91Ó°ÊÓ

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

role="math" localid="1660115689227" ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{v}}}^{\mathbf{0}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{v}}}^{\mathbf{âŠ\yen }}$, with respect to span(${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{…}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$ ).

Then ${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_1\right|\right|}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_2^{âŠ\yen }\right|\right|}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_j^{âŠ\yen }\right|\right|}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_m^{âŠ\yen }\right|\right|}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}^{\mathbf{âŠ\yen }}$ is an orthonormal basis of V. Then by the theorem 5.1.7 we have, as follows.

${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}^{\mathbf{║}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{-}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}^{\mathbf{║}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}\mathbf{-}\left({\stackrel{→}{u}}_1×{\stackrel{→}{v}}_1\right){\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{-}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{-}\left({\stackrel{→}{u}}_{j-1}×{\stackrel{→}{v}}_j\right){\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$.

Consider the basis of ${\mathrm{R}}^2$as follows.

${\stackrel{→}{v}}_1=\left[\left.\begin{array}{c}-3\\ 4\end{array}\right],{\stackrel{→}{v}}_2=\left[\begin{array}{c}1\\ 7\end{array}\right.\right]$

Now, to find $\left|\left|{\stackrel{→}{v}}_1\right|\right|$as follows.

$$\begin{gathered} \left|\left|{\stackrel{→}{v}}_1\right|\right|=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2} \\ \left|\left|{\stackrel{→}{v}}_1\right|\right|=\sqrt{9+16} \\ =\sqrt{25} \\ \left|\left|{\stackrel{→}{v}}_1\right|\right|=5 \end{gathered}$$

Now, to find ${\stackrel{→}{u}}_1$as follows.

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{\left|\left|{\stackrel{→}{v}}_1\right|\right|}{\stackrel{→}{v}}_1 \\ =\frac{1}{5}\left[\begin{array}{c}-3\\ 4\end{array}\right] \\ {\stackrel{→}{u}}_1=\frac{1}{5}\left[\begin{array}{c}-3\\ 4\end{array}\right] \end{gathered}$$

Now, to find ${\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2}{\left|\left|{\stackrel{→}{v}}_2\right|\right|}$as follows.

$$\begin{gathered} u_2=\left[\begin{array}{c}1\\ 7\end{array}\right]-pro{j}_{u_1}\left(v_2\right) \\ =\left[\begin{array}{c}1\\ 7\end{array}\right]-\stackrel{\text{'}}{a}\left[\begin{array}{c}-3\\ 4\end{array}\right]\left[\begin{array}{c}1\\ 7\end{array}\right]\tilde{n}\left[\begin{array}{c}-3\\ 4\end{array}\right] \\ =\left[\begin{array}{c}1\\ 7\end{array}\right]-\left[\begin{array}{c}9\\ 112\end{array}\right] \\ {u}_2=\left[\begin{array}{c}-8\\ -105\end{array}\right] \end{gathered}$$

Simplify further as follows.

$u_2=\left[\begin{array}{c}\frac{-8}{\sqrt{11089}}\\ \frac{-105}{\sqrt{11089}}\end{array}\right]$

Orthonormal basis as follows.

$\left\{\left[\begin{array}{c}\frac{-3}{5}\\ \frac{4}{5}\end{array}\right],\left[\begin{array}{c}\frac{-8}{\sqrt{11089}}\\ \frac{-105}{\sqrt{11089}}\end{array}\right]\right\}$

Hence, the length of $\stackrel{→}{v}$ is $\sqrt{170}$.