Question

Ortogonaliza a base V= {v1=(-1,1,0), v2=(2,1,1), v3=(1,1,1)}, usando o processo de Gram-Schmidt

Answer 1

seja $u_1=(-1,1,0)$ então 

$u_2=v_2-\dfrac{\langle v_2,u_1\rangle}{\langle u_1,u_1\rangle}\cdot u_1\\ \\ u_2=(2,1,1)-\dfrac{-1}{2}(-1,1,0)\\ \\ u_2=\left(\dfrac{3}{2},\dfrac{3}{2},1\right)\\ \\ \\ u_3=v_3-\dfrac{\langle v_3,u_1\rangle}{\langle u_1,u_1\rangle}\cdot u_1-\dfrac{\langle v_3,u_2\rangle}{\langle u_2,u_2\rangle}\cdot u_2\\ \\ \\ u_3=(1,1,1)-0-\dfrac{4}{\frac{11}{2}}\left(\dfrac{3}{2},\dfrac{3}{2},1\right)$


$u_3=(1,1,1)-\left(\dfrac{12}{11},\dfrac{12}{11},\dfrac{8}{11}\right)\\ \\ \\ u_3=\left(-\dfrac{1}{11},-\dfrac{1}{11},\dfrac{3}{11}\right)$

Por tanto uma base normal é

$R=\left\{(-1,1,0),\left(\dfrac{3}{2},\dfrac{3}{2},1\right),\left(-\dfrac{1}{11},-\dfrac{1}{11},\dfrac{3}{11}\right)\right\}\\ \\ \\ \text{e a base ortonormal \'e: }\\ \\ \\ R=\left\{\left(-\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2},0\right),\left(\dfrac{3}{\sqrt{22}},\dfrac{3}{\sqrt{22}},\dfrac{2}{\sqrt{22}}\right),\left(-\dfrac{1}{\sqrt{11}},-\dfrac{1}{\sqrt{11}},\dfrac{3}{\sqrt{11}}\right)\right\}$