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

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

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.

$\mathbf{T}\mathbf{:}{\mathbf{R}}^{\mathbf{2}\mathbf{脳}\mathbf{2}}\mathbf{鈫�}{\mathbf{R}}^{\mathbf{2}\mathbf{脳}\mathbf{2}}\mathbf{with}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}{\mathbf{S}}^{\mathbf{-}\mathbf{1}}\mathbf{AS}\mathbf{}\mathbf{where}\mathbf{}\mathbf{S}\mathbf{=}\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$.

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)={\mathrm{S}}^{-1}\mathrm{MS}$, where$S=\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$fromrole="math" localid="1659844771593" ${\mathrm{R}}^{2脳2}$to${\mathrm{R}}^{2脳2}$.

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

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

$\begin{array}{l}\mathrm{T}(\mathrm{A}+\mathrm{B})={\mathrm{S}}^{-1}(\mathrm{A}+\mathrm{B})\mathrm{S}\\ ={\mathrm{S}}^{-1}\mathrm{AS}+{\mathrm{S}}^{-1}\mathrm{BS}\\ \mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)\end{array}$

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.

$$\begin{gathered} \mathrm{T}\left(\mathrm{伪础}\right)={\mathrm{S}}^{-1}\left(\mathrm{伪础}\right)\mathrm{S} \\ ={\mathrm{伪厂}}^{-1}\mathrm{As} \\ \mathrm{T}\left(\mathrm{伪础}\right)=\mathrm{伪罢}\left(\mathrm{A}\right) \end{gathered}$$

Thus, T is a linear transformation.

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

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

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\left\{\mathrm{A}鈭�{\mathrm{R}}^{2脳2}\left|\mathrm{T}\right(\mathrm{A})=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\} \\ \mathrm{M}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right] \\ 鈬�{\mathrm{M}}^{-1}\frac{1}{\mathrm{detM}}\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right] \end{gathered}$$

Simplify as follows.

$$\begin{gathered} \mathrm{S}=\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right] \\ \mathrm{detS}=\left|\begin{array}{cc}3& 4\\ 5& 6\end{array}\right| \\ =3\left(6\right)-4\left(5\right) \\ =18-20 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{detS}=-2 \\ {\mathrm{S}}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}6& -4\\ -5& 3\end{array}\right] \end{gathered}$$

Consider a matrix A as follows.

role="math" localid="1659845529710" $\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$

The next equation as follows.

$\begin{array}{l}\mathrm{T}\left(\mathrm{A}\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ {\mathrm{S}}^{-1}\mathrm{AS}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Substitute the value $\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$for S and $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$for A and $\frac{1}{-2}\left[\begin{array}{cc}6& -4\\ -5& 3\end{array}\right]$for ${\mathrm{S}}^{-1}$in ${\mathrm{S}}^{-1}\mathrm{AS}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$as follows.

$\begin{array}{r}{S}^{鈭�1}AS=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \frac{1}{鈭�2}\left[\begin{array}{cc}6& 鈭�4\\ 鈭�5& 3\end{array}\right]\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\left[\begin{array}{ll}3& 4\\ 5& 6\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \frac{1}{鈭�2}\left[\begin{array}{cc}6a鈭�4c& 6b鈭�4d\\ 鈭�5a+3c& 鈭�5b+3d\end{array}\right]\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}3(6a鈭�4c)+5(6b鈭�4d)& 4(6b鈭�4d)+6(6b鈭�4d)\\ 3(鈭�5a+3c)+5(鈭�5b+3d)& 5(鈭�5a+3c)+6(鈭�5b+3d)\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Simplify further as follows.

role="math" localid="1659846304887" $\left[\begin{array}{cc}18\mathrm{a}-12\mathrm{c}+30\mathrm{b}-20\mathrm{d}& 24\mathrm{a}-16\mathrm{c}+36-24\mathrm{d}\\ -15+9\mathrm{c}-25\mathrm{b}+15\mathrm{d}& -20\mathrm{a}+12\mathrm{c}-30\mathrm{b}+18\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Equating the corresponding entries as follows.

$$\begin{gathered} 18\mathrm{a}-12\mathrm{c}+30\mathrm{b}-20\mathrm{d}=0 \\ 24\mathrm{a}-16\mathrm{c}+36-24\mathrm{d}=0 \\ -15+9\mathrm{c}-25\mathrm{b}+15\mathrm{d}=0 \\ -20\mathrm{a}+12\mathrm{c}-30\mathrm{b}+18\mathrm{d}=0 \end{gathered}$$

Since the determinant of the system differs from zero, so the system has a trivial solution.

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

Now, to check that if$\mathfrak{J}\left(\mathrm{T}\right)={\mathrm{R}}^{2脳2}$as follows.

$\mathfrak{J}\left(\mathrm{T}\right)=\left\{\mathrm{T}\left(\mathrm{A}\right)|\mathrm{A}鈭�{\mathrm{R}}^{2脳2}\right\}$

Considerrole="math" localid="1659846259471" $\mathrm{A}鈭�{\mathrm{R}}^{2脳2}$be an arbitrary matrix as follows.

role="math" localid="1659846343739" $$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{T}\left(\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\right) \\ =\left[\begin{array}{cc}18\mathrm{a}-12\mathrm{c}+30\mathrm{b}-20\mathrm{d}& 24\mathrm{a}-16\mathrm{c}+36-24\mathrm{d}\\ -15+9\mathrm{c}-25\mathrm{b}+15\mathrm{d}& -20\mathrm{a}+12\mathrm{c}-30\mathrm{b}+18\mathrm{d}\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}\right)=\mathrm{B} \end{gathered}$$

For each matrix B exist matrix as follows.

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]鈭�{\mathrm{R}}^{2脳2} \\ 鈬�\mathfrak{J}\left(\mathrm{T}\right)鈭�{\mathrm{R}}^{2脳2} \end{gathered}$$

Therefore, T is an isomorphism.

Thus, T is a linear transformation and is an isomorphism and ker (T) = {0} also $\mathfrak{J}\left(\mathrm{T}\right)鈭�{\mathrm{R}}^{2脳2}$.