91影视

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

A function D is called a linear transformation on$\mathbf{鈩浓划鈩浓划鈩浓划}$if the function D satisfies the following properties.

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

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

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

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

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

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

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

Assume$f鈭�P_2$and $a鈭�\mathrm{鈩�}$then$T\left(\left(伪f\right)\left(t\right)\right)=\left(伪f\right)\left(-t\right)$.

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

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

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

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

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

$$\begin{gathered} T\left\{t^2\right\}={\left(-t\right)}^2 \\ =t^2 \\ T\left\{t^2\right\}=0u_1+0.u_2+1.u_3 \end{gathered}$$

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

The kernel(T) is defined as follows.

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

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

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.

localid="1659758333166" $rank\left(T\right)=\left\{f\left(-t\right)=a_0+a_{1}t+a_2{t}^2|a_i鈭�R\right\}$

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

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

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

As the dimension of$kernel\left(T\right)$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.