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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 out which of the transformations in Exercises 1 through 50 are linear. For those that are linear, determine whether they are isomorphism, ${\mathbit{T}}\left(f\left(t\right)\right){\mathbf{=}}{\mathbit{F}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(t\right){\mathbit{f}}\left(t\right)$ from ${{\mathbit{P}}}_{{\mathbf{2}}}$ to ${{\mathbit{P}}}_{{\mathbf{2}}}$.

The transformation $T\left(\mathrm{f}\left(\mathrm{t}\right)\right)=F\text{'}\text{'}\left(\mathrm{t}\right)f\left(\mathrm{t}\right)$ 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,

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

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

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

$1.T\left(f+g\right)=T\left(f\right)+T\left(g\right)\phantom{\rule{0ex}{0ex}}2.T\left(kf\right)=kT\left(f\right)$

Verify the first condition.

Let $f\left(t\right)={a}_{1}{t}^{2}+{b}_{1}t+{c}_{1}$ and $g\left(t\right)={a}_{2}{t}^{2}+{b}_{2}t+{c}_{2}$ be two polynomial functions from ${P}_{2}$. Then, $T\left(\mathrm{f}\left(\mathrm{t}\right)\right)=F\text{'}\text{'}\left(\mathrm{t}\right)f\left(\mathrm{t}\right)$, $T\left(g\left(\mathrm{t}\right)\right)=g\text{'}\text{'}\left(\mathrm{t}\right)g\left(\mathrm{t}\right)$

Find $T\left(f\left(t\right)+g\left(t\right)\right)$.

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

It is clear that, the first condition $T\left(f+g\right)=T\left(f\right)+T\left(g\right)$ is not satisfied.

Thus, T is not a linear transformation.