91Ó°ÊÓ

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{m}$for we resolve the vector 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

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

Obtain the value of$u{⃗}_1,u{⃗}_{2}andu{⃗}_3$ according to Gram-Schmidt process.

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

$\frac{u{⃗}_3=v{⃗}_3-(u{⃗}_1.v{⃗}_3)u{⃗}_1-(u{⃗}_2.v{⃗}_3)u{⃗}_2}{\left|\right|v{⃗}_3-(u{⃗}_1.v{⃗}_3)u{⃗}_1-(u{⃗}_2.v{⃗}_3)u{⃗}_2\left|\right|}....\left(3\right)$

Now, find $u{⃗}_1=\frac{v{⃗}_1}{\left|\right|v⃗\left|\right|}$$$\begin{gathered} \stackrel{→}{u_1}=\frac{1}{\sqrt{4^2+0^2+3^2}}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \end{gathered}$$.

$$\begin{gathered} {\stackrel{→}{u}}_1·\stackrel{→}{v_2}=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right] \\ ⇒\frac{1}{5}\left(100-75\right)=5 \\ \stackrel{→}{v_2}-\left({\stackrel{→}{u}}_1·\stackrel{→}{v_2}\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-5·\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ =\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ =\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ ⇒||\stackrel{→}{v_2}-\left({\stackrel{→}{u}}_1·\stackrel{→}{v_2}\right){\stackrel{→}{u}}_1||=\sqrt{{21}^2+0^2+{\left(-28\right)}^2} \\ =\sqrt{1225} \\ =35 \\ Then, \\ {\stackrel{→}{u}}_2=\frac{1}{35}\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ =\left[\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right] \end{gathered}$$

Thus, the required vectors are, ${\stackrel{→}{u}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]and{\stackrel{→}{u}}_2=\left[\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right]$.

Now observe that, $v{⃗}_3â‹\dots u{⃗}_1=0andv{⃗}_3â‹\dots u{⃗}_2=0.$.

By normalizing it to obtain the value of $\stackrel{→}{u_3}$.

role="math" localid="1659434858573" $\stackrel{→}{u_3}=\frac{\stackrel{→}{v_3}}{||\stackrel{→}{v_3}||}=\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]$

Thus, the values ofrole="math" localid="1659434864823" ${\stackrel{→}{u}}_1=1/5\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right],{\stackrel{→}{u}}_2=1/5\left[\begin{array}{c}3\\ 0\\ -4\end{array}\right]and{\stackrel{→}{u}}_3=\frac{\stackrel{→}{v_3}}{||\stackrel{→}{v_3}||}=\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]$

Hence, the orthonormal vectors are $\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right],\left[\begin{array}{c}3/5\\ 0\\ -4/5\end{array}\right],\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]$.