## Elementary Linear Algebra 7th Edition

$B'=\{(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}) ,(1,0,0),(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})\}$ is orthonormal basis for $R^3$.
Let $B=\{ v_1=(0,1,2),v_2=(2,0,0), v_3=(1,1,1) \}$. Applying the Gram-Schmidt orthonormalization process produces $$w_1=v_1=(0,1,2)$$ $$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(2,0,0)- 0(0,1,2)=(2,0,0)$$ $$w_3=v_3- \frac{v_3\cdot w_1}{w_1\cdot w_1}w_1- \frac{v_3\cdot w_2}{w_2\cdot w_2}w_2\\ =(1,1,1) - \frac{3}{5}(0,1,2)-\frac{2}{4}(2,0,0)=(0,\frac{2}{5},-\frac{1}{5}).$$ Normalizing $w_1$, $w_2$ and $w_3$ produces the orthonormal set $$u_1=\frac{w_1}{\|w_1\|}=\frac{1}{5}(0,1,2)=(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$ $$u_2=\frac{w_2}{\|w_2\|}=\frac{1}{2}(2,0,0)=(1,0,0)$$ $$u_3=\frac{w_3}{\|w_3\|}=\sqrt{5}(0,\frac{1}{5},\frac{2}{5})=(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$ Hence, $B'=\{(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}) ,(1,0,0),(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})\}$ is orthonormal basis for $R^3$.