Considere o espaço vetorial IV = R³ e o produto interno usual de R³. Determinar uma base ortonormal de cada um dos seguintes subespaços de R4...

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a) Temos \(W=[(1,1,0,0),(0,1,2,0),(0,0,3,4)]\). Vamos aplicar o processo de Gram-Schmidt a esse subespaço. Chamaremos \(v_1=(1,1,0,0)\), \(v_2=(0,1,2,0)\) e \(v_3=(0,0,3,4)\) e encontraremos uma base ortonormal para o subespaço da forma \(B=\{u_1,u_2,u_3\}\).

O vetor \(v_1\) será usado como referência, precisamos apenas normaliza-lo. Logo \(u_1=\dfrac{v_1}{||v_1||}=\dfrac{(1,1,0,0)}{\sqrt{2}}=(\dfrac{\sqrt2}2, \dfrac{\sqrt2}2,0,0)\).

Vamos calcular agora \(u_2\) e \(u_3\):

\(u'_2=v_2-\dfrac{v_2\cdot u_1}{u_1\cdot u_1} u_1=(0,1,2,0)-\dfrac{\dfrac{\sqrt2}2\cdot0+\dfrac{\sqrt2}2\cdot1+2\cdot0+0\cdot0}{\dfrac{\sqrt2}2\cdot\dfrac{\sqrt2}2+\dfrac{\sqrt2}2\cdot\dfrac{\sqrt2}2+0+0}(\dfrac{\sqrt2}2, \dfrac{\sqrt2}2,0,0)=(0,1,2,0)-\dfrac{\sqrt2}2 \cdot (\dfrac{\sqrt2}2, \dfrac{\sqrt2}2,0,0)=(0,1,2,0)-(\dfrac{2}4,\dfrac{2}4,0,0)=(-\dfrac12,\dfrac12,-2,0)\). Mas precisamos normalizar \(u'_2\), então \(u_2=\dfrac{(-\dfrac12,\dfrac12,2,0)}{\sqrt{\dfrac14+\dfrac14+4}}=\dfrac{(-\dfrac12,\dfrac12,2,0)}{\dfrac{3\sqrt2}2}=(-\dfrac1{3\sqrt2},\dfrac1{3\sqrt2},\dfrac4{3\sqrt2},0)\).

\(u'_3=v_3-\dfrac{v_3\cdot u_1}{u_1\cdot u_1} u_1-\dfrac{v_3\cdot u_2}{u_2\cdot u_2} u_2=(0,0,3,4)-0\cdot(\dfrac{\sqrt2}2, \dfrac{\sqrt2}2,0,0)-\dfrac{4}{\sqrt2}\cdot(-\dfrac1{3\sqrt2},\dfrac1{3\sqrt2},\dfrac4{3\sqrt2},0)=(0,0,3,4)-(-\dfrac23,\dfrac23,\dfrac83,0)=(\dfrac23,-\dfrac23,\dfrac13,4)\).

Normalizando, \(u_3=\dfrac{(\dfrac23,-\dfrac23,\dfrac13,4)}{\sqrt{\dfrac49+\dfrac49+\dfrac19+16}}=\dfrac{(\dfrac23,-\dfrac23,\dfrac13,4)}{\sqrt{17}}=(\dfrac2{3\sqrt17},-\dfrac2{3\sqrt17},\dfrac1{3\sqrt17},\dfrac4{\sqrt17})\).

Portanto, temos a base ortonormal \(\boxed{B=\{(\dfrac{\sqrt2}2, \dfrac{\sqrt2}2,0,0),(-\dfrac1{3\sqrt2},\dfrac1{3\sqrt2},\dfrac4{3\sqrt2},0),(\dfrac2{3\sqrt17},-\dfrac2{3\sqrt17},\dfrac1{3\sqrt17},\dfrac4{\sqrt17}) \}}\)