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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 $T(\mathrm{x}+\mathrm{iy})=x^2+y^2$, from $\mathrm{鈩�}$to $\mathrm{鈩�}$.

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 $\mathrm{鈩�}$. Then,

$$\begin{gathered} T(A+B)=T(x_1+i{y}_1+x_2+i{y}_2) \\ =T\left(\left(x_1+x_2\right)+i\left(y_1+y_2\right)\right) \\ ={\left(x_1+x_2\right)}^2+{\left(y_1+y_2\right)}^2 \\ =x_1^2+x_2^2+2x_1{x}_2+y_1^2+y_2^2+2y_1{y}_2 \\ =\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)+2x_1{x}_2+2y_1{y}_2 \\ =T\left(A\right)+T\left(B\right)+2x_1{x}_2+2y_1{y}_2 \end{gathered}$$

It is clear that, the first condition$T\left(A+B\right)鈮�T\left(A\right)+T\left(B\right)$is not satisfied.

Thus, T is not a linear transformation.