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Expert-verified Found in: Page 184 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find out which of the transformations in Exercises 1 through 50 are linear. For those that are linear, determine whether they are isomorphism, ${\mathbit{T}}\left(x+iy\right){\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbit{y}}}^{{\mathbf{2}}}$ from ${\mathbf{ℂ}}$to ${\mathbf{ℂ}}$.

The transformation $T\left(\mathrm{x}+\mathrm{iy}\right)={x}^{2}+{y}^{2}$ is not a linear transformation.

See the step by step solution

## Step 1: Definition of Linear Transformation

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

${\mathbit{T}}\mathbf{\left(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{\right)}{\mathbf{=}}{\mathbit{T}}{\mathbf{\left(}}{\mathbit{f}}{\mathbf{\right)}}{\mathbf{+}}{\mathbit{T}}{\mathbf{\left(}}{\mathbit{g}}{\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbit{T}}{\mathbf{\left(}}{\mathbit{k}}{\mathbit{v}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{k}}{\mathbit{T}}{\mathbf{\left(}}{\mathbit{v}}{\mathbf{\right)}}$

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

An invertible linear transformation is called an isomorphism.

## Step 2: Check whether the given transformation is a linear or not.

Consider the transformation $T\left(\mathrm{x}+\mathrm{iy}\right)={x}^{2}+{y}^{2}$ , from $\mathrm{ℂ}$to $\mathrm{ℂ}$.

Check whether the transformation satisfies the below two conditions or not.

$1.T\left(A+B\right)=T\left(A\right)+T\left(B\right)\phantom{\rule{0ex}{0ex}}2.T\left(kA\right)=kT\left(A\right)$

Verify the first condition.

Let $A={x}_{1}+i{y}_{1}$ and $B={x}_{2}+i{y}_{2}$ be arbitrary complex numbers from $\mathrm{ℂ}$. Then,

$T\left(A+B\right)=T\left({x}_{1}+i{y}_{1}+{x}_{2}+i{y}_{2}\right)\phantom{\rule{0ex}{0ex}}=T\left(\left({x}_{1}+{x}_{2}\right)+i\left({y}_{1}+{y}_{2}\right)\right)\phantom{\rule{0ex}{0ex}}={\left({x}_{1}+{x}_{2}\right)}^{2}+{\left({y}_{1}+{y}_{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}_{1}^{2}+{x}_{2}^{2}+2{x}_{1}{x}_{2}+{y}_{1}^{2}+{y}_{2}^{2}+2{y}_{1}{y}_{2}\phantom{\rule{0ex}{0ex}}=\left({x}_{1}^{2}+{y}_{1}^{2}\right)+\left({x}_{2}^{2}+{y}_{2}^{2}\right)+2{x}_{1}{x}_{2}+2{y}_{1}{y}_{2}\phantom{\rule{0ex}{0ex}}=T\left(A\right)+T\left(B\right)+2{x}_{1}{x}_{2}+2{y}_{1}{y}_{2}$

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

Thus, T is not a linear transformation. ### Want to see more solutions like these? 