91Ó°ÊÓ

Consider the transformation as follows.

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

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

Consider the linear transformation as follows.

$T:P_2→P_{2}definedasT\left(f\right(t\left)\right)=t(f\text{'}(t\left)\right).$

$Since,f\left(t\right)∈P_2.$

Consider the equation as follows.

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

Simplify as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=t\left(f\text{'}\left(t\right)\right) \\ =t\left(\frac{df}{dt}\left(a{t}^2+bt+c\right)\right) \\ =t\left(2at+b\right) \\ =2a{t}^2+bt \end{gathered}$$

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 $\left\{t,t^2\right\}$.

Consider the transformation as follows.

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

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

Consider a constant polynomial as follows.

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

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.