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Found in: Page 185

### Linear Algebra With Applications

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

# T denotes the space of infinity sequence of real numbers,${\mathbit{T}}{\mathbf{\left(}}{\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{f}}{\mathbf{\left(}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{\right)}}$ from P to P.

The function T is linear but not isomorphism.

See the step by step solution

## Step 1: Determine the linearity of .

Consider the function $T\left(f\left(t\right)\right)=f\left({t}^{2}\right)$ from P to P.

A function D is called a linear transformation on ${\mathbit{R}}$ if the function D satisfies the following properties.

1. ${\mathbf{}}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{+}}{\mathbit{y}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{+}}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{y}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{f}}{\mathbit{o}}{\mathbit{r}}{\mathbf{}}{\mathbit{a}}{\mathbit{l}}{\mathbit{l}}{\mathbf{}}{\mathbit{x}}{\mathbf{,}}{\mathbit{y}}{\mathbf{\in }}{\mathbit{R}}{\mathbf{.}}$
2. ${\mathbf{}}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{\alpha }}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{\alpha }}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{f}}{\mathbit{o}}{\mathbit{r}}{\mathbf{}}{\mathbit{a}}{\mathbit{l}}{\mathbit{l}}{\mathbf{}}{\mathbit{c}}{\mathbit{o}}{\mathbit{n}}{\mathbit{s}}{\mathbit{t}}{\mathbit{a}}{\mathbit{n}}{\mathbit{t}}{\mathbf{}}{\mathbit{\alpha }}{\mathbf{\in }}{\mathbit{R}}{\mathbf{.}}$

An invertible linear transformation is called isomorphism or dimension of domain and co-domain is not same then the function is not isomorphism.

Assume $f,g\in PthenT\left(f\left(t\right)\right)=f\left({t}^{2}\right)andT\left(g\left(t\right)\right)=g\left({t}^{2}\right).$

Substitute the value $f\left({t}^{2}\right)forT\left(f\left(t\right)\right)andg\left({t}^{2}\right)forT\left(g\left(t\right)\right)inT\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$ for and for in as follows.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)=f\left({t}^{2}\right)+g\left({t}^{2}\right)$

Now, simplify role="math" localid="1659414024917" $T\left(\left\{f+g\right\}\left(t\right)\right)$as follows.

$T\left(\left\{f+g\right\}\left(t\right)\right)=\left\{f+g\right\}{\left(t\right)}^{2}\phantom{\rule{0ex}{0ex}}=f\left({t}^{2}\right)+g\left({t}^{2}\right)\phantom{\rule{0ex}{0ex}}T\left(\left\{f+g\right\}\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$

Assume $f\in Pand\alpha \in Rthen.$

Substitute the value $\alpha f\left({t}^{2}\right)forT\left(\alpha f\left(t\right)\right)$ as follows.

$T\left(\alpha f\left(t\right)\right)=\alpha f\left({t}^{2}\right)\phantom{\rule{0ex}{0ex}}T\left(\alpha f\left(t\right)\right)=\alpha T\left(f\left(t\right)\right)$

As $T\left(f+g\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)andT\left(\alpha f\left(t\right)\right)=\alpha T\left(f\left(t\right)\right)$ , by the definiti0on of linear transformation T is linear.

## Step 2: Determine the isomorphism of .

Assume $f\left(t\right)=timpliesf\left({t}^{2}\right)={t}^{2},$, substitute the value ${t}^{2}forf\left({t}^{2}\right)and\sqrt{t}forf\left(t\right)$ in the equation $T\left(f\left(t\right)\right)=f\left({t}^{2}\right)$ as follows.

$T\left(f\left(t\right)\right)=f\left({t}^{2}\right)\phantom{\rule{0ex}{0ex}}T\left(\sqrt{t}\right)={t}^{2}$

As the function T define from P to P, but $\sqrt{t}$does not belong to P.

By the definition of isomorphism, the function T is not isomorphism.

Hence, the transformation $T\left(f\left(t\right)\right)=f\left(t\right)+{f}^{\text{'}\text{'}}\left(t\right)+sin\left(t\right)$ is linear but not isomorphism.