91影视

Consider the function $T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$fromP toP.

A function is called a linear transformation on$\mathbf{鈩�}$ if the functionD satisfies the following properties.

(a)$\mathit{D}\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}\mathbf{)}\mathbf{=}\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{y}\mathbf{\right)}$for all $\mathit{x}\mathbf{,}\mathit{y}\mathbf{鈭�}\mathbf{鈩�}$.

(b)role="math" localid="1659869577736" $\mathit{D}\mathbf{\left(}\mathit{伪}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{伪}\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}$for all constant $\mathit{伪}\mathbf{鈭�}\mathbf{鈩�}$.

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

Assume$f,g鈭�P$then$T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$and $T\left(g\left(t\right)\right)=g\text{'}\left(t\right)$.

Substitute the value$f^{\text{'}}\left(t\right)$for$T\left(f\left(t\right)\right)$and$g^{\text{'}}\left(t\right)$for$T\left(g\left(t\right)\right)$in$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$as follows.

$T\left(f\right(t\left)\right)+T\left(g\right(t\left)\right)=f^{\text{'}}\left(t\right)+g^{\text{'}}\left(t\right)$

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

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

Substitute the value$\left\{伪f^{\text{'}}\left(t\right)\right\}$ for$T\left(伪f\right(t\left)\right)$ as follows.

$$\begin{gathered} T\left(伪f\right(t\left)\right)=\left\{伪f^{\text{'}}\left(t\right)\right\} \\ =伪\left\{f^{\text{'}}\left(t\right)\right\} \\ T\left(伪f\right(t\left)\right)=伪T\left(f\left(t\right)\right) \end{gathered}$$

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

Theorem: Consider a linear transformationT defined from$\mathit{T}\mathbf{:}\mathit{V}\mathbf{鈫�}\mathit{W}$then the transformationis an isomorphismT if and only if$\mathit{K}\mathit{e}\mathit{r}\left(T\right)\mathbf{=}\mathbf{0}$where$\mathit{K}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\left\{f\left(x\right)鈭�P:Tf\left(x\right)=0\right\}$and$\mathit{d}\mathit{i}\mathit{m}\mathbf{\left(}\mathit{K}\mathit{e}\mathit{r}\mathbf{\right(}\mathit{T}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\left\{f\left(x\right)鈭�P:T\left\{f\left(x\right)\right\}=0impliesf\left(x\right)=0\right\}$.

Consider a polynomial$f\left(t\right)鈭�P$then$f\left(t\right)=a_0+a_{1}t+...+a_n{t}^n$where $a_i鈭�\mathrm{鈩�}$.

As $T\left(f\left(t\right)\right)=f\text{'}\left(t\right)$, substitute the value$a_0+a_{1}t+...+a_n{t}^n$for$f\left(t\right)$in the equation$T\left(f\right(t\left)\right)=f^{\text{'}}\left(t\right)$as follows.

$$\begin{gathered} T\left(f\right(t\left)\right)=tf\left(t\right) \\ T\left(a_0+a_{1}t+...+a_n{t}^n\right)=\frac{d}{dt}\left\{a_0+a_{1}t+...+a_n{t}^n\right\} \\ T\left(a_0+a_{1}t+...+a_n{t}^n\right)=a_1+2a_{2}t+...+n{a}_n{t}^{n-1} \end{gathered}$$

Compare the equations $a_1+2a_{2}t鈥�+n{a}_n{t}^{n-1}$and $\left(0\right)+\left(0\right)t+鈥�+\left(0\right)t^t$as follows.

$a_1+2a_{2}t...+n{a}_n{t}^{n-1}=\left(0\right)+\left(0\right)t+...+\left(0\right)t^n$

Therefore, the values $a_1=0$,$a_2=0$ ,鈥�,$a_n=0$.

The definition of dimension of$Ker\left(T\right)$ is defined as follows.

$$\begin{gathered} \dim \left(\mathrm{Ker}\right(T\left)\right)=\left\{f\left(x\right)鈭�P:T\left\{f\right(x\left)\right\}=\left(0\right)+2\left(0\right)t鈥�+n\left(0\right)t^{n鈭�1}\mathrm{Where}f\left(x\right)={\mathrm{a}}_0+\left(0\right)t鈥�+\left(0\right)t^{n鈭�1}\right\} \\ =\left\{f\left(x\right)鈭�P:T\left\{f\right(x\left)\right\}=\left(0\right)+2\left(0\right)t鈥�+n\left(0\right)t^{n鈭�1}\mathrm{where}f\left(x\right)={\mathrm{a}}_0\right\} \\ \dim \left(\mathrm{Ker}\right(T\left)\right)鈮�\left\{0\right\} \end{gathered}$$

By the definition of isomorphism, the transformationt is not an isomorphism.

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