91Ó°ÊÓ

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

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

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

An invertible linear transformation is called an isomorphism.

Let’s define a transformation as follows.

$\mathbf{T}\mathbf{:}{\mathbf{R}}^{\mathbf{2}\mathbf{×}\mathbf{2}}\mathbf{→}{\mathbf{R}}^{\mathbf{2}\mathbf{×}\mathbf{2}}\mathbf{with}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{2}& \mathbf{0}\\ \mathbf{0}& \mathbf{3}\end{array}\mathbf{\right]}\mathbf{A}\mathbf{-}\mathbf{A}\mathbf{\left[}\begin{array}{cc}\mathbf{4}& \mathbf{0}\\ \mathbf{0}& \mathbf{5}\end{array}\mathbf{\right]}$

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{M}-\mathrm{M}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right],\mathrm{from}{\mathrm{R}}^{2×2}\mathrm{to}{\mathrm{R}}^{2×2}$

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left(\mathrm{A}+\mathrm{B}\right)-\left(\mathrm{A}+\mathrm{B}\right)\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right] \\ =\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{A}+\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{B}-\mathrm{A}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right]-\mathrm{B}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right] \\ =\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{B}-\mathrm{B}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right] \\ \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \end{gathered}$$

Similarly, to check the second condition as follows.

Let be an arbitrary scalar, and$\mathrm{A}∈{\mathrm{R}}^{2×2}$as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{Î\pm ´¡}\right)=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{Î\pm ´¡}-\mathrm{Î\pm ´¡}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right] \\ =\mathrm{Î\pm }\left(\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right]\right) \\ \mathrm{T}\left(\mathrm{Î\pm ´¡}\right)=\mathrm{Î\pm °Õ}\left(\mathrm{A}\right) \end{gathered}$$

Thus, T is a linear transformation.

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

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

$\begin{array}{r}\mathrm{T}\left(\mathrm{A}\right)=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]-\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right]=\left[\begin{array}{ll}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2\mathrm{a}+0& 2\mathrm{b}+0\\ 0+3\mathrm{c}& 0+3\mathrm{d}\end{array}\right]-\left[\begin{array}{cc}4\mathrm{a}+0& 0+5\mathrm{b}\\ 4\mathrm{c}+0& 0+5\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ \left[\begin{array}{cc}2\mathrm{a}-4\mathrm{a}& 2\mathrm{b}-5\mathrm{b}\\ 3\mathrm{c}-4\mathrm{c}& 3\mathrm{d}-5\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Simplify further as follows.

role="math" localid="1659856348694" $$\begin{gathered} \left[\begin{array}{cc}2\mathrm{a}-4\mathrm{a}& 2\mathrm{b}-5\mathrm{b}\\ 3\mathrm{c}-4\mathrm{c}& 3\mathrm{d}-5\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \left[\begin{array}{cc}-2\mathrm{a}& -2\mathrm{b}\\ -2\mathrm{c}& -2\mathrm{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Equating the corresponding entries as follows.

$$\begin{gathered} -2\mathrm{a}=0 \\ 2\mathrm{b}=0 \\ -2\mathrm{c}=0 \\ -2\mathrm{d}=0 \end{gathered}$$

Solve and find the values as follows.

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

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}\right)=\mathrm{T}\left(\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]-\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{cc}4& 0\\ 0& 5\end{array}\right]\right) \\ =\left[\begin{array}{cc}-2\mathrm{a}& -2\mathrm{b}\\ -2\mathrm{c}& -2\mathrm{d}\end{array}\right] \end{gathered}$$

T (A) = B

For each matrix B , $\mathrm{B}=\left[\begin{array}{cc}-2\mathrm{a}& -2\mathrm{b}\\ -2\mathrm{c}& -2\mathrm{d}\end{array}\right]∈{\mathrm{R}}^{2×2}$exist the matrix as follows.

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

Therefore, T is an isomorphism.

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