}+4^{2}} =\sqrt{1+16}=\sqrt{17},
e normalizamos:
e_{3} =\frac{u_{3}}{\|u_{3}\|} =\Bigl(\tfrac{2}{3\sqrt{17}},-\tfrac{2}{3\sqrt{17}},\tfrac{1}{3\sqrt{17}},\tfrac{4}{\sqrt{17}}\Bigr).

Resposta

Uma base ortonormal de W é
\left\{ \bigl(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0,0\bigr), \bigl(-\tfrac{1}{3\sqrt{2}},\tfrac{1}{3\sqrt{2}},\tfrac{4}{3\sqrt{2}},0\bigr), \bigl(\tfrac{2}{3\sqrt{17}},-\tfrac{2}{3\sqrt{17}},\tfrac{1}{3\sqrt{17}},\tfrac{4}{\sqrt{17}}\bigr) \right\}.

Exercícios relacionados