91影视

Consider the function $T\left(x+iy\right)=x-iyfromCtoC$.

A function D is called a linear transformation on if the function D satisfies the following properties.

1. $\mathbf{D}\left(X+y\right)\mathbf{=}\mathbf{D}\left(x\right)\mathbf{+}\mathbf{D}\left(y\right)$ for all localid="1659498894214" $\mathit{x}\mathbf{,}\mathit{y}\mathbf{鈭�}$.
2. localid="1659500565173" $\mathbf{D}\mathbf{\left(}\mathbf{伪虫}\mathbf{\right)}\mathbf{=}\mathbf{伪顿}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$for all constant $\mathit{伪}\mathbf{鈧�}\mathit{u}$.

An invertible linear transformation is called isomorphism or dimension of domain and co-domain is not same then the function is not an isomorphism.

Assume $x_1+i{y}_1,x_2+i{y}_2鈭�C$then $\mathit{T}\left(x_1+i{y}_1\right)=x_1-i{y}_1$and$\mathit{T}\left(x_2+i{y}_2\right)=x_2-i{y}_2$

Substitute the value $x_1-i{y}_1$for $\mathit{T}\left(x_1+i{y}_1\right)$and $x_2-i{y}_2$for $\mathit{T}\left(x_2+i{y}_2\right)$$\mathit{T}\left(x_1+i{y}_1\right)\mathbf{+}\mathit{T}({\mathrm{x}}_1+{\mathrm{iy}}_1)$in as follows.

$$\begin{gathered} \mathit{T}\left(\left(x_1+i{y}_1\right)+i(y_1+y_2)\right)=\left(x_1+x_2\right)-i\left(y_1+y_2\right) \\ =x_1+x_2-i{y}_1-i{y}_2 \\ =x_1-i{y}_1+x_1-i{y}_2 \\ =\mathit{T}\left(x_1+i{y}_1\right)\mathbf{+}\mathit{T}\left(x_2+i{y}_2\right) \end{gathered}$$

Assume $x+iy鈭�C$and then $T\left(伪\left\{x+iy\right\}\right)=\left(伪x\right)+i\left(伪y\right)$.

Simplify the equation $\mathit{T}\left(伪\left\{x+iy\right\}\right)=\left(伪x\right)+i\left(伪y\right)$as follows.

$$\begin{gathered} \mathit{T}\left(伪\left\{x+iy\right\}\right)=\left(伪x\right)+i\left(伪y\right) \\ =伪\left(x+iy\right) \\ =伪\mathit{T}\left(x+iy\right) \end{gathered}$$

As $\mathit{T}\left(\left(x_1+i{y}_1\right)+i(y_1+y_2)\right)=\mathit{T}\left(x_1+i{y}_1\right)+\mathit{T}\left(x_2+i{y}_2\right)$and $\mathit{T}\left(伪\left\{x+iy\right\}\right)=伪\mathit{T}\left(x+iy\right)$ by the definition of linear transformation T is linear.

By the definition of $L_B$transformation, the transformation of domain and codomain is defined as follows.

$$\begin{gathered} x+iy\stackrel{L_B}{鈫�}\left[\begin{array}{c}x\\ y\end{array}\right] \\ x-iy\stackrel{L_B}{鈫�}\left[\begin{array}{c}x\\ -y\end{array}\right] \end{gathered}$$

As the linear transformation transform $\left[\begin{array}{c}x\\ y\end{array}\right]\stackrel{L_B}{鈫�}\left[\begin{array}{c}x\\ -y\end{array}\right]$and $\left\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\right\}$is the basis element of C ,the matrix B is defined as follows.

$$\begin{gathered} T\left\{\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\right\} \\ T\left\{\left[\begin{array}{c}0\\ 1\end{array}\right]\right\}=\left[\begin{array}{c}0\\ -1\end{array}\right] \end{gathered}$$

Therefore, the matrix

Therefore, the matrix ..

The $\dim \left(\mathrm{kernel}\left(\mathrm{T}\right)\right)$is defined as follows.

$$\begin{gathered} \dim \left(\mathrm{T}\right)=\dim \left(\mathrm{kernel}\left(\mathrm{T}\right)\right)+\mathrm{rank}\left(\mathrm{T}\right) \\ 2=\dim \left(\mathrm{kernel}\left(\mathrm{T}\right)\right)+2 \\ \dim \left(\mathrm{kernel}\left(\mathrm{T}\right)\right)=0 \end{gathered}$$

Therefore, the dimension of $kernel\left(T\right)$is and $rank\left(T\right)$is 2 .

Theorem: Consider a linear transformation $\mathit{T}$defined from $\mathit{T}\mathbf{:}\mathit{V}\mathbf{鈫�}\mathit{W}$$\mathit{T}$then the transformation $\mathit{K}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{0}$is an isomorphism if and only if where .

$\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{鈭�}\mathbf{P}\mathbf{:}\mathbf{T}\mathbf{\left\{}\mathbf{f}\left(x\right)\mathbf{\right\}}\mathbf{=}\mathbf{0}\mathbf{\}}$

As the dimension $kernel\left(T\right)$of is $0$, by the theorem the function $T$is isomorphism.

Hence, the rank of transformation $T$is 2, kernel of the function $T$is $0$and linear transformation is isomorphism.