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Consider the function $T\left(\mathrm{M}\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$from $U^{2脳2}$to $U^{2脳2}$.

A function D is called a linear transformation onrole="math" localid="1659927484829" $\mathbf{鈩�}$if the functionD satisfies the following properties.

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

(b) $\mathit{D}\mathbf{\left(}\mathit{伪}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{伪}\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}$for all constant$\mathbf{}\mathit{伪}\mathbf{鈭�}\mathbf{鈩�}$.

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 $M,N鈭�U^{2脳2}$then$T\left(\mathrm{M}\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$and $T\left(\mathrm{N}\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}N\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$.

Substitute the value${\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$for $T\left(M\right)$and${\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}N\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$for $T\left(N\right)$in $T\left(M\right)+T\left(N\right)$as follows.

$$\begin{gathered} T\left(\mathrm{M}\right)+T\left(N\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]+{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}N\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \\ =\left\{{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}N\right\}\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \\ ={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left\{M+N\right\}\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \end{gathered}$$

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

$$\begin{gathered} T\left(M+N\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left\{M+N\right\}\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \\ =T\left(M\right)+T\left(N\right) \\ =T\left(M\right)+T\left(N\right) \end{gathered}$$

Assume$M鈭�U^{2脳2}$ and$伪鈭�\mathrm{鈩�}$ then $T\left(伪M\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left(伪M\right)\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$.

Simplify the equation $T\left(伪M\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left(伪M\right)\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$as follows.

$$\begin{gathered} T\left(伪M\right)={\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left(伪M\right)\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right] \\ =伪\left\{{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}M\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\right\} \\ =伪\left\{T\left(M\right)\right\} \end{gathered}$$

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

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

$$\begin{gathered} T\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\right\}=\left\{{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\right\} \\ =\frac{1}{3}\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{cc}3& -3\\ 0& 0\end{array}\right] \\ =3脳\frac{1}{3}\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

role="math" localid="1659928617985" $$\begin{gathered} T\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\right\}=3脳\frac{1}{3}\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right] \\ =\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right] \\ T\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\right\}=1.{渭}_1+0.{渭}_2+0.{渭}_3 \end{gathered}$$

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

$$\begin{gathered} T\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\right\}=\left\{{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\right\} \\ =\frac{1}{3}\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}0& 3\\ 0& 3\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{cc}0& 3\\ 0& 3\end{array}\right] \\ T\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]\right\}=3脳\frac{1}{3}\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} T\left\{\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\right\}=3脳\frac{1}{3}\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right] \\ T\left\{\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\right\}=0.{渭}_1+1.{渭}_2+0.{渭}_3 \end{gathered}$$

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

$$\begin{gathered} T\left\{\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\right\}=\left\{{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\right\} \\ =\frac{1}{3}\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{cc}0& 9\\ 0& 0\end{array}\right] \\ T\left\{\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right\}=9脳\frac{1}{3}\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} T\left\{\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right\}=9脳\frac{1}{3}\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right] \\ =3\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right] \\ T\left\{\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right\}=0.{渭}_1+0.{渭}_2+3.{渭}_3 \end{gathered}$$

Therefore, the matrix $B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 3\end{array}\right]$.

As all the rows and column are linear independent implies $rank\left(T\right)=3$.

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

$$\begin{gathered} \dim \left(T\right)=\dim \left(\mathrm{kernerl}\left(T\right)\right)+rank\left(T\right) \\ 4=\dim \left(\mathrm{kernel}\left(\mathrm{T}\right)\right)+3 \\ \dim \left(kernel\left(T\right)\right)=1 \end{gathered}$$

Therefore, the dimension of $\mathrm{kernel}\left(T\right)$is and $rank\left(T\right)$is 3 and the basis $rank\left(T\right)$is$\left\{\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right\}$.

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

As the dimension of$\mathrm{kernel}\left(T\right)$ is , by the theorem the functionT is not an isomorphism.

Hence, the rank of transformationT is 3, kernel of the functionT is 1 and linear transformation is not an isomorphism.