91Ó°ÊÓ

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

Then

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

Let the given vectors are $v{⃗}_1=\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right],v{⃗}_2=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]$.

Obtain the values of , according to the Gram-Schmidt process.

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

Find $u{⃗}_1$.

$$\begin{gathered} u{⃗}_1=\frac{1}{\sqrt{5^2+4^2+2^2+2^2}}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ =\frac{1}{7}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \end{gathered}$$

Now, here is need to find out the values of $v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_{1}and\left|\right|v{⃗}_2-(u{⃗}_1.v{⃗}_2)u{⃗}_1\left|\right|$to obtain the value of $u{⃗}_2$.

Consider the equations below.

role="math" localid="1659441637408" $$\begin{gathered} \stackrel{→}{u_1}·\stackrel{→}{v_1}=7 \\ \left(\stackrel{→}{u_1}·\stackrel{→}{v_1}\right)\stackrel{→}{u_1}=\left[\begin{array}{c}5\\ 4\\ 2\\ 3\end{array}\right] \\ \stackrel{→}{v_1}-\left(\stackrel{→}{u_1}·\stackrel{→}{v_1}\right)\stackrel{→}{u_1}=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]-\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ =\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right] \end{gathered}$$

Then,

$$\begin{gathered} ||\stackrel{→}{v_1}-\left(\stackrel{→}{u_1}·\stackrel{→}{v_1}\right)\stackrel{→}{u_1}||=\sqrt{4+4+25+16} \\ =\sqrt{49} \\ =7 \end{gathered}$$

Now find role="math" localid="1659441716297" $\stackrel{→}{u_2}$.

$\stackrel{→}{u_2}=\frac{1}{7}\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]$

Thus, the values of $u{⃗}_1,u{⃗}_2$are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.