91影视

Consider the function $T\left(f\left(t\right)\right)=\frac{f\left(t+2\right)-f\left(t\right)}{2}$fromP to P.

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)$\mathit{D}\mathbf{\left(}\mathit{伪}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{伪}\mathit{D}\mathbf{\left(}\mathit{x}\mathbf{\right)}$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$T\left(f\left(t\right)\right)=\frac{f\left(t+2\right)-f\left(t\right)}{2}$and $T\left(g\left(t\right)\right)=\frac{g\left(t+2\right)-g\left(t\right)}{2}$.

Substitute the value$\frac{f\left(t+2\right)-f\left(t\right)}{2}$for$T\left(f\left(t\right)\right)$and$\frac{g\left(t+2\right)-g\left(t\right)}{2}$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.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)=\frac{f(t+2)-f\left(t\right)}{2}\frac{g(t+2)-g\left(t\right)}{2}$

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

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

Assume$f鈭�P$ and$伪鈭�\mathrm{鈩�}$ then $T\left(f\left(t\right)\right)=\frac{f\left(t+2\right)-f\left(t\right)}{2}$.

Substitute the value$\frac{伪f\left(t+2\right)-伪f\left(t\right)}{2}$ for as follows.

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

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 transformationTis an isomorphism if and only if$\mathit{K}\mathit{e}\mathit{r}\left(T\right)\mathbf{=}\mathbf{0}$where $\mathit{K}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\left\{f\left(x\right)鈭�P:Tf\left(x\right)=0\right\}$.

Consider an arbitrary polynomial$f\left(t\right)鈭�P_2$as $f\left(t\right)=a_n{t}^n+a_{n-1}{t}^{n-1}+...+a_{1}t+a_0$.

Simplify $f\left(t+2\right)-f\left(t\right)$as follows.

$$\begin{gathered} \frac{f(t+2)鈭�f\left(t\right)}{2}=\frac{\left\{a_n(t+2{)}^n+a_{n鈭�1}(t+2{)}^{n鈭�1}+鈥�+a_1(t+2)+a_0\right\}鈭�\left\{a_n{t}^n+a_{n鈭�1}{t}^{n鈭�1}+鈥�+a_{1}t+a_0\right\}}{2} \\ =\frac{\left\{b_n{t}^n+b_{n鈭�1}{t}^{n鈭�1}+鈥�+b_{1}t+b_0\right\}鈭�\left\{a_n{t}^n+a_{n鈭�1}{t}^{n鈭�1}+鈥�+a_{1}t+a_0\right\}}{2} \\ =\frac{\left\{b_n鈭�a_n\right\}{t}^n+\left\{b_{n鈭�1}鈭�a_{n鈭�1}\right\}{t}^{n鈭�1}+鈥�+\left\{b_1鈭�a_1\right\}t}{2} \\ \frac{f(t+2)鈭�f\left(t\right)}{2}=a_n^{鈭�}{t}^n+a_{n鈭�1}^{鈭�}{t}^{n鈭�1}+鈥�+a_1^{鈭�}t\left\{\frac{b_n鈭�a_n}{2}=a_n^{鈭�}foralln\right\} \end{gathered}$$

Compare the equations$\frac{f\left(t+2\right)-f\left(t\right)}{2}=0$ and$\frac{f(t+2)鈭�f\left(t\right)}{2}=a_n^{鈭�}{t}^n+a_{n鈭�1}^{鈭�}{t}^{n鈭�1}+鈥�+a_1^{鈭�}t$ as follows.

$a_n^{鈭�}{t}^n+a_{n鈭�1}^{鈭�}{t}^{n鈭�1}+鈥�+a_1^{鈭�}t+=\left(0\right)t^n+\left(0\right)t^{n-1}+...+\left(0\right)t+a_0$ 鈥︹�� (1)

From the equations (1), the value$a_i=0$ for alli except$a_0$ means if$f\left(t\right)=a_0$ then$T\left\{f\left(t\right)\right\}=\frac{a_0-a_0}{2}=0$ but $f\left(t\right)鈮�0$.

As the dimension of $Kernal\left(T\right)鈮�0$, by the definition of isomorphism, the functionT is not isomorphism.

Hence, the transformation$T\left(f\left(t\right)\right)=\frac{f\left(t+2\right)-f\left(t\right)}{2}$is linear but not isomorphism.