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Expert-verified Found in: Page 185 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the image, kernel, rank, and nullity of the transformation T in ${\mathbf{T}}{\mathbf{\left(}}{\mathbf{f}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{t}}{\mathbf{\left(}}{\mathbf{f}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{from}}{\mathbf{}}{{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{.}}$

The image contains of all the linear polynomial of the form $2a{t}^{2}+bt$ and the nullity is 1.

See the step by step solution

## Step1: Definition of rank of T

Consider the transformation as follows.

${\mathbf{T}}{\mathbf{:}}{\mathbf{V}}{\mathbf{\to }}{\mathbf{W}}{\mathbf{}}{\mathbf{such}}{\mathbf{}}{\mathbf{that}}{\mathbf{}}{\mathbf{im}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}\left\{T\left(f\right):f\in V\right\}$

If the image of T is finite dimensional, then dim(imT) is called the rank of T .

## Step2: Explanation of the solution

Consider the linear transformation as follows.

$T:{P}_{2}\to {P}_{2}definedasT\left(f\left(t\right)\right)=t\left(f\text{'}\left(t\right)\right).$

$Since,f\left(t\right)\in {P}_{2}.$

Consider the equation as follows.

$f\left(t\right)=a{t}^{2}+bt+c$

Simplify as follows.

$T\left(f\left(t\right)\right)=t\left(f\text{'}\left(t\right)\right)\phantom{\rule{0ex}{0ex}}=t\left(\frac{df}{dt}\left(a{t}^{2}+bt+c\right)\right)\phantom{\rule{0ex}{0ex}}=t\left(2at+b\right)\phantom{\rule{0ex}{0ex}}=2a{t}^{2}+bt$

Therefore, the image contains of all the second-degree polynomial of the form $2a{t}^{2}+bt$.

Thus, the rank is 2 with the basis $\left\{t,{t}^{2}\right\}$.

## Step3: Definition of nullity of T

Consider the transformation as follows.

${\mathbit{T}}{\mathbf{:}}{\mathbit{V}}{\mathbf{\to }}{\mathbit{W}}{\mathbf{}}{\mathbit{s}}{\mathbit{u}}{\mathbit{c}}{\mathbit{h}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{a}}{\mathbit{t}}{\mathbf{}}{\mathbit{k}}{\mathbit{e}}{\mathbit{r}}{\mathbf{\left(}}{\mathbit{T}}{\mathbf{\right)}}{\mathbf{=}}\left\{f\in V|T\left(f\right)=0\right\}{\mathbf{.}}$

${\mathbit{I}}{\mathbit{f}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{k}}{\mathbit{e}}{\mathbit{r}}{\mathbit{n}}{\mathbit{e}}{\mathbit{l}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{T}}{\mathbf{}}{\mathbit{i}}{\mathbit{s}}{\mathbf{}}{\mathbit{f}}{\mathbit{i}}{\mathbit{n}}{\mathbit{i}}{\mathbit{t}}{\mathbit{e}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{m}}{\mathbit{e}}{\mathbit{n}}{\mathbit{s}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbit{a}}{\mathbit{l}}{\mathbf{,}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbit{n}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{m}}{\mathbf{\left(}}{\mathbit{k}}{\mathbit{e}}{\mathbit{r}}{\mathbit{T}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{i}}{\mathbit{s}}{\mathbf{}}{\mathbit{c}}{\mathbit{a}}{\mathbit{l}}{\mathbit{l}}{\mathbit{e}}{\mathbit{d}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{n}}{\mathbit{u}}{\mathbit{l}}{\mathbit{l}}{\mathbit{i}}{\mathbit{t}}{\mathbit{y}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{T}}$

Consider a constant polynomial as follows.

$T\left(f\left(t\right)\right)=0$

Then, f(t) is a constant polynomial

Therefore, the kernel consists of all the constant polynomial with the basis {1} .

The nullity is 1.

Hence, the image contains of all the linear polynomial of the form $2a{t}^{2}+bt$ and the rank is 2 whereas the kernel contains all the constant polynomial and the nullity is 1. ### Want to see more solutions like these? 