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 e \beta=\left\{\left[\begin{array}{l}1 \\\\ 2 \\\\ 3\end{array}\right],\left[\begin{array}{c}-1 \\\\ 2 \\\\ 1\end{array}\right],\left[\begin{array}{c}-4 \\\\ -4 \\\\ 0\end{array}\right]\right\}

Passo 1

O enunciado pede uma base ortonormal \beta' de V=\mathbb{R}^3 com o produto escalar canônico, a partir da base
\beta = \{v_1,v_2,v_3\}
onde
v_1=\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix},\quad v_2=\begin{pmatrix}-1\\\ 2\\\ 1\end{pmatrix},\quad v_3=\begin{pmatrix}-4\\\ -4\\\ 0\end{pmatrix}.
Para isso, usamos o processo de Gram–Schmidt: geramos vetores ortogonais u_i subtraindo projeções dos vetores originais e depois normalizamos para obter e_i = u_i/\|u_i\|.

Passo 2

Definimos
u_1 = v_1,\qquad \|u_1\| = \sqrt{1^2+2^2+3^2} = \sqrt{14},
logo
e_1 = \frac{u_1}{\|u_1\|} = \frac{1}{\sqrt{14}}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix}.

Passo 3

Calculemos
u_2 = v_2 - \mathrm{proj}_{u_1}(v_2), \quad \mathrm{proj}_{u_1}(v_2)=\frac{\langle v_2,u_1\rangle}{\langle u_1,u_1\rangle}\,u_1.
Aqui \langle v_2,u_1\rangle = -1\cdot1 +2\cdot2 +1\cdot3 = 6 e \langle u_1,u_1\rangle=14, então
\mathrm{proj}_{u_1}(v_2) = \frac{6}{14}u_1 = \frac{3}{7}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix},
u_2 = \begin{pmatrix}-1\\\ 2\\\ 1\end{pmatrix} - \frac{3}{7}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix} = \frac{1}{7}\begin{pmatrix}-10\\\ 8\\\ -2\end{pmatrix}.
A norma de u_2 é
\|u_2\| = \sqrt{\frac{(-10)^2 + 8^2 +(-2)^2}{7^2}} = \sqrt{\frac{168}{49}} = \sqrt{\frac{24}{7}},
portanto
e_2 = \frac{u_2}{\|u_2\|} = \frac{\tfrac{1}{7}(-10,8,-2)}{\sqrt{24/7}} = \frac{1}{\sqrt{42}}\begin{pmatrix}-5\\\ 4\\\ -1\end{pmatrix}.

Passo 4

Calculemos
u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3).
Temos
\langle v_3,u_1\rangle = -4\cdot1 +(-4)\cdot2 +0\cdot3 = -12, logo
\mathrm{proj}_{u_1}(v_3) = \frac{-12}{14}u_1 = -\frac{6}{7}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix}.
E
\langle v_3,u_2\rangle = \frac{1}{7}\bigl[(-4)(-10)+(-4)\cdot8+0\cdot(-2)\bigr] = \frac{8}{7}, \quad \langle u_2,u_2\rangle=\frac{24}{7},
\mathrm{proj}_{u_2}(v_3) = \frac{(8/7)}{(24/7)}\,u_2 = \frac{1}{3}\,\frac{1}{7}\begin{pmatrix}-10\\\ 8\\\ -2\end{pmatrix} = \frac{1}{21}\begin{pmatrix}-10\\\ 8\\\ -2\end{pmatrix}.
Então
u_3 = \begin{pmatrix}-4\\\ -4\\\ 0\end{pmatrix} - \Bigl(-\frac{6}{7}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix}\Bigr) - \frac{1}{21}\begin{pmatrix}-10\\\ 8\\\ -2\end{pmatrix} = \frac{8}{3}\begin{pmatrix}-1\\\ -1\\\ 1\end{pmatrix}.
Como \|u_3\| = \tfrac{8}{\sqrt{3}}, obtemos
e_3 = \frac{u_3}{\|u_3\|} = \frac{1}{\sqrt{3}}\begin{pmatrix}-1\\\ -1\\\ 1\end{pmatrix}.

Resposta

\displaystyle \beta' = \{\,e_1,e_2,e_3\} com
e_1 = \frac{1}{\sqrt{14}}\begin{pmatrix}1\\\ 2\\\ 3\end{pmatrix},\quad e_2 = \frac{1}{\sqrt{42}}\begin{pmatrix}-5\\\ 4\\\ -1\end{pmatrix},\quad e_3 = \frac{1}{\sqrt{3}}\begin{pmatrix}-1\\\ -1\\\ 1\end{pmatrix}.

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