91Ó°ÊÓ

Consider a basis of a subspace Vof${\mathit{R}}^{\mathbf{n}}$ for$\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{…}\mathbf{,}\mathit{m}$ 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

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

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

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

Find ${\stackrel{→}{u}}_1$.

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{\sqrt{100}}\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ =\frac{1}{\sqrt{10}}\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \end{gathered}$$

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 ${\stackrel{→}{u}}_2$.

$$\begin{gathered} {\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2=10 \\ \left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ \left|\left|{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|=\sqrt{2} \end{gathered}$$

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

role="math" localid="1659952093244" $$\begin{gathered} {\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2=10 \\ \left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ \left|\left|{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|=\sqrt{2} \end{gathered}$$

Thus,

${\stackrel{→}{u}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]$

Now, consider the equation below to obtain the value ofin 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 obtain the value of ${\stackrel{→}{u}}_3$.

$$\begin{gathered} {\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3=10 \\ \left({\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1=1/2\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{→}{v}}_3.{\stackrel{→}{u}}_2=0 \\ \left({\stackrel{→}{u}}_2·{\stackrel{→}{v}}_3\right)=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right] \end{gathered}$$

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

$$\begin{gathered} {\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2=\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]-\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right] \\ ⇒\left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2\right|\right|=\sqrt{2} \end{gathered}$$

Now, find ${\stackrel{→}{u}}_3$.

${\stackrel{→}{u}}_3=\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$

Thus, the values of ${\stackrel{→}{u}}_1$,${\stackrel{→}{u}}_2$ and${\stackrel{→}{u}}_3$ are $1/10\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$.

Hence, the orthonormal vectors are $1/10\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$.