Questão 2 Considere a base B de \mathbb{R}^{3} dada por B=\{(0,0,1),(2,0,1),(-1,1,2)\}.
(a) Determine uma base ortonormal para \mathbb{R}^{3} a partir da base ORDENADA B usando o processo de ortonormalização de Gram-Schmidt.
(b) Determine as coordenadas do vetor v=(-2,1,3) na base encontrada no item (a).

Passo 1

O enunciado dá a base ordenada \displaystyle B=\{b_{1},b_{2},b_{3}\}\subset\mathbb{R}^{3} com
b_{1}=(0,0,1),\quad b_{2}=(2,0,1),\quad b_{3}=(-1,1,2),
e pede:
(a) obter uma base ortonormal \{e_{1},e_{2},e_{3}\} via Gram–Schmidt;
(b) encontrar as coordenadas de v=(-2,1,3) nessa base.

Passo 2

Aplicamos Gram–Schmidt em b_{1}. Definimos
u_{1}=b_{1}=(0,0,1),
e
e_{1}=\frac{u_{1}}{\|u_{1}\|} =\frac{(0,0,1)}{\sqrt{0^{2}+0^{2}+1^{2}}} =(0,0,1).

Passo 3

Agora orthogonalizamos b_{2} a u_{1}:
\mathrm{proj}_{u_{1}}(b_{2}) =\frac{\langle b_{2},u_{1}\rangle}{\langle u_{1},u_{1}\rangle}\,u_{1} =\frac{(2,0,1)\cdot(0,0,1)}{1}\,(0,0,1) =(0,0,1).
Logo
u_{2}=b_{2}-\mathrm{proj}_{u_{1}}(b_{2}) =(2,0,1)-(0,0,1) =(2,0,0),
e
e_{2}=\frac{u_{2}}{\|u_{2}\|} =\frac{(2,0,0)}{\sqrt{4}} =(1,0,0).

Passo 4

Orthogonalizamos b_{3} a u_{1} e u_{2}:
\mathrm{proj}_{u_{1}}(b_{3}) =\frac{(-1,1,2)\cdot(0,0,1)}{1}\,(0,0,1) =(0,0,2),
\mathrm{proj}_{u_{2}}(b_{3}) =\frac{(-1,1,2)\cdot(2,0,0)}{4}\,(2,0,0) =\frac{-2}{4}\,(2,0,0) =(-1,0,0).
Então
u_{3} =b_{3}-\mathrm{proj}_{u_{1}}(b_{3})-\mathrm{proj}_{u_{2}}(b_{3}) =(-1,1,2)-(0,0,2)-(-1,0,0) =(0,1,0),
e_{3}=\frac{u_{3}}{\|u_{3}\|} =(0,1,0).

Passo 5

Como \{e_{1},e_{2},e_{3}\} é ortonormal, as coordenadas de \displaystyle v=(-2,1,3) são
c_{i}=\langle v,e_{i}\rangle,
logo
c_{1}=\langle v,e_{1}\rangle =(−2,1,3)\cdot(0,0,1)=3, \quad c_{2}=\langle v,e_{2}\rangle =(−2,1,3)\cdot(1,0,0)=-2, \quad c_{3}=\langle v,e_{3}\rangle =(−2,1,3)\cdot(0,1,0)=1.

Resposta

(a) Uma base ortonormal obtida é
\{\,e_{1}=(0,0,1),\;e_{2}=(1,0,0),\;e_{3}=(0,1,0)\}.
(b) As coordenadas de v nessa base são
(3,\,-2,\;1).

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