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

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

Let the given vectors are, ${\stackrel{→}{v}}_1=\left[\left.\begin{array}{c}4\\ 0\\ 3\end{array}\right],{\stackrel{→}{v}}_2=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right.\right]$.

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

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

Since,

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{\sqrt{4^2+0^2+3^2}}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ =\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \end{gathered}$$

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

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 $\left|\left|{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|$as below:

$$\begin{gathered} {\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right].\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right] \\ ⇒\frac{1}{5}\left(100-75\right)=5 \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1=\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-5.\frac{1\text{'}}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ =\left[\begin{array}{c}25\\ 0\\ -25\end{array}\right]-\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right] \\ =\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ ⇒\left|\left|{\stackrel{→}{v}}_2-\left({\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|=\sqrt{{21}^2+0^2+{\left(-28\right)}^2} \\ =\sqrt{1225} \\ =35 \end{gathered}$$

Then,

$$\begin{gathered} {\stackrel{→}{u}}_2=\frac{1}{35}\left[\begin{array}{c}21\\ 0\\ -28\end{array}\right] \\ =\left[\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right] \end{gathered}$$

Thus, the required vectors are, ${\stackrel{→}{u}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 3\end{array}\right]$ and ${\stackrel{→}{u}}_2=\left[\begin{array}{c}\frac{3}{5}\\ 0\\ \frac{4}{5}\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}4/5\\ 0\\ 3/5\end{array}\right],\left[\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right]$.