(3,0 pontos) Dada uma base do \mathbb{R}^{3}, B=\{(1,1,1),(1,-1,1),(1,-1,-1)\}. Determine uma base ortonormal U obtida de B pelo método de Gram-Schmidt

Passo 1

Temos a base B = \{v_1,v_2,v_3\} de \mathbb{R}^3 com
v_1 = (1,1,1),\quad v_2 = (1,-1,1),\quad v_3 = (1,-1,-1).
Vamos aplicar o processo de Gram–Schmidt para obter vetores ortogonais \{w_1,w_2,w_3\} e depois normalizá-los, gerando uma base ortonormal U=\{u_1,u_2,u_3\}.

Passo 2

Definimos
w_1 = v_1 = (1,1,1),
e normalizamos para obter
u_1 = \frac{w_1}{\|w_1\|} \quad\text{com}\quad \|w_1\| = \sqrt{1^2+1^2+1^2} = \sqrt{3}.
Logo
u_1 = \frac{1}{\sqrt{3}}\,(1,1,1).

Passo 3

Projetamos v_2 sobre u_1 e subtraímos de v_2:
\mathrm{proj}_{u_1}(v_2) = \bigl\langle v_2,u_1\bigr\rangle\,u_1 = \frac{1\cdot1 +(-1)\cdot1 + 1\cdot1}{\sqrt{3}}\,u_1 = \frac{1}{\sqrt{3}}\,u_1 = \frac{1}{3}(1,1,1).
Então
w_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = (1,-1,1) - \frac{1}{3}(1,1,1) = \Bigl(\tfrac{2}{3},-\tfrac{4}{3},\tfrac{2}{3}\Bigr).
A norma de w_2 é
\|w_2\| = \sqrt{\Bigl(\tfrac{2}{3}\Bigr)^2 + \Bigl(-\tfrac{4}{3}\Bigr)^2 + \Bigl(\tfrac{2}{3}\Bigr)^2} = \frac{\sqrt{24}}{3} = \frac{2\sqrt{6}}{3}.
Logo
u_2 = \frac{w_2}{\|w_2\|} = \frac{(2/3,\,-4/3,\,2/3)}{2\sqrt{6}/3} = \frac{1}{\sqrt{6}}\,(1,-2,1).

Passo 4

Projetamos v_3 sobre u_1 e u_2:
\mathrm{proj}_{u_1}(v_3) = \frac{\langle v_3,u_1\rangle}{1}\,u_1 = \frac{1-1-1}{\sqrt{3}}\,u_1 = -\frac{1}{3}(1,1,1),
\mathrm{proj}_{u_2}(v_3) = \frac{\langle v_3,u_2\rangle}{1}\,u_2 = \frac{1\cdot1 +(-1)\cdot(-2) +(-1)\cdot1}{\sqrt{6}}\,u_2 = \frac{2}{\sqrt{6}}\,u_2 = \frac{1}{3}(1,-2,1).
Então
w_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = (1,-1,-1) - \bigl(-\tfrac13,\,-\tfrac13,\,-\tfrac13\bigr) - \bigl(\tfrac13,-\tfrac23,\tfrac13\bigr) = (1,0,-1).
A norma de w_3 é \|w_3\|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}, logo
u_3 = \frac{w_3}{\|w_3\|} = \frac{1}{\sqrt{2}}\,(1,0,-1).

Passo 5

Reunindo os vetores normalizados, obtemos a base ortonormal
U = \bigl\{u_1,\,u_2,\,u_3\bigr\} = \left\{ \frac{1}{\sqrt{3}}(1,1,1),\; \frac{1}{\sqrt{6}}(1,-2,1),\; \frac{1}{\sqrt{2}}(1,0,-1) \right\}.

Resposta

\displaystyle U = \Bigl\{\tfrac{1}{\sqrt{3}}(1,1,1),\;\tfrac{1}{\sqrt{6}}(1,-2,1),\;\tfrac{1}{\sqrt{2}}(1,0,-1)\Bigr\}.

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