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

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

For all elements f,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$\mathbf{ker}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$ and$\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{W}$ or $\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{dm}\mathbf{\left(}\mathbf{W}\mathbf{\right)}$.

The given transformation as follows:

$\mathrm{T}\left(\mathrm{M}\right)=\mathrm{M}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{M}$, from${\mathrm{R}}^{2脳2}$to ${\mathrm{R}}^{2脳2}$.

By using the definition of linear transformation as follows.

role="math" localid="1659876823767" $$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \\ \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kT}\left(\mathrm{A}\right) \end{gathered}$$

Now, to check the first condition as follows:

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left(\mathrm{A}+\mathrm{B}\right)\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\left(\mathrm{A}+\mathrm{B}\right) \\ =\mathrm{A}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]+\mathrm{B}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{A}-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{B} \\ =\mathrm{A}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{A}+\mathrm{B}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{B} \\ =\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \end{gathered}$$

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.

$$\begin{gathered} \mathrm{T}\left(\mathrm{伪础}\right)=\mathrm{伪础}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{伪础} \\ =\mathrm{伪}\left(\mathrm{A}\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& \mathrm{k}\end{array}\right]\mathrm{A}\right) \\ =\mathrm{伪罢}\left(\mathrm{A}\right) \end{gathered}$$

Thus,T is a linear transformation.

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

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

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

Consider a matrix A as follows.

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

The next equation as follows:

$$\begin{gathered} T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& k\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}5a+0b& 0a+1b\\ 5c+0d& 0c+1d\end{array}\right]-\left[\begin{array}{cc}2a+0c& 2b\\ 0a+kc& 0d+kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}5a+0b-2a-0c& 0a+1b-2b\\ 5c+0d-0a+kc& 0c+1d-0d+kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} \left[\begin{array}{cc}5a+0b-2a-0c& 0a+1b-2b\\ 5c+0d-0a+kc& 0c+1d-0d+kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}5a-2a-0c& -b\\ 5c-kc& d-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}3a& -b\\ \left(5-k\right)c& \left(1-k\right)d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Equating the corresponding entries as follows.

$a=0$and$\mathrm{b}=0$

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.

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

For k=5 the solution as follows.

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

Therefore,T to be an isomorphism for $\mathrm{k}鈮�5$.

Similarly, for T to be an isomorphism for $\mathrm{k}鈮�1$.

Thus, T is a linear transformation and is an isomorphism when $\mathrm{k}鈭�\mathrm{R}-\{1,5\}$.