魅影直播

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

Then

$\mathit{u}{\mathbf{鈨�}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{\mathbf{鈨�}}_{\mathbf{1}}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{\mathbf{鈨�}}_{\mathbf{1}}\mathbf{,}\mathit{u}{\mathbf{鈨�}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{鈨�}}_{\mathbf{2}}}^{\mathbf{鈯�}}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{鈨�}}_{\mathbf{2}}}^{\mathbf{鈯�}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{u}{\mathbf{鈨�}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{鈨�}}_{\mathbf{j}}}^{\mathbf{鈯�}}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{鈨�}}_{\mathbf{j}}}^{\mathbf{鈯�}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{u}{\mathbf{鈨�}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}\mathbf{v}{{\mathbf{鈨�}}_{\mathbf{m}}}^{\mathbf{鈯�}}\mathbf{\left|}\mathbf{\right|}}\mathit{v}{{\mathbf{鈨�}}_{\mathbf{m}}}^{\mathbf{鈯�}}$

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

Obtain the values of , according to the Gram-Schmidt process.

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

Find $u{鈨�}_1$.

$$\begin{gathered} u{鈨�}_1=\frac{1}{\sqrt{5^2+4^2+2^2+2^2}}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ =\frac{1}{7}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \end{gathered}$$

Now, here is need to find out the values of $v{鈨�}_2-(u{鈨�}_1.v{鈨�}_2)u{鈨�}_{1}and\left|\right|v{鈨�}_2-(u{鈨�}_1.v{鈨�}_2)u{鈨�}_1\left|\right|$to obtain the value of $u{鈨�}_2$.

Consider the equations below.

role="math" localid="1659441637408" $$\begin{gathered} \stackrel{鈫�}{u_1}路\stackrel{鈫�}{v_1}=7 \\ \left(\stackrel{鈫�}{u_1}路\stackrel{鈫�}{v_1}\right)\stackrel{鈫�}{u_1}=\left[\begin{array}{c}5\\ 4\\ 2\\ 3\end{array}\right] \\ \stackrel{鈫�}{v_1}-\left(\stackrel{鈫�}{u_1}路\stackrel{鈫�}{v_1}\right)\stackrel{鈫�}{u_1}=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]-\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right] \\ =\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right] \end{gathered}$$

Then,

$$\begin{gathered} ||\stackrel{鈫�}{v_1}-\left(\stackrel{鈫�}{u_1}路\stackrel{鈫�}{v_1}\right)\stackrel{鈫�}{u_1}||=\sqrt{4+4+25+16} \\ =\sqrt{49} \\ =7 \end{gathered}$$

Now find role="math" localid="1659441716297" $\stackrel{鈫�}{u_2}$.

$\stackrel{鈫�}{u_2}=\frac{1}{7}\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]$

Thus, the values of $u{鈨�}_1,u{鈨�}_2$are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.