Question

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 1

Answer:

${\bf e}_{1}=\frac{(-1,1,0)}{\lVert (-1,1,0) \rVert}=\left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$

${\bf e}_{2}=\frac{{\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1}}{\lVert {\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1} \rVert}=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\sqrt{\frac{2}{3}}\right)$

${\bf e}_{3}=\frac{{\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2}}{\lVert {\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2} \rVert}=\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

Step-by-step explanation:

We have the basis of $\mathbb{R}^{3}$ ${\cal B}=\{(-1,1,0),(2,-1,1),(0,-1,1)\}$. From this basis we want to determine another orthonormal basis ${\cal B}'=\{{\bf e}_{1},{\bf e}_{2},{\bf e}_{3}\}$ of $\mathbb{R}^{3}$.

The first step is to define ${\bf e}_{1}$ as:

${\bf e}_{1}=\frac{(-1,1,0)}{\lVert (-1,1,0) \rVert}=\left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$

Now define ${\bf e}_{2}$ by:

${\bf e}_{2}=\frac{{\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1}}{\lVert {\bf v}_{2}-({\bf v}_{2}\cdot {\bf e}_{1}) {\bf e}_{1} \rVert}=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\sqrt{\frac{2}{3}}\right)$

Now define ${\bf e}_{3}$ by :

${\bf e}_{3}=\frac{{\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2}}{\lVert {\bf v}_{3}-({\bf v}_{3}\cdot{\bf e_{1}}){\bf e}_{1}-({\bf v}_{3}\cdot {\bf e}_{2}){\bf e}_{2} \rVert}=\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$