91Ó°ÊÓ

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

Then

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

Obtain the value of ${\stackrel{→}{u}}_1$,${\stackrel{→}{u}}_2$ and${\stackrel{→}{u}}_3$ according to Gram-Schmidt process.

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{{\stackrel{→}{v}}_1}{\left|\left|{\stackrel{→}{v}}_1\right|\right|}....\left(1\right) \\ {\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_2\right){\stackrel{→}{u}}_1}{\left|\left|{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_2\right){\stackrel{→}{u}}_1\right|\right|}....\left(2\right) \\ {\stackrel{→}{u}}_3=\frac{{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{u}}_3\right){\stackrel{→}{u}}_2}{\left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{u}}_2\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{u}}_3\right){\stackrel{→}{u}}_2\right|\right|}....\left(3\right) \end{gathered}$$

Find ${\stackrel{→}{u}}_1$.

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

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

$\begin{array}{l}{\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2=1\\ \left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=1/2\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\\ {\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]-\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\\ =1/2\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\end{array}$

Then,

$\left|\left|{\stackrel{→}{v}}_2·\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|=1$

Thus,

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

Now, consider the equation below to obtain the value of${\stackrel{→}{u}}_3$in equation (3).

Here is need to find out the values of${\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2$ and$\left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2\right|\right|$ to find out the value of ${\stackrel{→}{u}}_3$.

${\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3=1$

$$\begin{gathered} \left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1=1/2\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\stackrel{→}{v}}_3.{\stackrel{→}{u}}_2=2 \\ \left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2=1/2\left[\begin{array}{c}-1\\ 1\\ 1\\ -1\end{array}\right] \end{gathered}$$

Next, find ${\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2$and $\left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2\right|\right|$.

$$\begin{gathered} {\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2=\left[\begin{array}{c}0\\ 2\\ 1\\ -1\end{array}\right]-\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right] \\ =\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ 1-/2\end{array}\right] \\ =1/2\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \end{gathered}$$

Then,

$\left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2\right|\right|=1$

Find ${\stackrel{→}{u}}_3$.

${\stackrel{→}{u}}_3=1/2\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]$

Thus, the values of ${\stackrel{→}{u}}_1$,${\stackrel{→}{u}}_2$ and${\stackrel{→}{u}}_3$ are $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right]$.

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