91Ó°ÊÓ

Consider the transformation as follows.

$\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{→}\mathbf{W}$such that $\mathbf{ker}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{f}\mathbf{∈}\mathbf{V}\overline{)\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{=}\mathbf{0}}\mathbf{\}}$.

If the kernel of $\mathbf{T}$is finite dimensional, then $\mathbf{dim}\left(\mathrm{kerT}\right)$is called the nullity of $\mathbf{T}$.

Consider the linear transformation as follows.

$\mathrm{T}:{\mathrm{P}}_2→{\mathrm{R}}^2$defined as $T\left(f\left(t\right)\right)=\left[\begin{array}{c}f\left(7\right)\\ f\left(11\right)\end{array}\right]$.

Consider the equation as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=0 \\ \left[\begin{array}{c}f\left(7\right)\\ f\left(11\right)\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

Since, $f\left(t\right)∈P_2$.

Therefore,$f\left(t\right)=a\left(t-7\right)\left(t-11\right)$

Thus, the kernel contains all the functions of the form$a\left(t-7\right)\left(t-11\right)$ .

So, the nullity is 1 with the basis $\left\{\left(t-7\right)\left(t-11\right)\right\}$.

Consider the transformation as follows.

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

If the image of $\mathbf{T}$is finite dimensional, then $\mathbf{dim}\left(\mathrm{imT}\right)$is called the rankof$\mathbf{T}$.

Since,$T\left(f\left(t\right)\right)=\left[\begin{array}{c}f\left(7\right)\\ f\left(11\right)\end{array}\right]$ .

This shows that$f\left(7\right),f\left(11\right)∈R$

Therefore, the image contains$R^2$ .

So, the rank is 2.

Hence, the image contains $R^2$.and the rank is 2 whereas kernel contains the elements in the form $a\left(t-7\right)\left(t-11\right)$and the nullity is 1.