91Ó°ÊÓ

Consider the transformation as follows.

$\mathrm{T}:\mathrm{V}→\mathrm{W}\mathrm{such}\mathrm{time}\mathrm{im}\left(\mathrm{T}\right)=\left\{\mathrm{T}\left(\mathrm{f}\right):\mathrm{f}∈\mathrm{v}\right\}.$

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

Consider the linear transformation as follows.

$T:P_2→P_{2}definedasT\left(f\left(t\right)\right)=f\text{'}\text{'}\left(t\right)+4f\text{'}\text{'}\left(t\right)$

Since, $f\left(t\right)∈P_2$.

Consider the equation as follows.

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

Differentiate the equation $f\left(t\right)=a{t}^2+bt+c$with respect to t as follows.

$$\begin{gathered} f\left(t\right)=a{t}^2+bt+c \\ f\left(t\right)+2at+b \end{gathered}$$

Again differentiate the function $f\text{'}\text{'}\left(t\right)=2at+b$with respect to t as follows.

$$\begin{gathered} f\text{'}\text{'}\left(t\right)=2at+b \\ f\text{'}\text{'}\left(t\right)=2a \end{gathered}$$

Substitute the value 2a for f''(t) and 2at+b for f''(t) in T(f(t)) =f''(t)+4f''(t) as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=f\text{'}\text{'}\left(t+4f\text{'}\text{'}\left(t\right)\right) \\ =2a+4\left(2at+b\right) \\ =2a+8at+4b \\ =8at+2\left(a+2b\right) \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=8at+2\left(2a+b\right) \\ T\left(f\left(t\right)\right)=mt+c \end{gathered}$$

Therefore, the image contains of all the linear polynomial of the form mt+c .

There are two arbitrary constant m and c with the basis {1,t}.

Thus, the rank is 2.

Consider the transformation as follows.

$\mathrm{T}:\mathrm{V}→\mathrm{W}\mathrm{such}thatker\left(\mathrm{T}\right)=\left\{f∈v|T\left(f\right)=0\right\}$

If the kernel of T is finite dimensional, then dim (ker T) is called the nullity of T .

Consider a constant polynomial as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=f\text{'}\text{'}\left(t+4f\text{'}\text{'}\left(t\right)\right) \\ =0+0 \\ =0 \end{gathered}$$

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 .and the rank is 2 whereas the kernel contains all the constant polynomial and the nullity is 1.