91影视

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

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

For all elements $\mathbf{f}\mathbf{,}\mathbf{g}$of V and k is scalar.

A linear transformation$\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$ is said to be an isomorphism if and only if$\mathbf{ker}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\left\{0\right\}$ and$\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{W}$ or $\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{W}\mathbf{\right)}$.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{M}-\mathrm{M}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]$, from${\mathrm{R}}^{2脳2}$to ${\mathrm{R}}^{2脳2}$.

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\left(\mathrm{A}+\mathrm{B}\right)\left(\mathrm{A}-\mathrm{B}\right)\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right] \\ =\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{A}+\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{B}-\mathrm{A}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]-\mathrm{B}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right] \\ =\left[\begin{array}{cc}2& 3\\ 0& \mathrm{k}\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{B}-\mathrm{B}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \end{gathered}$$

Similarly, to check the second condition as follows.

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

$$\begin{gathered} T\left(伪A\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]伪A-伪A\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right] \\ =伪\left(\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]A-A\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]\right) \\ \mathrm{T}\left(\mathrm{伪础}\right)=\mathrm{伪罢}\left(\mathrm{A}\right) \end{gathered}$$

Thus,T is a linear transformation.

A linear transformation $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$is isomorphism if and only if $\mathbf{ker}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{0\right\}$and localid="1659426664071" $\mathbf{Im}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{W}\mathbf{}$

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

$$\begin{gathered} T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]-\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}2a+3c& 2b+3d\\ 0a+4c& 0d+4d\end{array}\right]-\left[\begin{array}{cc}3a+0b& 0a+kb\\ 3c+0d& 0c+kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}2a+3c& -3a-0d\\ 0a+4c& -3c-3d\end{array}\begin{array}{cc}2b+3d& -0a-kd\\ 0b+4d& -0c-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \left[\begin{array}{cc}2a+3c-3a-0d& 2b+3d-0a-kb\\ 0a+4c-3c-0d& 0d+4d-0c-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}a+3c& 2b-kb+3d\\ c& 4d-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}-a+3c& \left(2-k\right)b+3d\\ c& (4-k)d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Equating the corresponding entries as follows.

$c=0$and$a+3c=0$

Solve and find the values as follows.

$a=0,c=0$

Substitute the value 0 for and 0 for in $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$as follows.

$\begin{array}{l}A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}0& b\\ 0& d\end{array}\right]\end{array}$

For$k=4$the solution as follows.

$$\begin{gathered} T\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ ker\left(T\right)鈮�\left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\} \\ ker\left(T\right)鈮�\left\{0\right\} \end{gathered}$$

Therefore,T to be an isomorphism for $k鈮�4$.

Similarly, for T to be an isomorphism for $\mathrm{k}鈮�2$.

Thus, is a linear transformation and is an isomorphism when$k鈭�R-\left\{2,4\right\}$