91Ó°ÊÓ

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

localid="1659441812648" $$\begin{gathered} \mathbf{Then} \\ {\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{â‡\P Ä}{v}}_1||}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{â‡\P Ä}{v}}_1}^{âŠ\yen }||}{{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_2}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{â‡\P Ä}{v}}_j}^{âŠ\yen }||}{{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_2}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{â‡\P Ä}{v}}_j}^{âŠ\yen }||}{{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{1}}}^{\mathbf{âŠ\yen }} \end{gathered}$$

The given vectors are, ${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{1}}=\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{1}}=\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right].$.

Obtain the values of ${\stackrel{â‡\P Ä}{u}}_1,{\stackrel{â‡\P Ä}{u}}_2$ according to the Gram-Schmidt process.

$$\begin{gathered} {\stackrel{â‡\P Ä}{u}}_1=\frac{{\stackrel{â‡\P Ä}{\mathrm{v}}}_1}{||\stackrel{â‡\P Ä}{\mathrm{v}}||}....\left(1\right) \\ {\stackrel{â‡\P Ä}{u}}_2=\frac{{\stackrel{â‡\P Ä}{\mathrm{v}}}_2-\left({\stackrel{â‡\P Ä}{u}}_1-{\stackrel{â‡\P Ä}{\mathrm{v}}}_2\right){\stackrel{â‡\P Ä}{u}}_1}{||{\stackrel{â‡\P Ä}{\mathrm{v}}}_2-\left({\stackrel{â‡\P Ä}{u}}_1-{\stackrel{â‡\P Ä}{\mathrm{v}}}_2\right){\stackrel{â‡\P Ä}{u}}_1||}.....\left(2\right) \end{gathered}$$

localid="1659441845864" $$\begin{gathered} \mathrm{Find}. \\ {\stackrel{â‡\P Ä}{\mathrm{u}}}_1=\frac{1}{\sqrt{4^2+0^2+0^2+3^2}}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right] \\ =\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right] \\ \mathrm{Next},\mathrm{find}{\stackrel{â‡\P Ä}{\mathrm{u}}}_2. \\ {\stackrel{â‡\P Ä}{\mathrm{u}}}_2=\frac{1}{15}\left(\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]-\left[\begin{array}{c}8\\ 0\\ 0\\ 6\end{array}\right]\right) \\ {\stackrel{â‡\P Ä}{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{values}\mathrm{of}{\stackrel{â‡\P Ä}{\mathrm{u}}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{â‡\P Ä}{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{orthonormal}\mathrm{vectors}\mathrm{are}{\stackrel{â‡\P Ä}{\mathrm{u}}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{â‡\P Ä}{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \end{gathered}$$