91Ó°ÊÓ

Consider the function $T\left(f\right(t\left)\right)=f\left(t^2\right)$from P to P.

A function D is called a linear transformation on $\mathit{R}$if the function D satisfies the following properties.

1. $\mathbf{}\mathit{D}\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}\mathbf{)}\mathbf{=}\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{y}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{x}\mathbf{,}\mathit{y}\mathbf{∈}\mathit{R}\mathbf{.}$
   
2. $\mathbf{}\mathit{D}\mathbf{\left(}\mathit{Î\pm }\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{Î\pm }\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{}\mathit{Î\pm }\mathbf{∈}\mathit{R}\mathbf{.}$
   

An invertible linear transformation is called isomorphism or dimension of domain and co-domain is not same then the function is not isomorphism.

Assume $f,g∈PthenT\left(f\right(t\left)\right)=f\left(t^2\right)andT\left(g\right(t\left)\right)=g\left(t^2\right).$

Substitute the value $f\left(t^2\right)forT\left(f\right(t\left)\right)andg\left(t^2\right)forT\left(g\right(t\left)\right)inT\left(f\right(t\left)\right)+T\left(g\right(t\left)\right)$for and for in as follows.

$T\left(f\right(t\left)\right)+T\left(g\right(t\left)\right)=f\left(t^2\right)+g\left(t^2\right)$

Now, simplify role="math" localid="1659414024917" $T\left(\left\{f+g\right\}\right(t\left)\right)$as follows.

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

Assume $f∈Pand\mathrm{Î\pm }∈Rthen.$

Substitute the value $\mathrm{Î\pm }f\left(t^2\right)forT\left(\mathrm{Î\pm }f\right(t\left)\right)$as follows.

$$\begin{gathered} T\left(\mathrm{Î\pm }f\right(t\left)\right)=\mathrm{Î\pm }f\left(t^2\right) \\ T\left(\mathrm{Î\pm }f\right(t\left)\right)=\mathrm{Î\pm }T\left(f\right(t\left)\right) \end{gathered}$$

As $T(f+g(t\left)\right)=T\left(f\right(t\left)\right)+T\left(g\right(t\left)\right)andT\left(\mathrm{Î\pm }f\right(t\left)\right)=\mathrm{Î\pm }T\left(f\right(t\left)\right)$, by the definiti0on of linear transformation T is linear.

Assume $f\left(t\right)=timpliesf\left(t^2\right)=t^2,$, substitute the value $t^{2}forf\left(t^2\right)and\sqrt{t}forf\left(t\right)$in the equation $T\left(f\right(t\left)\right)=f\left(t^2\right)$as follows.

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

As the function T define from P to P, but $\sqrt{t}$does not belong to P.

By the definition of isomorphism, the function T is not isomorphism.

Hence, the transformation $T\left(f\right(t\left)\right)=f\left(t\right)+f^{\text{'}\text{'}}\left(t\right)+sin\left(t\right)$is linear but not isomorphism.