Assinale a alternativa que contém a base de \mathbb{R}^{3} gerada a partir da base B=\{(1,0,0),(0,2,1),(-1,-1,-2)\} utilizando 0 processo de Gram Schimidt.
a. \left\{(-1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0, \frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)\right\}
b. \left\{(1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0, \frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)\right\}
c. \left\{(1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}
d. \quad\left\{(-1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}
e. \left\{(-1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}
f. \left\{(1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}

Passo 1

O enunciado dá a base \displaystyle B = \{\,v_{1}=(1,0,0),\;v_{2}=(0,2,1),\;v_{3}=(-1,-1,-2)\}
de \mathbb{R}^{3} e pede uma base ortonormal obtida pelo processo de Gram–Schmidt. Vamos usar o produto interno padrão \langle x,y\rangle = x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}
e a fórmula de projeção \displaystyle \mathrm{proj}_{u}(v)=\frac{\langle v,u\rangle}{\langle u,u\rangle}\,u.

Passo 2

Definimos \displaystyle u_{1}=v_{1}=(1,0,0).
O seu comprimento é \displaystyle \|u_{1}\|=\sqrt{\langle u_{1},u_{1}\rangle}=\sqrt{1}=1,
logo \displaystyle e_{1}=\frac{u_{1}}{\|u_{1}\|}=(1,0,0).

Passo 3

Calculamos \displaystyle u_{2}=v_{2}-\mathrm{proj}_{u_{1}}(v_{2}).
Como \langle v_{2},u_{1}\rangle=0, a projeção é zero e \displaystyle u_{2}=v_{2}=(0,2,1).
O comprimento é \displaystyle \|u_{2}\|=\sqrt{2^{2}+1^{2}}=\sqrt{5},
portanto \displaystyle e_{2}=\frac{u_{2}}{\|u_{2}\|}=\Bigl(0,\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\Bigr).

Passo 4

Agora \displaystyle u_{3}=v_{3}-\mathrm{proj}_{u_{1}}(v_{3})-\mathrm{proj}_{u_{2}}(v_{3}).
Primeiro \langle v_{3},u_{1}\rangle=-1 e \langle u_{1},u_{1}\rangle=1, então \mathrm{proj}_{u_{1}}(v_{3})=(-1,0,0),
logo v_{3}-\mathrm{proj}_{u_{1}}(v_{3})=(0,-1,-2).
Depois \langle v_{3},u_{2}\rangle=(0)(0)+(-1)(2)+(-2)(1)=-4 e \langle u_{2},u_{2}\rangle=5,
portanto \mathrm{proj}_{u_{2}}(v_{3})=\frac{-4}{5}(0,2,1)=(0,-\tfrac{8}{5},-\tfrac{4}{5}).
Subtraindo, \displaystyle u_{3}=\Bigl(0,-1,-2\Bigr)-\Bigl(0,-\tfrac{8}{5},-\tfrac{4}{5}\Bigr)=\Bigl(0,\tfrac{3}{5},-\tfrac{6}{5}\Bigr).
O comprimento é \displaystyle \|u_{3}\|=\sqrt{\bigl(\tfrac{3}{5}\bigr)^{2}+\bigl(-\tfrac{6}{5}\bigr)^{2}} =\frac{3\sqrt{5}}{5},
assim \displaystyle e_{3}=\frac{u_{3}}{\|u_{3}\|} =\Bigl(0,\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\Bigr).

Passo 5

A base ortonormal resultante é
\Bigl\{(1,0,0),\;\bigl(0,\tfrac{2}{\sqrt{5}},\tfrac{1}{\sqrt{5}}\bigr),\;\bigl(0,\tfrac{1}{\sqrt{5}},-\tfrac{2}{\sqrt{5}}\bigr)\Bigr\},
o que corresponde à alternativa (b).

Resposta

(b)

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