91Ó°ÊÓ

Consider a basis of a subspace Vof ${\mathbf{R}}^{\mathbf{n}}$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 ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$,

Then localid="1659439775013" ${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{âˆ\yen {\stackrel{→}{v}}_1âˆ\yen }{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{âˆ\yen {\stackrel{→}{v}}_2^{âŠ\yen }âˆ\yen }{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{…}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{âˆ\yen {\stackrel{→}{v}}_j^{âŠ\yen }âˆ\yen }{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{…}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{âˆ\yen {\stackrel{→}{v}}_m^{âŠ\yen }âˆ\yen }{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}^{\mathbf{âŠ\yen }}$

The given vectors are${\stackrel{→}{v}}_1=\left[\begin{array}{l}1\\ 1\\ 1\\ 1\end{array}\right],{\stackrel{→}{v}}_2=\left[\begin{array}{l}1\\ 9\\ −5\\ 3\end{array}\right]$.

Obtain the values of $u_1,u_2$according toGram-Schmidt process.

localid="1659955449914" $\begin{array}{r}{\stackrel{→}{u}}_1=\frac{{\stackrel{→}{v}}_1}{âˆ\yen \stackrel{→}{v}âˆ\yen }…\left(1\right)\\ {\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1}{∥{\stackrel{→}{v}}_2−\left({\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1∥}…\left(2\right)\end{array}$

Find ${\stackrel{→}{u}}_1$.

localid="1659955512481" $$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Now, here is need to find out the values of ${\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_2\right)\stackrel{→}{u_1}$and$||{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_2\right)\stackrel{→}{u_1}||$to obtain the value of ${\stackrel{→}{u}}_2$.

Consider the equations below.

$\begin{array}{c}{\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2=4\\ \left({\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{l}2\\ 2\\ 2\\ 2\end{array}\right]\end{array}$

And,

$\begin{array}{r}{\stackrel{→}{v}}_2−\left({\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 9\\ −5\\ 3\end{array}\right]−\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right]\\ =\left[\begin{array}{c}−1\\ 7\\ −7\\ 1\end{array}\right]\end{array}$

Then,

localid="1659955715369" $$\begin{gathered} ∥{\stackrel{→}{v}}_2−\left({\stackrel{→}{u}}_1â‹\dots {\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1∥=\sqrt{1+49+49+1} \\ =\sqrt{100} \\ =10 \end{gathered}$$

Now find, ${\stackrel{→}{u}}_2$.

${\stackrel{→}{u}}_2=\frac{1}{10}\left[\begin{array}{c}-1\\ 7\\ -7\\ 1\end{array}\right]$

Thus, the values of ${\stackrel{→}{u}}_1$,${\stackrel{→}{u}}_2$ are $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]$.