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

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

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.

$\mathbf{T}\mathbf{:}{\mathbf{R}}^{\mathbf{2}\mathbf{脳}\mathbf{2}}\mathbf{鈫�}{\mathbf{R}}^{\mathbf{2}\mathbf{脳}\mathbf{2}}\mathbf{with}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{2}& \mathbf{3}\\ \mathbf{5}& \mathbf{7}\end{array}\mathbf{\right]}\mathbf{A}\mathbf{-}\mathbf{A}\mathbf{\left[}\begin{array}{cc}\mathbf{2}& \mathbf{3}\\ \mathbf{5}& \mathbf{7}\end{array}\mathbf{\right]}$

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{M}-\mathrm{M}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{from}{\mathrm{R}}^{2脳2}\mathrm{to}{\mathrm{R}}^{2脳2}$

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left(\mathrm{A}+\mathrm{B}\right)-\left(\mathrm{A}+\mathrm{B}\right)\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right] \\ =\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{A}+\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{B}-\mathrm{A}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]-\mathrm{B}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right] \\ =\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{B}-\mathrm{B}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right] \end{gathered}$$

$\mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$

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.

$$\begin{gathered} \mathrm{T}\left(\mathrm{伪础}\right)=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{伪础}-\mathrm{伪础}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right] \\ =\mathrm{伪}\left(\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\right) \\ \mathrm{T}\left(\mathrm{伪础}\right)=\mathrm{伪罢}\left(\mathrm{A}\right) \end{gathered}$$

Thus, is a linear transformation.

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

Now, check if ker (t) = {0} as follows.

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

Consider a matrixas follows.

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

The next equation as follows.

$\begin{array}{r}\mathrm{T}\left(\mathrm{A}\right)=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]-\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2\mathrm{a}+3\mathrm{c}& 2\mathrm{b}+3\mathrm{d}\\ 5\mathrm{a}+7\mathrm{c}& 5\mathrm{b}+7\mathrm{d}\end{array}\right]-\left[\begin{array}{cc}2\mathrm{a}+5\mathrm{b}& 3\mathrm{a}+7\mathrm{b}\\ 2\mathrm{c}+5\mathrm{d}& 3\mathrm{c}+7\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2\mathrm{a}+3\mathrm{c}-2\mathrm{a}-5\mathrm{b}& 2\mathrm{b}+3\mathrm{d}-3\mathrm{a}-7\mathrm{b}\\ 5\mathrm{a}+7\mathrm{c}-2\mathrm{c}-5\mathrm{d}& 5\mathrm{b}+7\mathrm{d}-3\mathrm{c}-7\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Simplify further as follows.

$$\begin{gathered} \left[\begin{array}{cc}2\mathrm{a}+3\mathrm{c}-2\mathrm{a}-5\mathrm{b}& 2\mathrm{b}+3\mathrm{d}-3\mathrm{a}-7\mathrm{b}\\ 5\mathrm{a}+7\mathrm{c}-2\mathrm{c}-5\mathrm{d}& 5\mathrm{b}+7\mathrm{d}-3\mathrm{c}-7\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}3\mathrm{c}-5\mathrm{b}& -3\mathrm{a}-5\mathrm{b}-3\mathrm{d}\\ 5\mathrm{a}+5\mathrm{c}+5\mathrm{d}& 5\mathrm{b}-3\mathrm{c}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Equating the corresponding entries as follows.

$$\begin{gathered} 3\mathrm{c}-5\mathrm{b}=0 \\ -3\mathrm{a}-5\mathrm{b}+3\mathrm{d}=0 \\ 5\mathrm{a}+5\mathrm{c}-5\mathrm{d}=0 \\ 5\mathrm{b}-3\mathrm{c}=0 \end{gathered}$$

Solve and find the values as follows.

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

Therefore,T is not an isomorphism.

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