91Ó°ÊÓ

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

Then

$\mathit{u}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{,}\mathit{u}{\mathbf{⃗}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{⃗}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{⃗}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{u}{\mathbf{⃗}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{⃗}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{⃗}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{u}{\mathbf{⃗}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{⃗}}_{\mathbf{m}}}^{\mathbf{âŠ\yen }}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{⃗}}_{\mathbf{m}}}^{\mathbf{âŠ\yen }}$

Obtain the value of $u{⃗}_{1}andu{⃗}_2$ according to the Gram-Schmidt process.

$$\begin{gathered} u{⃗}_1=\frac{v{⃗}_1}{\left|\right|v⃗\left|\right|}.........\left(1\right) \\ u{⃗}_2=\frac{v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1}{\left|\right|v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1\left|\right|}..........\left(2\right) \end{gathered}$$

Since,

localid="1659436309006" $$\begin{gathered} {\stackrel{â‡\P Ä}{u}}_1=\frac{1}{2^2+2^2+1}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \end{gathered}$$

As, the formula for $u{⃗}_{2}is,u{⃗}_2=\frac{v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1}{\left|\right|v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1\left|\right|}.$

Now, find the values of role="math" localid="1659436285499" $u{⃗}_1.v{⃗}_2,v{⃗}_2-(u{⃗}_1â‹\dots v{⃗}_2)u{⃗}_{1}and‖v{⃗}_2-(u{⃗}_1â‹\dots v{⃗}_2)u{⃗}_1‖asbelow:$

role="math" localid="1659436642617" $$\begin{gathered} u{⃗}_1.v{⃗}_2=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]·\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right] \\ =\frac{1}{3}\left(2+2+5\right) \\ =3 \\ v{⃗}_2-\left(u{⃗}_1.v{⃗}_2\right)u{⃗}_1=\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]-3·\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\left[\begin{array}{c}1\\ 1\\ 5\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right] \\ ||v{⃗}_2-\left(u{⃗}_1.v{⃗}_2\right)u{⃗}_1||=\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+4^2} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

Now, find $u{⃗}_2=\frac{v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1}{\left|\right|v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1\left|\right|}$.

${\stackrel{â‡\P Ä}{u}}_2=\frac{1}{3\sqrt{2}}\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]$

Thus, the orthonormal vectors are${\stackrel{â‡\P Ä}{u}}_1=1/3\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$ and role="math" localid="1659436905459" ${\stackrel{â‡\P Ä}{u}}_2=\frac{1}{3\sqrt{2}}\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]$.

Hence, the values of the orthonormal vectors arerole="math" localid="1659436914679" ${\stackrel{â‡\P Ä}{u}}_1=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],{\stackrel{â‡\P Ä}{u}}_2=\frac{1}{3\sqrt{2}}\left[\begin{array}{c}-1\\ -1\\ 4\end{array}\right]$