91Ó°ÊÓ

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

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

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

An invertible linear transformation is called an isomorphism.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\mathrm{M}\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\mathrm{from}{\mathrm{R}}^{2×2}\mathrm{to}{\mathrm{R}}^{2×2}$.

By using the definition of linear transformation as follows.

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

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

$A=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]$

Now, simplify for the transformation as follows.

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

Similarly, simplify for second condition as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kA}\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right] \\ \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kT}\left(\mathrm{A}\right) \end{gathered}$$

This shows that is a linear transformation.

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

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

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

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

$A=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]$

Now, set the equation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}\mathrm{a}+3\mathrm{b}& 2\mathrm{a}+6\mathrm{b}\\ \mathrm{c}+3\mathrm{d}& 2\mathrm{c}+6\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Now, equating the corresponding entries as follows.

$\begin{array}{l}\mathrm{a}+3\mathrm{b}=0\\ 2\mathrm{a}+6\mathrm{b}=0\\ \mathrm{c}+3\mathrm{d}=0\\ 2\mathrm{c}+6\mathrm{d}=0\end{array}$

Simplify the 1^st and 3^rd equation as follows.

role="math" localid="1659844040312" $\begin{array}{l}\mathrm{a}+3\mathrm{b}=0\\ \mathrm{a}=-3\mathrm{b}\\ \mathrm{c}+3\mathrm{d}=0\\ \mathrm{c}=-3\mathrm{d}\end{array}$

Now, substitute the values in the ker (T) $=\left\{\mathrm{A}∈{\mathrm{R}}^{2×2}\left|\mathrm{T}\right(\mathrm{A})=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}$as follows.

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\left\{\mathrm{A}∈{\mathrm{R}}^{2×2}\left|\mathrm{T}\right(\mathrm{A})=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\} \\ \ker \left(\mathrm{T}\right)=\left\{\left[\begin{array}{cc}-3\mathrm{b}& \mathrm{b}\\ -3\mathrm{d}& \mathrm{d}\end{array}\right]|\mathrm{b},\mathrm{d}∈\mathrm{R}\right\}≠\left\{0\right\} \end{gathered}$$

Since, $\mathrm{kert}≠\left\{0\right\}$, T is not an isomorphism.

Hence, the transformation T is linear and not isomorphism and .$\ker \left(\mathrm{T}\right)=\left\{\left[\begin{array}{cc}-3\mathrm{b}& \mathrm{b}\\ -3\mathrm{d}& \mathrm{d}\end{array}\right]|\mathrm{b},\mathrm{d}∈\mathrm{R}\right\}≠\left\{0\right\}$ .