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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, ${\mathbit{T}}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{i}\mathbf{y}\mathbf{\right)}{\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbit{y}}}^{{\mathbf{2}}}$ from $\mathrm{ℂ}$ to $\mathrm{ℂ}$.

The transformation $T\left(\mathrm{x}+\mathrm{iy}\right)={x}^{2}+{y}^{2}$ is a linear transformation and is an isomorphism.

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 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,

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

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

Verify the second condition.

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

$T\left(kA\right)=T\left(k\left({x}_{1}+i{y}_{1}\right)\right)\phantom{\rule{0ex}{0ex}}=ik\left({x}_{1}+i{y}_{1}\right)\phantom{\rule{0ex}{0ex}}=k\left(i\left({x}_{1}+i{y}_{1}\right)\right)\phantom{\rule{0ex}{0ex}}=kT\left({x}_{1}+i{y}_{1}\right)\phantom{\rule{0ex}{0ex}}T\left(kA\right)=kT\left(A\right)$

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

Thus,T is a linear transformation.

## Step 3: Properties of isomorphism

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

Now, check whether ker (T) $=\left\{0\right\}$.

According to the definition of the kernel of a transformation,

localid="1659422383534" $ker\left(T\right)=\left\{A\in \mathrm{ℂ},T\left(A\right)=0\right\}$.

Consider a complex number A as A = x + iy

Then,

$\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}T\left(A\right)=0\phantom{\rule{0ex}{0ex}}\mathit{}T\left(x+iy\right)=0\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}i\left(x+iy\right)=0\phantom{\rule{0ex}{0ex}}\mathit{}ix\left(x+iy\right)=0\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}ix\mathit{+}{i}^{\mathit{2}}y=0+0i\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}ix\mathit{-}y=\mathit{0}+0i\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{-}y\mathit{+}ix=\mathit{0}+0i$

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

Clearly, $ker\left(T\right)=\left\{0\right\}$.

Now, check whether Im (T) $=\mathrm{ℂ}$.

According to the definition of the image of a transformation,

localid="1659416713747" $Im\left(T\right)=\left\{T\left(A\right):A\in \right\}$

Let $A=x+iy$ is in .

Then,

$T\left(A\right)=T\left(x+iy\right)\phantom{\rule{0ex}{0ex}}=i\left(x+iy\right)\phantom{\rule{0ex}{0ex}}=ix+{i}^{2}y\phantom{\rule{0ex}{0ex}}=ix-y\phantom{\rule{0ex}{0ex}}=-y+ix$

, say (B = - Y + ix).

T (A) = B

It is clear that, $T\left(A\right)=-y+ix$.

This means for any localid="1659422430378" $B\in \mathrm{ℂ}$ there exists a localid="1659422483414" $A\in \mathrm{ℂ}$such that $T\left(A\right)=B$.

So, Im (T) = $\mathrm{ℂ}$

Therefore, the transformation T is an isomorphism.

Thus, the transformation T is a linear transformation and T is an isomorphism. ### Want to see more solutions like these? 