91影视

Consider the transformation $T:{\mathrm{鈩�}}^n鈫�{\mathrm{鈩�}}^n$ defined as $T\left(\stackrel{鈫�}{x}\right)={\stackrel{鈫�}{x}}^{\left|\right|}=\underset{i=1}{\overset{}{鈭�}}\left({\stackrel{鈫�}{u}}_i路\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_i$ where, $\left\{{\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_2,...,{\stackrel{鈫�}{u}}_m\right\}$.

If the vector $\stackrel{鈫�}{x}$ is perpendicular $\stackrel{鈫�}{y}$ then $\stackrel{鈫�}{x}路\stackrel{鈫�}{y}=0$.

Assume $\stackrel{鈫�}{x},\stackrel{鈫�}{y}鈭�{\mathrm{鈩�}}^n$ and $伪,尾鈭�\mathrm{鈩�}$, simplify $T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)={\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)}^{\left|\right|}$ as follows.

$\begin{array}{rcl}T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& {\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)}^{\left|\right|}\\ T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& \underset{i=1}{\overset{}{鈭�}}\left({\stackrel{鈫�}{u}}_i路\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)\right){\stackrel{鈫�}{u}}_i\\ & =& \underset{i=1}{\overset{}{鈭�}}\left({\stackrel{鈫�}{u}}_i路伪\stackrel{鈫�}{x}+{\stackrel{鈫�}{u}}_i路尾\stackrel{鈫�}{y}\right){\stackrel{鈫�}{u}}_i\\ T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& \underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路伪\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_i+\left({\stackrel{鈫�}{u}}_i路尾\stackrel{鈫�}{y}\right){\stackrel{鈫�}{u}}_i\right)\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& \underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路伪\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_i+\left({\stackrel{鈫�}{u}}_i路尾\stackrel{鈫�}{y}\right){\stackrel{鈫�}{u}}_i\right)\\ T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& \underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路伪\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_i\right)+\underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路尾\stackrel{鈫�}{y}\right){\stackrel{鈫�}{u}}_i\right)\\ T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& 伪\left\{\underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路\stackrel{鈫�}{x}\right){\stackrel{鈫�}{u}}_i\right)\right\}+尾\left\{\underset{i=1}{\overset{}{鈭�}}\left(\left({\stackrel{鈫�}{u}}_i路\stackrel{鈫�}{y}\right){\stackrel{鈫�}{u}}_i\right)\right\}\\ T\left(伪\stackrel{鈫�}{x}+尾\stackrel{鈫�}{y}\right)& =& 伪\left\{T\left(\stackrel{鈫�}{x}\right)\right\}+尾\left\{T\left(\stackrel{鈫�}{y}\right)\right\}\\ & & \end{array}$

By the definition of linear transformation, the transformation $T$is linear.

Hence, the transformation $T:{\mathrm{鈩�}}^n鈫�{\mathrm{鈩�}}^n$with respect to the basis $\left\{{\stackrel{鈫�}{u}}_1,{\stackrel{鈫�}{u}}_2,...,{\stackrel{鈫�}{u}}_m\right\}$is linear.