Considere o subespaço U=L\{(1,1,1,1),(1,0,1,0)\} de R4 .
a) Obtenha uma base ortonormada para U.
b) Determine a projeção ortogonal de (0,1,0,2) sobre U.
c) Qual a distância de (0,1,0,2) a U?

Passo 1

O enunciado nos dá o subespaço
U = \mathrm{span}\{(1,1,1,1),(1,0,1,0)\}\subset\mathbb{R}^4
e pede:
– (a) uma base ortonormada de U;
– (b) a projeção ortogonal de p=(0,1,0,2) em U;
– (c) a distância de p a U.
O conteúdo envolvido é o produto interno em \mathbb{R}^4, o processo de Gram–Schmidt para obter base ortonormal, a fórmula de projeção ortogonal e a norma de um vetor.

Passo 2

Aplicamos Gram–Schmidt aos geradores
\,v_1=(1,1,1,1) e v_2=(1,0,1,0).
Calculemos primeiro
\|v_1\| = \sqrt{1^2+1^2+1^2+1^2} = 2,
logo
e_1 = \frac{v_1}{\|v_1\|} = \Bigl(\tfrac12,\tfrac12,\tfrac12,\tfrac12\Bigr).
Em seguida projetamos v_2 sobre e_1:
\langle v_2,e_1\rangle = 1\cdot\tfrac12+0\cdot\tfrac12+1\cdot\tfrac12+0\cdot\tfrac12 = 1,
u_2 = v_2 - \langle v_2,e_1\rangle\,e_1 = (1,0,1,0) - (\,\tfrac12,\tfrac12,\tfrac12,\tfrac12) = \Bigl(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12\Bigr).
Como
\|u_2\| = \sqrt{4\cdot(\tfrac12)^2} =1,
definimos
e_2 = u_2 = \Bigl(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12\Bigr).
Portanto uma base ortonormada de U é
\{\,e_1,e_2\} =\Bigl\{\;(\tfrac12,\tfrac12,\tfrac12,\tfrac12),\;(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12)\Bigr\}.

Passo 3

Para projetar
\,p=(0,1,0,2) em U usamos
\mathrm{proj}_U(p) = \langle p,e_1\rangle\,e_1 + \langle p,e_2\rangle\,e_2.
Calculando os escalares,
\langle p,e_1\rangle = 0\cdot\tfrac12+1\cdot\tfrac12+0\cdot\tfrac12+2\cdot\tfrac12 = \tfrac32,
\langle p,e_2\rangle = 0\cdot\tfrac12 +1\cdot(-\tfrac12)+0\cdot\tfrac12+2\cdot(-\tfrac12) = -\tfrac12 -1 = -\tfrac32.
Logo
\mathrm{proj}_U(p) = \tfrac32\,(\tfrac12,\tfrac12,\tfrac12,\tfrac12) - \tfrac32\,(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12) = (0,\tfrac32,0,\tfrac32).

Passo 4

A distância de p a U é
\|\,p - \mathrm{proj}_U(p)\| = \bigl\|(0,1,0,2)-(0,\tfrac32,0,\tfrac32)\bigr\| = \bigl\|(0,-\tfrac12,0,\tfrac12)\bigr\| = \sqrt{\bigl(-\tfrac12\bigr)^2+\bigl(\tfrac12\bigr)^2} = \frac{\sqrt2}{2}.

Resposta

(a) \{(\tfrac12,\tfrac12,\tfrac12,\tfrac12),(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12)\}
(b) \mathrm{proj}_U(0,1,0,2)=(0,\tfrac32,0,\tfrac32)
(c) \mathrm{dist}((0,1,0,2),U)=\tfrac{\sqrt2}{2}

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