91Ó°ÊÓ

Consider a set$W=span\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$ and$\stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$ is perpendicular to W .

If the vector$\stackrel{\mathbf{→}}{\mathbf{x}}$is perpendicular $\stackrel{\mathbf{→}}{\mathbf{y}}$then$\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{=}\mathbf{0}$ .

As is perpendicular$\stackrel{→}{x}$ to , by the definition of orthogonal$W.\stackrel{→}{x}=0$ .

Simplify the equation$\stackrel{→}{x}.\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0$ as follows.

$$\begin{gathered} \stackrel{→}{x}.W=0 \\ \left[\begin{array}{cccc}{\stackrel{→}{x}}_1& {\stackrel{→}{x}}_2& {\stackrel{→}{x}}_3& {\stackrel{→}{x}}_4\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0 \\ \begin{array}{cccc}{\stackrel{→}{x}}_1+& 2{\stackrel{→}{x}}_2+& 3{\stackrel{→}{x}}_3+& 4{\stackrel{→}{x}}_4\end{array}=0 \end{gathered}$$

Simplify the equation$\stackrel{→}{x}.\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]=0$ as follows.

$$\begin{gathered} \left[\begin{array}{cccc}{\stackrel{→}{x}}_1& {\stackrel{→}{x}}_2& {\stackrel{→}{x}}_3& {\stackrel{→}{x}}_4\end{array}\right].W=0 \\ \left[\begin{array}{cccc}{\stackrel{→}{x}}_1& {\stackrel{→}{x}}_2& {\stackrel{→}{x}}_3& {\stackrel{→}{x}}_4\end{array}\right]\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]=0 \\ \begin{array}{cccc}5{\stackrel{→}{x}}_1+& 6{\stackrel{→}{x}}_2+& 7{\stackrel{→}{x}}_3+& 8{\stackrel{→}{x}}_4\end{array}=0 \end{gathered}$$

Simplify the equations $5{\stackrel{→}{x}}_1+6{\stackrel{→}{x}}_2+7{\stackrel{→}{x}}_3+8{\stackrel{→}{x}}_4=0$ and${\stackrel{→}{x}}_1+2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0$ as follows.

role="math" localid="1659928785801" $$\begin{gathered} {\stackrel{→}{x}}_1+2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0 \\ -\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}={\stackrel{→}{x}}_1 \end{gathered}$$

Substitute the value$\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}$ for${\stackrel{→}{x}}_1$ in the equation$5{\stackrel{→}{x}}_1+6{\stackrel{→}{x}}_2+7{\stackrel{→}{x}}_3+8{\stackrel{→}{x}}_4=0$ as follows.

$$\begin{gathered} 5{\stackrel{→}{x}}_1+6{\stackrel{→}{x}}_2+7{\stackrel{→}{x}}_3+8{\stackrel{→}{x}}_4=0 \\ 5.-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}+6{\stackrel{→}{x}}_2+7{\stackrel{→}{x}}_3+8{\stackrel{→}{x}}_4=0 \\ -\left\{10{\stackrel{→}{x}}_2+15{\stackrel{→}{x}}_3+20{\stackrel{→}{x}}_4\right\}+6{\stackrel{→}{x}}_2+7{\stackrel{→}{x}}_3+8{\stackrel{→}{x}}_4=0 \\ -4{\stackrel{→}{x}}_2-8{\stackrel{→}{x}}_3-12{\stackrel{→}{x}}_4=0 \end{gathered}$$

Simplify the equation as follows:

$$\begin{gathered} -4{\stackrel{→}{x}}_2-8{\stackrel{→}{x}}_3-12{\stackrel{→}{x}}_4=0 \\ -{\stackrel{→}{x}}_2-{\stackrel{→}{x}}_3-{\stackrel{→}{x}}_4=0 \\ -2{\stackrel{→}{x}}_3-3{\stackrel{→}{x}}_4={\stackrel{→}{x}}_2 \end{gathered}$$

Substitute the valuerole="math" localid="1659929137563" $-2{\stackrel{→}{x}}_3-3{\stackrel{→}{x}}_4$ for${\stackrel{→}{x}}_2$ in the equation${\stackrel{→}{x}}_1+2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0$ as follows:

$$\begin{gathered} {\stackrel{→}{x}}_1+2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0 \\ {\stackrel{→}{x}}_1+2\left\{-2{\stackrel{→}{x}}_3+3{\stackrel{→}{x}}_4\right\}+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0 \\ {\stackrel{→}{x}}_1+\left\{-4{\stackrel{→}{x}}_3+6{\stackrel{→}{x}}_4\right\}+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0 \\ {\stackrel{→}{x}}_1+{\stackrel{→}{x}}_3+2{\stackrel{→}{x}}_4=0 \end{gathered}$$

Further, simplify the equation as follows:

$$\begin{gathered} {\stackrel{→}{x}}_1+{\stackrel{→}{x}}_3+2{\stackrel{→}{x}}_4=0 \\ {\stackrel{→}{x}}_1={\stackrel{→}{x}}_3+2{\stackrel{→}{x}}_4 \end{gathered}$$

Substitute the values$-2{\stackrel{→}{x}}_3-3{\stackrel{→}{x}}_4$ for${\stackrel{→}{x}}_2$ and${\stackrel{→}{x}}_3+2{\stackrel{→}{x}}_4$ for${\stackrel{→}{x}}_1$ in the equation$\stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$ as follows:

$$\begin{gathered} \stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right] \\ \stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_3+2{\stackrel{→}{x}}_4\\ -2{\stackrel{→}{x}}_3-3{\stackrel{→}{x}}_4\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right] \\ \stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_3\\ -2{\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_3\\ 0\end{array}\right]+\left[\begin{array}{c}2{\stackrel{→}{x}}_4\\ -3{\stackrel{→}{x}}_4\\ 0\\ {\stackrel{→}{x}}_4\end{array}\right] \\ \stackrel{→}{x}={\stackrel{→}{x}}_3\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right]+{\stackrel{→}{x}}_4\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right] \end{gathered}$$

Therefore, the orthogonal basis is$\stackrel{→}{x}=span\left\{\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right]\right\}$ .

Hence, the orthogonal basis is$\stackrel{→}{x}=span\left\{\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ -3\\ 0\\ 1\end{array}\right]\right\}$ for$W=span\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right]\right\}$ .