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

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

For all elements of and is scalar.

An invertible linear transformation is called an isomorphism.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)\mathrm{M}+{\mathrm{I}}_2\mathrm{from}{\mathrm{R}}^{2脳2}\mathrm{to}{\mathrm{R}}^{2脳2}$

By using the definition of linear transformation as follows.

$$\begin{gathered} \mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \\ \mathrm{A}+\mathrm{B}+{\mathrm{I}}_2=\mathrm{A}+{\mathrm{I}}_2+\mathrm{B}+{\mathrm{I}}_2 \end{gathered}$$

Consider the matrix for and as follows.

$$\begin{gathered} A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right] \\ B=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Substitute the matrix $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$for A and $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$for B in $\mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{A}+\mathrm{B}+{\mathrm{I}}_2$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{A}+\mathrm{B}+{\mathrm{I}}_2 \\ =\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right] \end{gathered}$$

Now, simplify for $\mathrm{T}(\mathrm{A}+\mathrm{B})\mathrm{and}\mathrm{A}+\mathrm{B}+{\mathrm{I}}_2$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{A}+{\mathrm{I}}_2 \\ =\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Similarly, simplify for $\mathrm{T}\left(\mathrm{B}\right)=\mathrm{B}+{\mathrm{I}}_2$as follows.

role="math" localid="1659842505826" $$\begin{gathered} \mathrm{T}\left(\mathrm{B}\right)=\mathrm{B}+{\mathrm{I}}_2 \\ =\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{T}\left(\mathrm{B}\right)=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right] \end{gathered}$$

Then adding the transformations as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=\mathrm{A}+{\mathrm{I}}_2+\mathrm{B}+{\mathrm{I}}_2 \\ \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \end{gathered}$$

Now, substitute the value $\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$for T (A + B) and $\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$for T (A) + T (B) in the transformation $\mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$as follows.

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

Therefore, the transformation is $\mathrm{T}(\mathrm{A}+\mathrm{B})鈮�\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$.

Hence, the transformation is not linear.