91Ó°ÊÓ

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}$for j = 2, ...,m we resolve the vector ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}$into its components parallel and perpendicular to the span of the preceding vectors ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$,

Then

${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{→}{v}}_1||}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_2}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_j}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_m}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{âŠ\yen }}$

Let the given vectors are, ${\stackrel{→}{v}}_1=\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]\mathrm{and}{\stackrel{→}{v}}_2=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$and .

Obtain the value of ${\stackrel{→}{u}}_1$and ${\stackrel{→}{u}}_2$ according to the Gram-Schmidt process.

role="math" localid="1659948138442" $$\begin{gathered} {\stackrel{→}{u}}_1=\frac{{\stackrel{→}{v}}_1}{||\stackrel{→}{v}||}....\left(1\right) \\ {\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1}{||{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1||}.....\left(2\right) \end{gathered}$$

Find ${\stackrel{→}{u}}_1$as follows:

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{\sqrt{6^2+3^2+2^2}}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right] \\ =\frac{1}{7}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right] \end{gathered}$$

As, the formula for ${\stackrel{→}{u}}_2$is, ${\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1}{||{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1||}$.

Now, find the values of${\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2,{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1\mathrm{and}||{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1||$, and as below:

role="math" localid="1659948520118" $$\begin{gathered} {\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2=\frac{1}{7}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ =\frac{1}{7}\left(12-18+6\right) \\ =0 \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{v}}_2({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ ||{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1||=\sqrt{2^2+{(-6)}^2+3^2} \\ =7 \end{gathered}$$

Then, use the obtained values in ${\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1}{||{\stackrel{→}{v}}_2-({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2){\stackrel{→}{u}}_1||}$.

${\stackrel{→}{u}}_2=\frac{1}{7}\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$

Thus, the required vectors are${\stackrel{→}{u}}_1=\frac{1}{7}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]$and${\stackrel{→}{u}}_2=\frac{1}{7}\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right]\left[\begin{array}{c}2/7\\ -6/7\\ 3/7\end{array}\right]$.