91影视

Consider a set $span\left\{{\stackrel{鈫�}{v}}_1=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],{\stackrel{鈫�}{v}}_2=\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right],{\stackrel{鈫�}{v}}_3=\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\right\}$spanned ${\mathrm{鈩�}}^4$and a vector $\stackrel{鈫�}{x}=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$.

If the vector localid="1659950736693" $\stackrel{\mathbf{鈫�}}{\mathbf{x}}$is perpendicular $\stackrel{\mathbf{鈫�}}{\mathbf{y}}$then $\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{y}}\mathbf{=}\mathbf{0}$.

By the theorem of orthogonal basis, the orthogonal ${\stackrel{鈫�}{u}}_1$is defined as follows:

${\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||}$

Substitute the value $\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$for ${\stackrel{鈫�}{v}}_1$in the equation ${\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||}$as follows:

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||} \\ {\stackrel{鈫�}{u}}_1=\frac{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]||} \\ {\stackrel{鈫�}{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{{\stackrel{鈫�}{v}}_1}{||{\stackrel{鈫�}{v}}_1||} \\ {\stackrel{鈫�}{u}}_1=\frac{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]||} \\ {\stackrel{鈫�}{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

${\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$

Similarly, the orthogonal${\stackrel{鈫�}{u}}_2$ is defined as follows.

${\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||}$

Substitute the value$\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]$ for${\stackrel{鈫�}{v}}_2$ in the equation${\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||}$ as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{{\stackrel{鈫�}{v}}_2}{||{\stackrel{鈫�}{v}}_2||} \\ {\stackrel{鈫�}{u}}_2=\frac{\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]||} \\ {\stackrel{鈫�}{u}}_2=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right] \end{gathered}$$

Similarly, the orthogonal${\stackrel{鈫�}{u}}_3$ is defined as follows.

${\stackrel{鈫�}{u}}_3=\frac{{\stackrel{鈫�}{v}}_3}{||{\stackrel{鈫�}{v}}_3||}$

Substitute the value$\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]$ for${\stackrel{鈫�}{v}}_3$ in the equation${\stackrel{鈫�}{u}}_3=\frac{{\stackrel{鈫�}{v}}_3}{||{\stackrel{鈫�}{v}}_3||}$ as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_3=\frac{{\stackrel{鈫�}{v}}_3}{||{\stackrel{鈫�}{v}}_3||} \\ {\stackrel{鈫�}{u}}_3=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \\ {\stackrel{鈫�}{u}}_3=\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_3=\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \\ {\stackrel{鈫�}{u}}_3=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right] \end{gathered}$$

Therefore, the orthogonal basis is$span\left\{{\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right],{\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right],{\stackrel{鈫�}{u}}_3=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right]\right\}$ .

By the theorem of orthogonal projection, the orthogonal projection of vector$\stackrel{鈫�}{x}$is defined as follows.

$pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2+\left({\stackrel{鈫�}{u}}_3.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_3$

Substitute the value$\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]$for${\stackrel{鈫�}{u}}_1$, $\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]$for${\stackrel{鈫�}{u}}_2$, $\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right]$for${\stackrel{鈫�}{u}}_3$and$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$for$\stackrel{鈫�}{x}$in the equation$pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2+\left({\stackrel{鈫�}{u}}_3.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_3$as follows.

localid="1659952933885" $$\begin{gathered} pro{j}_v\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{u}}_1.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_1+\left({\stackrel{鈫�}{u}}_2.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_2+\left({\stackrel{鈫�}{u}}_3.\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_3 \\ pro{j}_v\stackrel{鈫�}{x}=\left\{\left(\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]+\left(\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right] \\ \left(\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right]\right\} \\ pro{j}_v\stackrel{鈫�}{x}=\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]+\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]+\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \\ pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{cccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \end{gathered}$$

Hence, the orthogonal projection of the vector$\stackrel{鈫�}{x}$ is$pro{j}_v\stackrel{鈫�}{x}=\left[\begin{array}{cccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right]$ .