91Ó°ÊÓ

Consider two linear spaces and . A transformation is said to be a linear transformation if it satisfies the properties,

$$\begin{gathered} \mathit{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathit{T}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{+}\mathit{T}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{T}\mathbf{\left(}\mathit{k}\mathit{v}\mathbf{\right)}\mathbf{=}\mathit{k}\mathit{T}\mathbf{\left(}\mathit{v}\mathbf{\right)} \end{gathered}$$

For all elements f,g of v and k is scalar.

An invertible linear transformation is called an isomorphism.

Consider the transformation $T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$, from$P_2$ to $P_2$.

Check whether the transformationsatisfies the below two conditions or not.

$$\begin{gathered} 1.T(\mathrm{f}+\mathrm{g})=T\left(\mathrm{f}\right)+T\left(\mathrm{g}\right) \\ 2.T\left(kf\right)=kT\left(f\right) \end{gathered}$$

Verify the first condition.

Let$f\left(t\right)$and$g\left(t\right)$be two polynomial functions from $P_2$. Then,$T\left(f\right(t\left)\right)=f\text{'}\left(t\right),T\left(g\left(t\right)\right)=g\text{'}\left(t\right)$

Find $T\left(f\left(t\right)+g\left(t\right)\right)$.

$$\begin{gathered} T\left(f\left(t\right)+g\left(t\right)\right)=T\left(\left(f+g\right)\left(t\right)\right) \\ =\left(f+g\right)\text{'}\left(2t\right) \\ =\left(f\text{'}+g\text{'}\right)+g\text{'}\left(t\right) \\ =f\text{'}\left(t\right)+g\text{'}\left(t\right) \\ T\left(f\right(t)+g(t\left)\right)=T\left(f\right(t\left)\right)+T\left(g\right(t\left)\right) \end{gathered}$$

It is clear that, the first condition$T\left(f+g\right)=T\left(f\right)+T\left(g\right)$ is satisfied.

Verify the second condition.

Let k be an arbitrary scalar, and$f\left(t\right)∈P_2$ as follows.

$$\begin{gathered} T\left(\left(kf\right)\left(t\right)\right)=\left(kf\right)\text{'}\left(t\right) \\ =k(f\text{'}(t\left)\right) \\ =kT\left(f\left(t\right)\right) \end{gathered}$$

It is clear that, the second condition$T\left(kf\right)=kT\left(f\right)$ is also satisfied.

Thus, T is a linear transformation.

A linear transformation $\mathit{T}\mathbf{:}\mathit{V}\mathbf{→}\mathit{W}$ is said to be an isomorphism if and only if $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and$\mathbf{lm}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\left\{0\right\}$.

Now, check whether $\ker \left(T\right)=\left\{0\right\}$.

According to the definition of the kernelof a transformation,

$ker\left(T\right)=\left\{f∈P_2,T\left(f\right(t\left)\right)=0\right\}$.

Consider a polynomial function $f\left(t\right)∈P_2$asrole="math" localid="1659855689876" $f\left(t\right)=a+bt+c{t}^2$

Then,

$$\begin{gathered} T\left(f\right(t\left)\right)=0 \\ f\text{'}\left(t\right)=0 \\ \frac{d}{dt}(a+bt+c{t}^2)=0 \\ 0+b+2ct=0+0t+0t^2 \\ b+2ct=0t+0t^2 \end{gathered}$$

Comparing both sides, it can be concluded that$b=0,c=0$

Here, the variableis a free variable. This means, the kernel of the transformation T will be $ker\left(T\right)=\left\{a+0t+0t^2|a∈\mathrm{ℂ}\right\}$.

Clearly, $\ker \left(T\right)≠\left\{0\right\}$.

It can be concluded that, one of the property of the isomorphism definition is not satisfied.

Therefore, the transformation T is not an isomorphism.

Thus, the transformation T is a linear transformation and is not an isomorphism.