91Ó°ÊÓ

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

Then

localid="1659954506203" ${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{→}{v}}_1||}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_2}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_2}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_m}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{âŠ\yen }}$

Let the given vectors are,${\stackrel{→}{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right],{\stackrel{→}{v}}_2=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right],{\stackrel{→}{v}}_3=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$.

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

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

Find ${\stackrel{→}{u}}_1$.

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{\stackrel{→}{v_1}}{||\stackrel{→}{v_1}||} \\ =\frac{1}{\sqrt{2^2+0^2+0^2}}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \end{gathered}$$

As, the formula for${\stackrel{→}{u}}_2$is, ${\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1}{||{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right)\stackrel{→}{u_1}||}$.

Now, find the values of $\stackrel{→}{u_1}.\stackrel{→}{V_2}$, $\stackrel{→}{v_2}-\left(\stackrel{→}{u_1}.\stackrel{→}{v_2}\right){\stackrel{→}{u}}_1$and$||\stackrel{→}{v_2}-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right)\stackrel{→}{u_1}||$as below:

Calculate $\stackrel{→}{u_1}.\stackrel{→}{v_2}$.

$$\begin{gathered} \stackrel{→}{u_1}.\stackrel{→}{v_2}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right] \\ =3+0+0 \\ =3 \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right)\stackrel{→}{u_1}=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]-3.\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]-\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \end{gathered}$$

$$\begin{gathered} ||{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1||=\sqrt{0^2+4^2+0^2} \\ =4 \end{gathered}$$

Now, find ${\stackrel{→}{u}}_2$

$$\begin{gathered} {\stackrel{→}{u}}_2=\frac{1}{4}\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \end{gathered}$$

Now, to obtain the value of ${\stackrel{→}{u}}_3$consider the equation is given below and then put the values obtained in the step 2.

Now find,role="math" localid="1659954011731" $\stackrel{→}{v_3}.\stackrel{→}{u_1}$and $\stackrel{→}{v_3}.\stackrel{→}{u_2}$.

$$\begin{gathered} \stackrel{→}{v_3}.\stackrel{→}{u_1}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =5 \end{gathered}$$

$$\begin{gathered} \stackrel{→}{v_3}-\stackrel{→}{u_2}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =6 \end{gathered}$$

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

$$\begin{gathered} {\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]-\left[\begin{array}{c}5\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 6\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \end{gathered}$$

$$\begin{gathered} ||{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2.{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2||=\sqrt{0^2+0^2+7^2} \\ =7 \end{gathered}$$

Finally, find $\stackrel{→}{u_3}$.

$$\begin{gathered} \stackrel{→}{u_3}=\frac{1}{7}\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right] \end{gathered}$$

Thus, the values of ${\stackrel{→}{u}}_1$,${\stackrel{→}{u}}_2$and ${\stackrel{→}{u}}_3$are ${\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\stackrel{→}{\mathrm{μ}}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],{\stackrel{→}{u}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Hence, the orthonormal vectors are ${\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\stackrel{→}{\mathrm{μ}}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],{\stackrel{→}{u}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$.