Considere os seguintes conjuntos de vetores. (i) Mostre que o conjunto é linearmente independente; (ii) use o processo de Gram-Schmidt para encontrar um conjunto ortogonal de vetores; (iii) determine um conjunto ortonormal de vetores a partir dos vetores em (ii).
\mathbf{v}_{1}=(1,1,1,1)^{t}, \mathbf{v}_{2}=(0,2,2,2)^{t}, \mathbf{v}_{3}=(1,0,0,1)^{t}

Passo 1

Temos os vetores em \mathbb{R}^4:
\mathbf{v}_{1} = (1,1,1,1)^{t},\quad \mathbf{v}_{2} = (0,2,2,2)^{t},\quad \mathbf{v}_{3} = (1,0,0,1)^{t}.
O que pedem:
  1. Mostrar que \{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\} é linearmente independente.
  2. Aplicar o processo de Gram–Schmidt para obter um conjunto ortogonal \{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}.
  3. Normalizar esse conjunto para obter um conjunto ortonormal.

Passo 2

Para checar independência linear, consideramos
c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3} = \mathbf{0}.
Escrevendo coordenada a coordenada:
\begin{cases} c_{1} + 0\cdot c_{2} + c_{3} = 0,\\\ c_{1} + 2c_{2} + 0\cdot c_{3} = 0,\\\ c_{1} + 2c_{2} + 0\cdot c_{3} = 0,\\\ c_{1} + 2c_{2} + c_{3} = 0. \end{cases}
Da segunda equação vem c_{1}=-2c_{2}. Substituindo na primeira, obtemos -2c_{2}+c_{3}=0\Rightarrow c_{3}=2c_{2}. Na quarta equação:
c_{1}+2c_{2}+c_{3} = -2c_{2}+2c_{2}+2c_{2}=2c_{2}=0
logo c_{2}=0, e daí c_{1}=c_{3}=0. Concluímos que são linearmente independentes.

Passo 3

Aplicamos Gram–Schmidt com o produto interno padrão \langle x,y\rangle = x^{t}y.
Definimos
\mathbf{u}_{1} = \mathbf{v}_{1}.
Em seguida
\mathbf{u}_{2} = \mathbf{v}_{2} - \frac{\langle \mathbf{v}_{2},\mathbf{u}_{1}\rangle}{\langle \mathbf{u}_{1},\mathbf{u}_{1}\rangle}\,\mathbf{u}_{1},
onde
\langle \mathbf{v}_{2},\mathbf{u}_{1}\rangle = 0\cdot1 + 2\cdot1 + 2\cdot1 + 2\cdot1 = 6, \quad \langle \mathbf{u}_{1},\mathbf{u}_{1}\rangle = 4.
Logo
\mathbf{u}_{2} = (0,2,2,2) - \frac{6}{4}(1,1,1,1) = \bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr).
Agora
\mathbf{u}_{3} = \mathbf{v}_{3} - \frac{\langle \mathbf{v}_{3},\mathbf{u}_{1}\rangle}{\langle \mathbf{u}_{1},\mathbf{u}_{1}\rangle}\,\mathbf{u}_{1} - \frac{\langle \mathbf{v}_{3},\mathbf{u}_{2}\rangle}{\langle \mathbf{u}_{2},\mathbf{u}_{2}\rangle}\,\mathbf{u}_{2}.
Calculamos
\langle \mathbf{v}_{3},\mathbf{u}_{1}\rangle = 1+0+0+1 = 2, \quad \langle \mathbf{v}_{3},\mathbf{u}_{2}\rangle = (1,0,0,1)\cdot\bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr) = -\tfrac32 + \tfrac12 = -1,
\langle \mathbf{u}_{2},\mathbf{u}_{2}\rangle = \bigl(-\tfrac32\bigr)^{2} + 3\bigl(\tfrac12\bigr)^{2} = \tfrac{9}{4} + \tfrac{3}{4} = 3.
Então
\mathbf{u}_{3} = (1,0,0,1) - \tfrac{2}{4}(1,1,1,1) - \bigl(-\tfrac{1}{3}\bigr)\bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr) = \bigl(0,-\tfrac13,-\tfrac13,\tfrac23\bigr).
Dessa forma \{\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}\} é ortogonal.

Passo 4

Para o conjunto ortonormal normalizamos cada \mathbf{u}_{i}:
\|\mathbf{u}_{1}\| = \sqrt{4}=2,\quad \|\mathbf{u}_{2}\| = \sqrt{3},\quad \|\mathbf{u}_{3}\| = \sqrt{\tfrac{2}{3}}.
Assim
\mathbf{e}_{1} = \frac{\mathbf{u}_{1}}{\|\mathbf{u}_{1}\|} = \tfrac12(1,1,1,1), \quad \mathbf{e}_{2} = \frac{\mathbf{u}_{2}}{\|\mathbf{u}_{2}\|} = \frac{1}{\sqrt{3}}\bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr), \quad \mathbf{e}_{3} = \frac{\mathbf{u}_{3}}{\|\mathbf{u}_{3}\|} = \sqrt{\tfrac{3}{2}}\bigl(0,-\tfrac13,-\tfrac13,\tfrac23\bigr).

Resposta

Conjunto linearmente independente: \{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\}.
Conjunto ortogonal (Gram–Schmidt):
\mathbf{u}_{1} = (1,1,1,1),\quad \mathbf{u}_{2} = \bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr),\quad \mathbf{u}_{3} = \bigl(0,-\tfrac13,-\tfrac13,\tfrac23\bigr).
Conjunto ortonormal:
\mathbf{e}_{1} = \tfrac12(1,1,1,1),\quad \mathbf{e}_{2} = \frac{1}{\sqrt{3}}\bigl(-\tfrac32,\tfrac12,\tfrac12,\tfrac12\bigr),\quad \mathbf{e}_{3} = \sqrt{\tfrac{3}{2}}\bigl(0,-\tfrac13,-\tfrac13,\tfrac23\bigr).

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