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Q32E

Expert-verified
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 T is not a liner transformation.

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.

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

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

A linear transformation ${\mathbit{T}}{\mathbf{:}}{\mathbit{V}}{\mathbf{\to }}{\mathbit{W}}$ is said to be an isomorphism if and only if ${\mathbit{k}}{\mathbit{e}}{\mathbit{r}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\left\{}}{\mathbf{0}}{\mathbf{\right\}}}$ and ${\mathbit{i}}{\mathbit{m}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{W}}$ or localid="1659415485656" ${\mathbit{d}}{\mathbit{i}}{\mathbit{m}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{d}}{\mathbit{i}}{\mathbit{m}}{\mathbf{\left(}}{\mathbit{W}}{\mathbf{\right)}}$.

## Step2: Explanation of the solution

The given transformation as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\left(\mathrm{t}\right)+{\mathrm{t}}^{2}$ , from ${P}_{2}$ to ${P}_{2}$.

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 the polynomials as follows.

f( x ) and g( x ) from ${P}_{2}$ and k is scalar.

Now, simplify as follows.

$\mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)=\left(\mathrm{f}+\mathrm{g}\right)\text{'}+{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{f}\text{'}+\mathrm{g}\text{'}+{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}\ne \mathrm{f}\text{'}+{\mathrm{t}}^{2}+\mathrm{g}\text{'}+{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)\ne \mathrm{T}\left(\mathrm{f}\right)+\mathrm{T}\left(\mathrm{g}\right)$

Thus, the function T is not a linear transformation.

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