(3.0) Seja B=\{(1,1,0) ;(1,0,1) ;(0,2,0)\} uma base do \mathbb{R}^{3}. Use o processo de GramSchmidt para achar uma base ortonormal \beta.

Passo 1

O enunciado dá a base
B = \{v_{1}=(1,1,0),\;v_{2}=(1,0,1),\;v_{3}=(0,2,0)\}
em \mathbb{R}^{3} e pede uma base ortonormal \beta obtida pelo processo de Gram–Schmidt. Para isso, vamos construir vetores ortogonais u_{1},u_{2},u_{3} a partir de v_{1},v_{2},v_{3} e então normalizá-los para obter e_{1},e_{2},e_{3}.

Passo 2

Definimos
u_{1} = v_{1} = (1,1,0).
Calculamos sua norma:
\|u_{1}\| = \sqrt{1^{2}+1^{2}+0^{2}} = \sqrt{2}.
Logo o primeiro vetor ortonormal é
e_{1} = \frac{u_{1}}{\|u_{1}\|} = \Bigl(\frac{1}{\sqrt{2}},\,\frac{1}{\sqrt{2}},\,0\Bigr).

Passo 3

Agora
u_{2} = v_{2} - \frac{\langle v_{2},u_{1}\rangle}{\langle u_{1},u_{1}\rangle}\,u_{1}.
Calculamos
\langle v_{2},u_{1}\rangle = 1 e \langle u_{1},u_{1}\rangle = 2, de onde
u_{2} = (1,0,1) - \tfrac12\,(1,1,0) = \Bigl(\tfrac12,\,-\tfrac12,\,1\Bigr).
A norma de u_{2} é
\|u_{2}\| = \sqrt{\tfrac14 + \tfrac14 + 1} = \sqrt{\tfrac32} = \frac{\sqrt{3}}{\sqrt{2}},
e assim
e_{2} = \frac{u_{2}}{\|u_{2}\|} = \Bigl(\frac{1}{\sqrt{6}},\,-\frac{1}{\sqrt{6}},\,\frac{2}{\sqrt{6}}\Bigr).

Passo 4

Finalmente
u_{3} = v_{3} - \frac{\langle v_{3},u_{1}\rangle}{\langle u_{1},u_{1}\rangle}\,u_{1} - \frac{\langle v_{3},u_{2}\rangle}{\langle u_{2},u_{2}\rangle}\,u_{2}.
Temos \langle v_{3},u_{1}\rangle = 2 e \langle v_{3},u_{2}\rangle = -1. Portanto
u_{3} = (0,2,0) - (1,1,0) - \bigl(-\tfrac23\bigr)\bigl(\tfrac12,-\tfrac12,1\bigr) = \Bigl(-\tfrac23,\;\tfrac23,\;\tfrac23\Bigr).
A norma de u_{3} é
\|u_{3}\| = \frac{2}{3}\,\sqrt{3} = \frac{2\sqrt{3}}{3},
logo
e_{3} = \frac{u_{3}}{\|u_{3}\|} = \Bigl(-\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\Bigr).

Resposta

A base ortonormal pedida é
\beta = \Bigl\{ \bigl(\tfrac1{\sqrt2},\tfrac1{\sqrt2},0\bigr),\; \bigl(\tfrac1{\sqrt6},-\tfrac1{\sqrt6},\tfrac2{\sqrt6}\bigr),\; \bigl(-\tfrac1{\sqrt3},\tfrac1{\sqrt3},\tfrac1{\sqrt3}\bigr) \Bigr\}.

Exercícios relacionados