91Ó°ÊÓ

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

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

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

An invertible linear transformation is called an isomorphism.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\mathrm{M}\mathrm{from}{\mathrm{R}}^{2×2}\mathrm{to}{\mathrm{R}}^{2×2}$

By using the definition of linear transformation as follows.

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

Consider the $2×2$matrix for A as follows.

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

Now, simplify for the transformation as follows.

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

Similarly, simplify for second condition as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kA}\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kT}\left(\mathrm{A}\right) \end{gathered}$$

This shows that is T a linear transformation.

Now, to show that T is isomorphism.

That is to solve that T(A) = B for A.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{B} \\ \left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\mathrm{A}=\mathrm{B} \\ \mathrm{A}={\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]}^{-1}\mathrm{B} \\ \mathrm{A}=\frac{1}{4-6}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]\mathrm{B} \end{gathered}$$

Simplify further as follows.

role="math" localid="1659417570457" $\mathrm{A}=\frac{1}{-2}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]\mathrm{B}$

This, shows that the inverse transformation is $\mathrm{T}\left(\mathrm{B}\right)=\frac{1}{2}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]\mathrm{A}$and T is an isomorphism.

Thus, T is a linear transformation and is an isomorphism.