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Q33E

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Found in: Page 184

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the transformation is linear and determine whether the transformation is an isomorphism.

The solution is that T is a liner transformation and not an isomorphism.

See the step by step solution

## Step1: Definition of Linear Transformation

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

${\mathbf{T}}\mathbf{\left(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{\right)}{\mathbf{=}}{\mathbf{T}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}{\mathbf{+}}{\mathbf{T}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{T}}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}{\mathbf{=}}{\mathbf{kT}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}$

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

A linear transformation ${\mathbf{T}}{\mathbf{:}}{\mathbf{V}}{\mathbf{\to }}{\mathbf{W}}$ is said to be an isomorphism if and only if ${\mathbf{ker}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\left\{}}{\mathbf{0}}{\mathbf{\right\}}}$ and ${\mathbf{im}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{W}}$ or ${\mathbf{dim}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{dim}}{\mathbf{\left(}}{\mathbf{W}}{\mathbf{\right)}}$.

## Step2: Explanation of the solution

The given transformation as follows.

$\mathrm{T}\left({\mathrm{x}}_{0,}{\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3},...\right)=\left({\mathrm{x}}_{0,}{\mathrm{x}}_{2},{\mathrm{x}}_{4},...\right)$, from V to V.

By using the definition of linear transformation as follows.

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

Now, to check the first condition as follows.

Consider two infinite sequences V from as follows.

$x=\left({x}_{0}.{x}_{1},...\right)$ and $y=\left({y}_{0}.{y}_{1},...\right)$ be an infinite sequences from V.

Now, simplify as follows.

$\mathrm{T}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{T}\left({\mathrm{x}}_{0}+{\mathrm{y}}_{0},{\mathrm{x}}_{1}+{\mathrm{y}}_{1}+...\right)\phantom{\rule{0ex}{0ex}}=\left({\mathrm{x}}_{0}+{\mathrm{y}}_{0},{\mathrm{x}}_{2}+{\mathrm{y}}_{2}+...\right)\phantom{\rule{0ex}{0ex}}=\left({\mathrm{x}}_{0}+{\mathrm{x}}_{2},...\right)+\left({\mathrm{y}}_{0}+{\mathrm{y}}_{2},...\right)\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{T}\left(\mathrm{x}\right)+\mathrm{T}\left(\mathrm{y}\right)$

Simplify for the second condition as follows.

$\mathrm{T}\left(\mathrm{kx}\right)=\mathrm{T}\left({\mathrm{kx}}_{0},{\mathrm{kx}}_{1},{\mathrm{kx}}_{2}+...\right)\phantom{\rule{0ex}{0ex}}=\left({\mathrm{kx}}_{0},{\mathrm{kx}}_{2},{\mathrm{kx}}_{4}+...\right)\phantom{\rule{0ex}{0ex}}=\mathrm{k}\left({\mathrm{x}}_{0},{\mathrm{x}}_{2},{\mathrm{x}}_{4}\right)\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{kx}\right)=\mathrm{kT}\left(\mathrm{x}\right)$

Thus, T is a linear transformation.

## Step3: Properties of isomorphism

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

Now, it easy to show that $\mathrm{T}\left(0,1,0,1,...\right)=\left(0,0,0,...\right)$ and $\mathrm{T}\left(\left(0,2,0,2,...\right)\right)=\left(0,0,0,...\right)$.

That is the inverse of role="math" localid="1659416754005" $\left(0,0,0,\dots \right)$ does not have unique vector to map to.

This means that T is not an isomorphism.

Thus, T is a linear transformation and not an isomorphism.

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