O conjunto B=\left\{\vec{u}_{1}, \vec{u}_{2}, \vec{u}_{3}\right\}, onde \vec{u}_{1}=(1,1,1), \vec{u}_{2}=(1,2,2) e \vec{u}_{3}=(1,1,0) é una base para \mathbb{R}^{3}. Transforme B em uma base orto mormal B^{H}.

Passo 1

O enunciado dá a base B=\{\vec u_{1},\vec u_{2},\vec u_{3}\} de \mathbb R^3 com \vec u_{1}=(1,1,1), \vec u_{2}=(1,2,2), \vec u_{3}=(1,1,0), e pede uma base ortonormal B^{H}. Vamos usar o processo de Gram–Schmidt, que consiste em gerar vetores ortogonais v_{1},v_{2},v_{3} e, em seguida, normalizá-los para obter e_{1},e_{2},e_{3}.

Passo 2

Definimos v_{1}=\vec u_{1}. Calculamos sua norma e o vetor unitário correspondente:
\|v_{1}\|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3},
logo
e_{1}=\frac{v_{1}}{\|v_{1}\|}=\frac{1}{\sqrt{3}}(1,1,1).

Passo 3

Subtraímos de \vec u_{2} sua projeção em e_{1}:
\vec u_{2}\cdot e_{1}=\frac{1+2+2}{\sqrt{3}}=\frac{5}{\sqrt{3}}, \quad \mathrm{proj}_{e_{1}}(\vec u_{2})=\frac{5}{\sqrt{3}}\,e_{1}=\frac{5}{3}(1,1,1).
Assim
v_{2}=\vec u_{2}-\mathrm{proj}_{e_{1}}(\vec u_{2}) =\Bigl(1,2,2\Bigr)-\Bigl(\tfrac{5}{3},\tfrac{5}{3},\tfrac{5}{3}\Bigr) =\Bigl(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{1}{3}\Bigr).
A norma de v_{2} é
\|v_{2}\|=\sqrt{\bigl(-\tfrac{2}{3}\bigr)^{2}+\bigl(\tfrac{1}{3}\bigr)^{2}+\bigl(\tfrac{1}{3}\bigr)^{2}} =\frac{\sqrt{6}}{3},
portanto
e_{2}=\frac{v_{2}}{\|v_{2}\|} =\biggl(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\biggr).

Passo 4

Subtraímos de \vec u_{3} as projeções em e_{1} e e_{2}:
\vec u_{3}\cdot e_{1}=\frac{1+1+0}{\sqrt{3}}=\frac{2}{\sqrt{3}}, \quad \vec u_{3}\cdot e_{2} =-\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{6}}+0=-\frac{1}{\sqrt{6}}.
Logo
\mathrm{proj}_{e_{1}}(\vec u_{3})=\frac{2}{3}(1,1,1), \quad \mathrm{proj}_{e_{2}}(\vec u_{3}) =-\frac{1}{\sqrt{6}}\,e_{2} =\Bigl(\tfrac{1}{3},-\tfrac{1}{6},-\tfrac{1}{6}\Bigr),
e
v_{3} =\vec u_{3} -\mathrm{proj}_{e_{1}}(\vec u_{3}) -\mathrm{proj}_{e_{2}}(\vec u_{3}) =(1,1,0)-\Bigl(\tfrac{2}{3},\tfrac{2}{3},\tfrac{2}{3}\Bigr) -\Bigl(\tfrac{1}{3},-\tfrac{1}{6},-\tfrac{1}{6}\Bigr) =(0,\tfrac{1}{2},-\tfrac{1}{2}).
A norma de v_{3} é \|v_{3}\|=1/\sqrt{2}, daí
e_{3}=\frac{v_{3}}{\|v_{3}\|} =\bigl(0,\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\bigr).

Resposta

A base ortonormal B^{H} é
B^{H}=\bigl\{\,e_{1},\,e_{2},\,e_{3}\bigr\}
com
e_{1}=\frac{1}{\sqrt{3}}(1,1,1), \quad e_{2}=\bigl(-\tfrac{2}{\sqrt{6}},\tfrac{1}{\sqrt{6}},\tfrac{1}{\sqrt{6}}\bigr), \quad e_{3}=\bigl(0,\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\bigr).

Exercícios relacionados