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Consider the function $T\left(M\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M$from $R^{2脳2}\mathrm{to}{R}^{2脳2}$.

A function D is called a linear transformation on $\mathbf{鈩�}$if the function D satisfies the following properties.

1. $\mathit{D}\left(x+y\right)\mathbf{=}\mathit{D}\left(x\right)\mathbf{+}\mathit{D}\left(y\right)$for all$\mathit{x}\mathbf{,}\mathit{y}\mathbf{}\mathbf{鈭�}\mathbf{}\mathbf{鈩�}$ .
2. $\mathit{D}\left(伪x\right)\mathbf{=}\mathit{伪}\mathit{D}\left(x\right)$for all constant$\mathit{伪}\mathbf{}\mathbf{鈭�}\mathbf{}\mathbf{鈩�}$ .

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$M,N鈭�R^{2脳2}$ thenrole="math" localid="1665137642901" $T\left(M\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M$ and $T\left(N\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]N$.

Substitute the value$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M$ for$T\left(M\right)$ and$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]N$ for$T\left(N\right)$ in$T\left(M\right)+T\left(N\right)$ as follows.

$T\left(M\right)+T\left(N\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M+\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]N$

Now, simplify$T\left(M+N\right)$ as follows.

$$\begin{gathered} T\left(M+N\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]\left(M+N\right) \\ =\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M+\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]N \\ =T\left(M\right)+T\left(N\right) \end{gathered}$$

Assume$M鈭�R^{2脳2}$ and$伪鈭�\mathrm{鈩�}$ then $T\left(伪M\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]伪M$.

Simplify the equation$T\left(伪M\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]伪M$ as follows.

$$\begin{gathered} T\left(伪M\right)=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]伪M \\ =伪\left\{\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]M\right\} \\ =伪\left\{T\left(M\right)\right\} \end{gathered}$$

As$T\left(M+N\right)=T\left(M\right)+T\left(N\right)$ and $T\left(伪M\right)=伪\left\{T\left(M\right)\right\}$, by the definition of linear transformation$T$ is linear.

As$\left\{\left[\begin{array}{cc}1& 0\\ -1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& -1\end{array}\right],\left[\begin{array}{cc}1& 0\\ 2& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 2\end{array}\right]\right\}$ is the basis element of$R^{2脳2}$ , the matrix$B$ is defined as follows.

$\begin{array}{r}T\left\{\left[\begin{array}{ll}1& 0\\ 0& 0\end{array}\right]\right\}=\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 鈭�1& 0\end{array}\right]\\ =\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ =0鈰�u_1+0鈰�u_2+0鈰�u_3+0鈰�u_4\end{array}$

The image of T at point $\left[\begin{array}{ll}0& 1\\ 0& -1\end{array}\right]$is defined as follows.

$\begin{array}{r}T\left\{\left[\begin{array}{ll}0& 1\\ 0& 0\end{array}\right]\right\}=\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 鈭�1\end{array}\right]\\ =\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ =0鈰�u_1+0鈰�u_2+0鈰�u_3+0鈰�u_4\end{array}$

The image of T at point$\left[\begin{array}{ll}1& 0\\ 2& 0\end{array}\right]$ is defined as follows.

$\begin{array}{r}T\left\{\left[\begin{array}{ll}0& 0\\ 1& 0\end{array}\right]\right\}=\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\left[\begin{array}{ll}1& 0\\ 2& 0\end{array}\right]\\ =\left[\begin{array}{ll}3& 0\\ 6& 0\end{array}\right]\\ =3\left[\begin{array}{ll}1& 0\\ 2& 0\end{array}\right]\\ =0鈰�u_1+0鈰�u_2+3鈰�u_3+0鈰�u_4\end{array}$

The image of T at point is defined as follows.

$\begin{array}{r}T\left\{\left[\begin{array}{ll}0& 0\\ 0& 1\end{array}\right]\right\}=\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\left[\begin{array}{ll}0& 1\\ 0& 2\end{array}\right]\\ =\left[\begin{array}{ll}0& 3\\ 0& 6\end{array}\right]\\ =3\left[\begin{array}{ll}0& 1\\ 0& 2\end{array}\right]\\ =0鈰�u_1+0鈰�u_2+0鈰�u_3+3鈰�u_4\end{array}$

Therefore, the matrix $B=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 3\end{array}\right]$.

As two rows are linearly dependent and two rows are zero implies$rank\left(T\right)=2$

The$kernel\left(T\right)$is defined as follows.

$$\begin{gathered} dim\left(T\right)=dim\left(kernel\left(T\right)\right)+rank\left(T\right) \\ 4=dim\left(kernel\left(T\right)\right)+2 \\ dim\left(kernel\left(T\right)\right)=2 \end{gathered}$$

Therefore, the dimension of $kernel\left(T\right)$is 2 and$rank\left(T\right)$ is 2 and the basis$rank\left(T\right)$ is $\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}$.

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

As the dimension of kernel$\left(T\right)$is 2 , by the theorem the function$T$is not an isomorphism.

Hence, the rank of transformation$T$is 2, kernel of the function$T$is 2 and linear transformation is not an isomorphism.