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Consider the function $T\left(f\left(t\right)\right)=f\left(2t\right)from{P}_{2}to{P}_2$.

A function D is called a linear transformation onrole="math" localid="1659760842355" $\mathbf{鈩浓划鈩�}$if the function D satisfies the following properties.

(a) $\mathit{D}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{=}\mathit{D}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{x}\mathbf{,}\mathit{y}\mathbf{鈭�}\mathbf{鈩�}\mathbf{}\mathit{x}\mathbf{,}\mathit{y}\mathbf{鈭�}\mathbf{鈩�}$.

(b) $\mathit{D}\mathbf{\left(}\mathbf{ax}\mathbf{\right)}\mathbf{=}\mathit{a}\mathit{D}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{}\mathit{a}\mathbf{鈭�}\mathbf{鈩�}$.

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

Assume $f,g鈭�P_{2}thenT\left(f\left(t\right)\right)andT\left(g\left(t\right)\right)=g\left(2t\right)$.

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

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

Now, simplify$T\left(\left\{f+g\right\}\left(t\right)\right)$ as follows.

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

Assume$f鈭�P_2$ and $a鈭�Ra鈭�Ra鈭�Ra鈭�R$then$T\left(\left(伪f\right)\left(t\right)\right)=\left(伪f\right)\left(2t\right)$ .

Simplify the equation $T\left(\left(伪f\right)\left(t\right)\right)=\left(伪f\right)\left(2t\right)$as follows.

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

As$T\left(\left\{f+g\right\}\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)andT\left(伪f\left(t\right)\right)=伪T\left(f\left(t\right)\right)$ and , by the definition of linear T transformation is linear.

As$\left\{1,t,t^2\right\}$ is the basis element of$p_2$ , the matrix B is defined as follows.

$$\begin{gathered} T\left\{1\right\}=1 \\ T\left\{1\right\}=1.u_1+0.u_2+0.u_3 \end{gathered}$$

The image of at point is defined as follows.

role="math" localid="1659761824055" $$\begin{gathered} T\left\{t\right\}=2t \\ T\left\{t\right\}=0.u_1+2.u_2+0.u_3 \end{gathered}$$

The image of T at point $t^2$is defined as follows.

$$\begin{gathered} T\left\{t^2\right\}={\left(2t\right)}^2 \\ =4t^2 \\ T\left\{t^2\right\}=0u_1+0.u_2+4.u_3 \end{gathered}$$

Therefore, the matrix$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 4\end{array}\right]$ .

The kernel(T) is defined as follows.

$kernal\left(T\right)=\left\{f\left(t\right)=\overline{)0}t鈭�R\right\}$

As Kernel(T) is spanned by$\left\{0\right\}$ implies$\left(kernel\left(T\right)\right)=0$ .

The rank(T) is defined as follows.

$$\begin{gathered} dim\left(T\right)=dim\left(kernel\left(T\right)\right)+rank\left(T\right) \\ 3=0+rank\left(T\right) \\ rank\left(T\right)=3 \end{gathered}$$

The rank(T) is defined as follows.

$rank\left(T\right)=\left\{f\left(-t\right)=a_0+a_{1}t+a_2{t}^2|a_i鈭�R\right\}$

As rank(T) is spanned by$\left\{1,t,t^2\right\}$ implies the rank(T)=3 .

Therefore, the dimension of kernel(T) is 0 and rank(T) is 3 and the basis of kernel(T) is$\left\{0\right\}$ and the basis of rank(T) is$\left\{1,t,t^2\right\}$ .

Theorem: Consider a linear transformation T defined from $\mathit{T}\mathbf{:}\mathit{V}\mathbf{鈫�}\mathit{W}$then the transformation is an isomorphism if and only if ker(T) where $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{鈭�}\mathbf{P}\mathbf{:}\mathbf{T}\mathbf{\left\{}\mathbf{f}\left(x\right)\mathbf{\right\}}\mathbf{=}\mathbf{0}\mathbf{\}}$.

As the dimension of kernel(T) is 0, by the theorem the function T is an isomorphism.

Hence, the rank of transformation T is 3, kernel of the function T is 0 and linear transformation is an isomorphism.