Convert the basis v1 = (-1,1,0)^T, V2 = (2,-1, 1)^T, V3 = (0, -1, 1)^T for Rinto an orthonormal basis, using the Gram Schmidt process and the standard inner product in R?

Answer:[tex]{\bf e}_{1}=\frac{(-1,1,0)}{\lVert (-1,1,0) \rVert}=\left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)[/tex][tex]{\bf e}_{2}=\frac{{\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1}}{\lVert {\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1} \rVert}=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\sqrt{\frac{2}{3}}\right)[/tex][tex]{\bf e}_{3}=\frac{{\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2}}{\lVert {\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2} \rVert}=\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)[/tex]Step-by-step explanation:We have the basis of [tex]\mathbb{R}^{3}[/tex] [tex]{\cal B}=\{(-1,1,0),(2,-1,1),(0,-1,1)\}[/tex]. From this basis we want to determine another orthonormal basis [tex]{\cal B}'=\{{\bf e}_{1},{\bf e}_{2},{\bf e}_{3}\}[/tex] of [tex]\mathbb{R}^{3}[/tex].The first step is to define [tex]{\bf e}_{1}[/tex] as:[tex]{\bf e}_{1}=\frac{(-1,1,0)}{\lVert (-1,1,0) \rVert}=\left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)[/tex]Now define [tex]{\bf e}_{2}[/tex] by:[tex]{\bf e}_{2}=\frac{{\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1}}{\lVert {\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1} \rVert}=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\sqrt{\frac{2}{3}}\right)[/tex]Now define [tex]{\bf e}_{3}[/tex] by :[tex]{\bf e}_{3}=\frac{{\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2}}{\lVert {\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2} \rVert}=\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)[/tex]