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Consider the transformation as follows.

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

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

Consider the linear transformation as follows.

$T:P鈫�P$defined as $T\left(f\left(t\right)\right)=t\left(f\left(t\right)\right)$.

Consider the equation as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=0 \\ t\left(f\left(t\right)\right)=0 \end{gathered}$$

Since, $f\left(t\right)$ is a polynomial.

Thus, the kernel contains the zero function only.

Consider the transformation as follows.

$\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$such that $\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{:}\mathbf{f}\mathbf{鈭�}\mathbf{V}\mathbf{\}}$.

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

Since, $T\left(f\left(t\right)\right)=T\left(f\left(t\right)\right)$.

So, there will be no constant in the image.

Therefore, the image contains all the polynomial in which constant terms is zero.

That is $F\left(0\right)=0$.

Hence, the kernel contains the zero function only and the image contains all the polynomial in which constant term is zero.