91Ó°ÊÓ

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

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

For all elements of and is scalar.

An invertible linear transformation is called an isomorphism.

Let’s define a transformation as follows.

$\mathbf{T}\mathbf{:}{\mathbf{R}}^{\mathbf{2}\mathbf{×}\mathbf{2}}\mathbf{→}{\mathbf{R}}^{\mathbf{2}\mathbf{×}\mathbf{2}}\mathbf{with}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{PAQ}\mathbf{}\mathbf{}\mathbf{where}\mathbf{}\mathbf{P}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{2}& \mathbf{3}\\ \mathbf{5}& \mathbf{7}\end{array}\mathbf{\right]}\mathbf{and}\mathbf{}\mathbf{Q}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{3}& \mathbf{5}\\ \mathbf{7}& \mathbf{11}\end{array}\mathbf{\right]}$

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\mathrm{PMQ},\mathrm{where}\mathrm{P}=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\mathrm{and}\mathrm{Q}=\left[\begin{array}{cc}3& 5\\ 7& 11\end{array}\right]\mathrm{from}{\mathrm{R}}^{2×2}\mathrm{to}{\mathrm{R}}^{2×2}$

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

Let and be arbitrary matrices from and as follows.

$\begin{array}{l}\mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{P}(\mathrm{A}+\mathrm{B})\mathrm{QS}\\ =\mathrm{PAQ}+\mathrm{PBQ}\\ \mathrm{T}(\mathrm{A}+\mathrm{B})=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)\end{array}$

Similarly, to check the second condition as follows.

Let be an arbitrary scalar, and as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{Î\pm ´¡}\right)={\mathrm{S}}^{-1}\left(\mathrm{Î\pm ´¡}\right)\mathrm{S} \\ =\mathrm{Î\pm \pm Ê}\left(\mathrm{Î\pm ´¡}\right)\mathrm{Q} \\ =\mathrm{Î\pm \pm ÊAQ} \\ \mathrm{T}\left(\mathrm{Î\pm ´¡}\right)=\mathrm{Î\pm °Õ}\left(\mathrm{A}\right) \end{gathered}$$

Thus, T is a linear transformation.

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

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

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

Consider a matrix A as follows.

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

The next equation as follows.

$\begin{array}{l}\mathrm{T}\left(\mathrm{A}\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \mathrm{PAQ}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Substitute the value $\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]$for P and $\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]$for A and $\left[\begin{array}{cc}3& 5\\ 7& 11\end{array}\right]$for in $\mathrm{PAQ}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$as follows.

$\begin{array}{r}\mathrm{PAQ}=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{ll}3& 5\\ 7& 11\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2\mathrm{a}+3\mathrm{c}& 2\mathrm{b}−3\mathrm{d}\\ 5\mathrm{a}+7\mathrm{c}& 5\mathrm{b}+7\mathrm{d}\end{array}\right]\left[\begin{array}{cc}3& 5\\ 7& 11\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}3(2\mathrm{a}+3\mathrm{c})+7(2\mathrm{b}+3\mathrm{d})& 5(2\mathrm{a}+3\mathrm{c})+11(2\mathrm{a}+3\mathrm{c})\\ 3(5\mathrm{a}+7\mathrm{c})+7(5\mathrm{a}+7\mathrm{c})& 5(5\mathrm{b}+7\mathrm{d})+11(5\mathrm{b}+7\mathrm{d})\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Simplify further as follows.

$\left[\begin{array}{cc}6\mathrm{a}+9\mathrm{c}+14\mathrm{b}+21\mathrm{d}& 10\mathrm{a}+15\mathrm{c}+22\mathrm{b}+33\mathrm{d}\\ 15\mathrm{a}+21\mathrm{c}+35\mathrm{b}+49\mathrm{d}& 25\mathrm{a}+35\mathrm{c}+55\mathrm{b}+77\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Equating the corresponding entries as follows.

$$\begin{gathered} 6\mathrm{a}+9\mathrm{c}+14\mathrm{b}+21\mathrm{d}=0 \\ 10\mathrm{a}+15\mathrm{c}+22\mathrm{b}+33\mathrm{d}=0 \\ 15\mathrm{a}+21\mathrm{c}+35\mathrm{b}+49\mathrm{d}=0 \\ 25\mathrm{a}+35\mathrm{c}+55\mathrm{b}+77\mathrm{d}=0 \end{gathered}$$

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 $\mathfrak{J}\left(\mathrm{T}\right)={\mathrm{R}}^{2×2}$if as follows.

$\mathfrak{J}\left(\mathrm{T}\right)=\left\{\mathrm{T}\left(\mathrm{A}\right)|\mathrm{A}∈{\mathrm{R}}^{2×2}\right\}$

Consider $\mathrm{A}∈{\mathrm{R}}^{2×2}$be an arbitrary matrix as follows.

role="math" localid="1659850907182" $$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{T}\left(\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\right) \\ =\left[\begin{array}{cc}6\mathrm{a}+9\mathrm{c}+14\mathrm{b}+21\mathrm{d}& 10\mathrm{a}+15\mathrm{c}+22\mathrm{b}+33\mathrm{d}\\ 15\mathrm{a}+21\mathrm{c}+35\mathrm{b}+49\mathrm{d}& 25\mathrm{a}+35\mathrm{c}+55\mathrm{b}+77\mathrm{d}\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}\right)=\mathrm{B} \end{gathered}$$

For each matrix B exist matrix as follows.

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

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