91影视

Consider two linear spaces$V$and$W$. A function$T$is said to be linear transformation if the following holds.

$\begin{array}{c}T(f+g)=T\left(f\right)+T\left(g\right)\\ T\left(kf\right)=kT\left(f\right)\end{array}$

For all elements$f,g$of$V$and$k$is scalar.

An invertible linear transformation is called an isomorphism.

Let鈥檚 define a transformation as follows.

$T:R^{2脳2}鈫�R^{2脳2}$with$T\left(A\right)=A\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]A$.

The given transformation as follows.

$T\left(M\right)=M\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]M$, from$R^{2脳2}$to$R^{2脳2}$.

By using the definition of linear transformation as follows.

$\begin{array}{c}T(A+B)=T\left(A\right)+T\left(B\right)\\ T\left(kA\right)=kT\left(A\right)\end{array}$

Now, to check the first condition as follows.

Let $A$and$B$be arbitrary matrices from$R^{2脳2}$and as follows.

$\begin{array}{c}T(A+B)=(A+B)\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right](A+B)\\ =A\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]+B\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]A-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]B\\ =A\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]A+B\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]B\\ T(A+B)=T\left(A\right)+T\left(B\right)\end{array}$

Similarly, to check the second condition as follows.

Let$伪$ be an arbitrary scalar, and$A鈭�R^{2脳2}$as follows.

$\begin{array}{c}T\left(伪A\right)=伪A\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]伪A\\ =伪(A\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]A)\\ T\left(伪A\right)=伪T\left(A\right)\end{array}$

Thus,$T$ is a linear transformation.

A linear transformation$T:V鈫�W$is isomorphism if and only if$ker\left(t\right)=\left\{0\right\}$and$Im\left(t\right)=W$

Now, check if$ker\left(t\right)=\left\{0\right\}$as follows.

$ker\left(T\right)=\left\{A鈭�R^{2脳2}|T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}$

Consider a matrix$A$as follows.

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

The next equation as follows.

$\begin{array}{c}T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}a+0& 2a+b\\ c+d& 2c+d\end{array}\right]-\left[\begin{array}{cc}a+2c& b+2d\\ 0+c& 0+d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}a-a-2c& 2a+b-b-2d\\ c-c& 2c+d-d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Simplify further as follows.

$\left[\begin{array}{cc}-2c& 2a-2d\\ 0& 2c\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Equating the corresponding entries as follows.

$\begin{array}{c}2c=0\\ 2a-2d=0\\ 0=0\\ 2c=0\end{array}$

Solve and find the values as follows.

$\begin{array}{l}a=d,c=0,b鈭�R\\ ker\left(T\right)鈮�\left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}\\ ker\left(T\right)鈮�\left\{0\right\}\end{array}$

Therefore,$T$is not an isomorphism.

Thus, $T$is a linear transformation and is not an isomorphism and$ker\left(T\right)鈮�\left\{0\right\}$