91Ó°ÊÓ

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

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

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

An invertible linear transformation is called an isomorphism.

Consider the transformation $T\left(f\left(t\right)\right)=f\left(2t\right)$, from$P_2$to $P_2$.

Check whether the transformationsatisfies the below two conditions or not.

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

Verify the first condition.

Let$f\left(t\right)$and$g\left(t\right)$be two polynomial functions from $P_2$. Then,$T\left(f\right(t\left)\right)=f(-t),T\left(g\left(t\right)\right)=g(-t)$

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

$$\begin{gathered} T\left(f\left(t\right)+g\left(t\right)\right)=T\left(\left(f+g\right)\left(t\right)\right) \\ =\left(f+g\right)\left(2t\right) \\ =f\left(2t\right)+g\left(2t\right) \\ =T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right) \end{gathered}$$

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

Verify the second condition.

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

$$\begin{gathered} T\left(\left(kf\right)\left(t\right)\right)=\left(kf\right)\left(-t\right) \\ =k\left(f\right(-t\left)\right) \\ =kT\left(f\left(t\right)\right) \end{gathered}$$

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 ifrole="math" localid="1659782544447" $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$ and$\mathit{I}\mathit{m}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathit{W}$.

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

According to the definition of the kernel of a transformation,

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

Consider a polynomial function $f\left(t\right)∈P_2$as$f\left(t\right)=a+bt+c^2$

Then,

$$\begin{gathered} T\left(f\left(t\right)\right)=0 \\ f\left(2t\right)=0 \\ a+b\left(2t\right)+c{\left(2t\right)}^2=0 \\ a+2bt+c{t}^2=0+0t+0t^2 \end{gathered}$$

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

This means, the kernel of the transformation T will be $\ker \left(T\right)=\left\{0\right\}$.

Now, check whether$\mathrm{Im}\left(T\right)=\mathrm{â„‚}$.

According to the definition of the image of a transformation, $\mathrm{lm}\left(T\right)=\left\{T\left(f\left(t\right)\right):f\left(t\right)∈P_2\right\}$

Let$f\left(t\right)=a+bt+c^2$is in $P_2$.

Then,

role="math" localid="1659782181604" $$\begin{gathered} T\left(f\left(t\right)\right)=f\left(2t\right) \\ =a+b\left(2t\right)+c{\left(2t\right)}^2 \\ =a+2bt+4c{t}^2 \\ =g\left(t\right) \end{gathered}$$

Say, $\left(g\right(t)=a+2bt+4c{t}^2)$.

It is clear that, role="math" localid="1659781881759" $T\left(f\left(t\right)\right)=g\left(t\right)∈P_2$.

This means for any $g\left(t\right)∈P_2$, there exists$f\left(t\right)∈P_2$ a such that$T\left(f\left(t\right)\right)=g\left(t\right)$

So,$\mathrm{Im}\left(T\right)=P_2$

Therefore, the transformation T is an isomorphism.

Thus, the transformation T is a linear transformation and T is an isomorphism.