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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, ${\mathbit{T}}{\mathbf{\left(}}{\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{=}}\mathbf{\left[}\begin{array}{c}\mathbf{f}\left(5\right)\\ \mathbf{f}\left(7\right)\\ \mathbf{f}\left(11\right)\end{array}\mathbf{\right]}{\mathbf{}}{\mathbf{}}{\mathbit{f}}{\mathbit{r}}{\mathbit{o}}{\mathbit{m}}{\mathbf{}}{{\mathbit{P}}}_{\mathbf{2}\mathbf{}}{\mathbit{t}}{\mathbit{o}}{\mathbf{}}{{\mathbit{R}}}^{{\mathbf{3}}}{\mathbf{.}}$ .

The function T is linear and isomorphism.

See the step by step solution

## Step 1: Determine the linearity of T.

Consider the function $T\left(f\left(t\right)\right)=\left[\begin{array}{c}\mathrm{f}\left(5\right)\\ \mathrm{f}\left(7\right)\\ \mathrm{f}\left(11\right)\end{array}\right]from{P}_{2}to{R}^{3}.$

A function is called a linear transformation on if the function 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. ${\mathbf{}}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{\alpha }}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{\alpha }}{\mathbit{D}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{f}}{\mathbit{o}}{\mathbit{r}}{\mathbf{}}{\mathbit{a}}{\mathbit{l}}{\mathbit{l}}{\mathbf{}}{\mathbit{c}}{\mathbit{o}}{\mathbit{n}}{\mathbit{s}}{\mathbit{t}}{\mathbit{a}}{\mathbit{n}}{\mathbit{t}}{\mathbf{}}{\mathbit{\alpha }}{\mathbf{\in }}{\mathbit{R}}{\mathbf{.}}$

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

Assume $f,g\in {P}_{2}t$ then $T\left(f\left(t\right)\right)=\left[\begin{array}{c}\mathrm{f}\left(5\right)\\ \mathrm{f}\left(7\right)\\ \mathrm{f}\left(11\right)\end{array}\right]$ and $T\left(g\left(t\right)\right)=\left[\begin{array}{c}g\left(5\right)\\ g\left(7\right)\\ g\left(11\right)\end{array}\right]$ .

Substitute the value role="math" localid="1659411768440" $\left[\begin{array}{c}\mathrm{f}\left(5\right)\\ \mathrm{f}\left(7\right)\\ \mathrm{f}\left(11\right)\end{array}\right]forT\left(f\left(t\right)\right)and\left[\begin{array}{c}g\left(5\right)\\ g\left(7\right)\\ g\left(11\right)\end{array}\right]forT\left(g\left(t\right)\right)inT\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$ as follows.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)=\left[\begin{array}{c}\mathrm{f}\left(5\right)\\ \mathrm{f}\left(7\right)\\ \mathrm{f}\left(11\right)\end{array}\right]+\left[\begin{array}{c}g\left(5\right)\\ g\left(7\right)\\ g\left(11\right)\end{array}\right]$

## Step 2: Simplify further

Now, simplify $T\left(\left\{f+g\right\}\left(t\right)\right)$as follows.

$T\left(\left\{f+g\right\}\left(t\right)\right)=\left[\begin{array}{cc}\left\{f+g\right\}& \left(5\right)\\ \left\{f+g\right\}& \left(7\right)\\ \left\{f+g\right\}& \left(11\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}f\left(5\right)& g\left(5\right)\\ f\left(7\right)& g\left(7\right)\\ f\left(11\right)& g\left(11\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\\ f\left(11\right)\end{array}\end{array}\right]+\left[\begin{array}{c}\begin{array}{c}g\left(5\right)\\ g\left(7\right)\\ g\left(11\right)\end{array}\end{array}\right]\phantom{\rule{0ex}{0ex}}T\left(\left\{f+g\right\}\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$

Assume $f\in {P}_{2}and\alpha \in R$ then role="math" localid="1659412647658" $T\left(f\left(t\right)\right)=\left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\\ f\left(11\right)\end{array}\end{array}\right]$.

Substitute the value $\left[\begin{array}{c}\begin{array}{c}\alpha f\left(5\right)\\ \alpha f\left(7\right)\\ \alpha f\left(11\right)\end{array}\end{array}\right]forT\left(\alpha f\left(t\right)\right)$for as follows.

$T\left(\alpha f\left(t\right)\right)=\left[\begin{array}{c}\begin{array}{c}\alpha f\left(5\right)\\ \alpha f\left(7\right)\\ \alpha f\left(11\right)\end{array}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\alpha \left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\\ f\left(11\right)\end{array}\end{array}\right]\phantom{\rule{0ex}{0ex}}T\left(\alpha f\left(t\right)=\alpha T\left(f\left(t\right)$

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 definition of linear transformation T is linear.

## Step 3: Determine the isomorphism of .

As the function define from ${P}_{2}to{R}^{3}and{P}_{2}$ is spanned by $\left\{1,t,{t}^{2}\right\}$ means dimension of ${P}_{2}$ is 3 and dimension of ${R}^{3}$ is 3.

By the definition of polynomial, every polynomial form ${P}_{2}$is described in a unique way means for every $\left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\\ f\left(11\right)\end{array}\end{array}\right]\in {R}^{3}$ there exist a unique $f\left(t\right)\in {P}_{2}$ such that $T\left(f\left(t\right)\right)=\left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\\ f\left(11\right)\end{array}\end{array}\right]$ .

By the definition of isomorphism, the function is isomorphism.

Hence, the transformation localid="1662122176677" $T\left(f\left(t\right)\right)=\left[\begin{array}{c}\begin{array}{c}f\left(5\right)\\ f\left(7\right)\end{array}\end{array}\right]\phantom{\rule{0ex}{0ex}}$ is linear and isomorphism.