91Ó°ÊÓ

Consider two linear spaces V and W. A transformation T is said to be a linear transformation if it satisfies the properties,

$$\begin{gathered} \mathit{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathit{T}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{+}\mathit{T}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathit{T}\mathbf{\left(}\mathit{k}\mathit{v}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{T}\mathbf{\left(}\mathit{v}\mathbf{\right)} \end{gathered}$$

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

An invertible linear transformation is called an isomorphism.

Consider the transformation $T(\mathrm{x}+\mathrm{iy})=y+ix$, from$\mathrm{â„‚}$to $\mathrm{â„‚}$.

Check whether the transformationsatisfies the below two conditions or not.

$$\begin{gathered} 1.T(A+B)=T\left(A\right)+T\left(B\right) \\ 2.T\left(kA\right)=kT\left(A\right) \end{gathered}$$

Verify the first condition.

Let $A=x_1+i{y}_1$and$B=x_2+i{y}_2$be arbitrary complex numbers from$\mathrm{â„‚}$. Then,

$$\begin{gathered} T(A+B)=T(x_1+i{y}_1+x_2+i{y}_2) \\ =T\left(\left(x_1+x_2\right)+i\left(y_1+y_2\right)\right) \\ =\left(y_1+y_2\right)+i\left(x_1+x_2\right) \\ =y_1+i{x}_1+y_2+i{x}_2 \\ =T\left(x_1+i{y}_1\right)+T\left(x_2+i{y}_2\right) \\ T(A+B)=T\left(A\right)+T\left(B\right) \end{gathered}$$

It is clear that, the first condition $T(A+B)=T\left(A\right)+T\left(B\right)$is satisfied.

Verify the second condition.

Let k be an arbitrary scalar, and$A∈\mathrm{ℂ}$as follows.

$$\begin{gathered} T\left(kA\right)=T\left(k\left(x_1+i{y}_1\right)\right) \\ =ik\left(x_1+ik{y}_1\right) \\ =k{y}_1+ik{x}_1 \\ =kT\left(x_1+i{y}_1\right) \\ T\left(kA\right)=kT\left(A\right) \end{gathered}$$

It is clear that, the second condition$T\left(kA\right)=kT\left(A\right)$ is also satisfied.

Thus, T is a linear transformation.

A linear transformation $\mathit{T}\mathbf{:}\mathit{V}\mathbf{→}\mathit{W}$is said to be an isomorphism if and only $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and $\mathit{I}\mathit{m}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\mathit{W}$.

Now, check whether ker (T) = {0}.

According to the definition of thekernel of a transformation,

$ker\left(T\right)=\{A∈\mathrm{ℂ},T(A\left)0\right\}$.

Consider a complex number A as A = x + iy

Then,

$$\begin{gathered} T\left(A\right)=0 \\ T\left(x+iy\right)=0 \\ y+ix=0+0i \end{gathered}$$

Comparing both sides, it can be concluded that x = 0, y = 0

Clearly, ker (T) = {0}.

Now, check whether $Im\left(T\right)=\mathrm{â„‚}$.

According to the definition of the imageof a transformation,

$Im\left(T\right)=\left\{T\left(A\right):A∈\right\}$

Let $A=x+iy$is in $\mathrm{â„‚}$.

Then,

$$\begin{gathered} T\left(A\right)=T(x+iy) \\ =y+ix,\mathrm{say}(B=y+ix) \\ =B \end{gathered}$$

It is clear that,$T\left(A\right)=B∈\mathrm{ℂ}$.

This means for any $B∈\mathrm{ℂ}$, there exists a$A∈\mathrm{ℂ}$such that $T\left(A\right)=B$.

So, $Im\left(T\right)=\mathrm{â„‚}$

Therefore, the transformation T is an isomorphism.

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