Seja V=\mathbb{R}^{4} e W=[(1,1,0,0),(0,1,1,0),(0,0,1,1)].
a) Determinar uma base ortonormal de W com relação ao produto interno canônico.
b) Determinar uma base ortonormal de W com relação ao produto interno definido por
\left\langle\left(x_{1}, y_{1}, z_{1}, w_{1}\right),\left(x_{2}, y_{2}, z_{2}, w_{2}\right)\right\rangle=x_{1} x_{2}+y_{1} x_{2}+x_{1} y_{2}+2 y_{1} y_{2}+z_{1} z_{2}+2 w_{1} w_{2}

Passo 1

O enunciado pede duas bases ortonormais de
W=\operatorname{Span}\{v_{1},v_{2},v_{3}\}\subset\mathbb{R}^{4},
onde
v_{1}=(1,1,0,0),\quad v_{2}=(0,1,1,0),\quad v_{3}=(0,0,1,1).
Para (a) usamos o produto interno canônico \langle x,y\rangle=x_{1}y_{1}+\cdots+x_{4}y_{4}
e para (b) usamos
\bigl\langle(x,y,z,w),(x',y',z',w')\bigr\rangle =x\,x'+y\,x'+x\,y'+2\,y\,y'+z\,z'+2\,w\,w'.
Em ambos os casos aplicamos o processo de Gram–Schmidt em \{v_{1},v_{2},v_{3}\} e depois normalizamos.

Passo 2

(Parte a) Gram–Schmidt com o produto canônico
  1. u_{1}=\dfrac{v_{1}}{\|v_{1}\|}=\dfrac{(1,1,0,0)}{\sqrt{1^{2}+1^{2}}}=(\tfrac1{\sqrt2},\tfrac1{\sqrt2},0,0).
  2. Subtrair de v_{2} a projeção em u_{1}:
\mathrm{proj}_{u_{1}}v_{2} =\bigl\langle v_{2},u_{1}\bigr\rangle\,u_{1} =\frac{0\cdot\tfrac1{\sqrt2}+1\cdot\tfrac1{\sqrt2}}{1}\,u_{1} =\frac1{\sqrt2}\,u_{1} =(\tfrac12,\tfrac12,0,0).
Logo
w_{2}=v_{2}-\mathrm{proj}_{u_{1}}v_{2} =(-\tfrac12,\tfrac12,1,0), \|w_{2}\|=\sqrt{\tfrac14+\tfrac14+1}=\sqrt{\tfrac32},
u_{2}=\frac{w_{2}}{\|w_{2}\|} =\Bigl(-\tfrac1{\sqrt6},\,\tfrac1{\sqrt6},\,\tfrac2{\sqrt6},\,0\Bigr).

Passo 3

Continuando para v_{3}: \mathrm{proj}_{u_{1}}v_{3}=0 pois \langle v_{3},u_{1}\rangle=0.
\mathrm{proj}_{u_{2}}v_{3} =\bigl\langle v_{3},u_{2}\bigr\rangle u_{2} =\frac{2/\sqrt6}{1}\;u_{2} =\frac2{\sqrt6}\;u_{2} =(-\tfrac13,\tfrac13,\tfrac23,0).
Logo
w_{3}=v_{3}-\mathrm{proj}_{u_{2}}v_{3} =(\tfrac13,-\tfrac13,\tfrac13,1), \quad \|w_{3}\|=\sqrt{\tfrac13+1}=\tfrac{2}{\sqrt3},
u_{3}=\frac{w_{3}}{\|w_{3}\|} =\Bigl(\tfrac{\sqrt3}{6},-\tfrac{\sqrt3}{6},\tfrac{\sqrt3}{6},\tfrac{\sqrt3}{2}\Bigr).

Passo 4

(Parte b) Agora o produto interno é \langle x,x\rangle=x^{T}Gx com
G=\begin{pmatrix} 1&1&0&0\\\\ 1&2&0&0\\\\ 0&0&1&0\\\\ 0&0&0&2 \end{pmatrix}.
Começamos por v_{1}: \langle v_{1},v_{1}\rangle=5,
u_{1}'=\frac{v_{1}}{\sqrt5}=(\tfrac1{\sqrt5},\tfrac1{\sqrt5},0,0).

Passo 5

Projete v_{2} em u_{1}': \langle v_{2},u_{1}'\rangle=3/\sqrt5,
w_{2}'=v_{2}-\frac{3}{5}v_{1} =\bigl(-\tfrac35,\tfrac25,1,0\bigr), \quad \langle w_{2}',w_{2}'\rangle=\tfrac{6}{5},
u_{2}'=\frac{w_{2}'}{\sqrt{6/5}} =\Bigl(-\tfrac{3}{\sqrt{30}},\,\tfrac{2}{\sqrt{30}},\,\tfrac{5}{\sqrt{30}},\,0\Bigr).

Passo 6

Projete v_{3} em u_{1}' e u_{2}': \langle v_{3},u_{1}'\rangle=0,\quad \langle v_{3},u_{2}'\rangle=\sqrt{5/6}.
w_{3}'=v_{3}-\sqrt{\tfrac{5}{6}}\,u_{2}' =(\tfrac12,-\tfrac13,\tfrac16,1), \quad \langle w_{3}',w_{3}'\rangle=\tfrac{13}{6},
u_{3}'=\frac{w_{3}'}{\sqrt{13/6}} =\Bigl(\tfrac{\sqrt6}{2\sqrt{13}},\,-\tfrac{\sqrt6}{3\sqrt{13}},\, \tfrac{\sqrt6}{6\sqrt{13}},\,\tfrac{\sqrt6}{\sqrt{13}}\Bigr).

Resposta

(a) Uma base ortonormal de W para o produto canônico é
u_{1}=(\tfrac1{\sqrt2},\tfrac1{\sqrt2},0,0),\quad u_{2}=(-\tfrac1{\sqrt6},\tfrac1{\sqrt6},\tfrac2{\sqrt6},0),\quad u_{3}=(\tfrac{\sqrt3}{6},-\tfrac{\sqrt3}{6},\tfrac{\sqrt3}{6},\tfrac{\sqrt3}{2}).
(b) Para o produto interno dado, uma base ortonormal de W é
u'_{1}=(\tfrac1{\sqrt5},\tfrac1{\sqrt5},0,0),\quad u'_{2}=(-\tfrac{3}{\sqrt{30}},\tfrac{2}{\sqrt{30}},\tfrac{5}{\sqrt{30}},0),\quad u'_{3}=(\tfrac{\sqrt6}{2\sqrt{13}},-\tfrac{\sqrt6}{3\sqrt{13}},\tfrac{\sqrt6}{6\sqrt{13}},\tfrac{\sqrt6}{\sqrt{13}}).

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