Sejam V um espaço vetorial com produto interno \langle, \rangle e \beta uma base de V, use o processo de Gram-Schmidt para achar uma base ortonormal \beta^{\prime} de V, nos seguintes casos.
V=\mathbb{R}^{3} munido do produto escalar com pesos p_{1}=1, p_{2}=2, p_{3}=1 e \beta= \left\{\left[\begin{array}{l}1 \\\\ 1 \\\\ 0\end{array}\right],\left[\begin{array}{l}1 \\\\ 0 \\\\ 1\end{array}\right],\left[\begin{array}{l}0 \\\\ 2 \\\\ 0\end{array}\right]\right\}

Passo 1

O espaço
V = \mathbb{R}^3
está equipado com o produto interno pesado
\langle x,y\rangle = x_1y_1 + 2\,x_2y_2 + x_3y_3.
A base original é
\beta = \Bigl\{v_1,v_2,v_3\Bigr\}, \quad v_1 = \begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix}, \; v_2 = \begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix}, \; v_3 = \begin{pmatrix}0\\\ 2\\\ 0\end{pmatrix}.
Vamos aplicar o processo de Gram–Schmidt para produzir uma base ortonormal \beta'.

Passo 2

Definimos
u_1 = v_1.
Calculamos
\|u_1\|^2 = \langle v_1,v_1\rangle = 1^2 + 2\cdot1^2 + 0^2 = 3,
logo
e_1 = \frac{u_1}{\|u_1\|} = \frac1{\sqrt{3}}\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix}.

Passo 3

Para o segundo vetor, primeiro projetamos v_2 sobre u_1:
\langle v_2,u_1\rangle = 1\cdot1 + 2\cdot0\cdot1 + 1\cdot0 = 1, \quad \mathrm{proj}_{u_1}(v_2)=\frac{1}{3}\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix}.
Então
u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = \begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix} - \frac13\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix} = \begin{pmatrix}2/3\\\ -1/3\\\ 1\end{pmatrix}.
Calculamos a norma:
\|u_2\|^2 = \Bigl(\tfrac23\Bigr)^2 + 2\Bigl(-\tfrac13\Bigr)^2 + 1^2 = \frac{4}{9} + \frac{2}{9} + 1 = \frac{5}{3},
portanto
e_2 = \frac{u_2}{\|u_2\|} = \sqrt{\frac{3}{5}}\begin{pmatrix}2/3\\\ -1/3\\\ 1\end{pmatrix} = \begin{pmatrix}\tfrac{2\sqrt3}{3\sqrt5}\\[6pt]-\tfrac{\sqrt3}{3\sqrt5}\\[6pt]\tfrac{\sqrt3}{\sqrt5}\end{pmatrix}.]

Passo 4

Agora projetamos v_3 nas direções de u_1 e u_2:
\langle v_3,u_1\rangle = 0\cdot1 + 2\cdot1\cdot1 + 0\cdot0 = 4, \quad \mathrm{proj}_{u_1}(v_3)=\frac{4}{3}\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix},
\langle v_3,u_2\rangle = 0\cdot\tfrac23 + 2\cdot(-\tfrac13)\cdot2 + 0\cdot1 = -\tfrac{4}{3}, \quad \mathrm{proj}_{u_2}(v_3) = \frac{-4/3}{5/3}\,u_2 = -\tfrac{4}{5}\begin{pmatrix}2/3\\\ -1/3\\\ 1\end{pmatrix} = \begin{pmatrix}-\tfrac{8}{15}\\[4pt]\tfrac{4}{15}\\[4pt]-\tfrac{4}{5}\end{pmatrix}.
Então
u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = \begin{pmatrix}0\\\ 2\\\ 0\end{pmatrix} - \begin{pmatrix}\tfrac{4}{3}\\tfrac{4}{3}\\\ 0\end{pmatrix} - \begin{pmatrix}-\tfrac{8}{15}\\tfrac{4}{15}\\\ -\tfrac{4}{5}\end{pmatrix} = \begin{pmatrix}-\tfrac{4}{5}\\[4pt]\tfrac{2}{5}\\[4pt]\tfrac{4}{5}\end{pmatrix}.
Calculamos
\|u_3\|^2 = \Bigl(-\tfrac45\Bigr)^2 + 2\Bigl(\tfrac25\Bigr)^2 + \Bigl(\tfrac45\Bigr)^2 = \frac{16}{25} + \frac{8}{25} + \frac{16}{25} = \frac{40}{25} = \frac{8}{5},
logo
e_3 = \frac{u_3}{\|u_3\|} = \sqrt{\frac{5}{8}}\begin{pmatrix}-\tfrac{4}{5}\\[4pt]\tfrac{2}{5}\\[4pt]\tfrac{4}{5}\end{pmatrix} = \begin{pmatrix}-\tfrac{\sqrt{10}}{5}\\[4pt]\tfrac{\sqrt{10}}{10}\\[4pt]\tfrac{\sqrt{10}}{5}\end{pmatrix}.]

Resposta

A base ortonormal \beta'=\{e_1,e_2,e_3\} é
e_1 = \begin{pmatrix}\tfrac{1}{\sqrt{3}}\\[4pt]\tfrac{1}{\sqrt{3}}\\[4pt]0\end{pmatrix}, \quad e_2 = \begin{pmatrix}\tfrac{2\sqrt3}{3\sqrt5}\\[4pt]-\tfrac{\sqrt3}{3\sqrt5}\\[4pt]\tfrac{\sqrt3}{\sqrt5}\end{pmatrix}, \quad e_3 = \begin{pmatrix}-\tfrac{\sqrt{10}}{5}\\[4pt]\tfrac{\sqrt{10}}{10}\\[4pt]\tfrac{\sqrt{10}}{5}\end{pmatrix}.]

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