91影视

Consider the function from to .

A functionD 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)$\mathit{D}\mathbf{\left(}\mathit{伪}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{伪}\mathit{D}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$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 $T\left(f\right(t\left)\right)=(t-1)f\left(t\right)$and $T\left(g\right(t\left)\right)=(t-1)g\left(t\right)$.

Substitute the value $\left(t-1\right)f\left(t\right)$for$T\left(f\left(t\right)\right)$and $\left(t-1\right)g\left(t\right)$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.

$T\left(f\right(t\left)\right)+T\left(g\right(t\left)\right)=(t-1)f\left(t\right)+(t-1)g\left(t\right)$

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

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

Assume$f鈭�P$and$伪鈭�\mathrm{鈩�}$then $T\left(f\right(t\left)\right)=(t-1)f\left(t\right)$.

Substitute the value $\left(t-1\right)\left\{伪f\left(t\right)\right\}$for $T\left(伪f\left(t\right)\right)$as follows.

$$\begin{gathered} T\left(伪f\right(t\left)\right)=\left(t-1\right)t\left\{伪f\left(t\right)\right\} \\ =伪\left\{\left(t-1\right)f\left(t\right)\right\} \\ T\left(伪f\right(t\left)\right)=伪T\left(f\left(t\right)\right) \end{gathered}$$

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 transformation is linear.

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$T\left(f\right(t\left)\right)=(t-1)f\left(t\right)$, substitute the value$a_0+a_{1}t+...+a_n{t}^n$ for$f\left(t\right)$ in the equation$T\left(f\left(t\right)\right)=\left(t-1\right)f\left(t\right)$ as follows.

$$\begin{gathered} T\left(f\right(t\left)\right)=tf\left(t\right) \\ T\left(a_0+a_{1}t+...+a_n{t}^n\right)=\left(t-1\right)\left\{a_0+a_{1}t+...+a_n{t}^n\right\} \\ =t\left\{a_0+a_{1}t+...+a_n{t}^n\right\}-\left\{a_0+a_{1}t+...+a_n{t}^n\right\} \\ =a_0+a_{1}t+...+a_n{t}^n-a_0+a_{1}t+...+a_n{t}^n \\ T\left(a_0+a_{1}t+...+a_n{t}^n\right)=-a_0+\left(a_0-a_1\right)t+\left(a_1-a_2\right)t^2+...+\left(a_{n-1}-a_n\right)t^n+a_n{t}^{n+1} \end{gathered}$$

From the equation, there is no zero degree polynomial in the codomain means for a polynomial$g\left(t\right)鈭�P$ there is no$f\left(t\right)鈭�P$ such that $T\left(f\left(t\right)\right)=g\left(t\right)$.

Hence, the transformation$T\left(f\left(t\right)\right)=\left(t-1\right)f\left(t\right)$ is linear but not isomorphism.