91影视

Consider the function from to .

A function is called a linear transformation on if the function 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 $\mathit{伪}\mathbf{鈭�}\mathit{R}$.

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$T\left(f\left(t\right)\right)={鈭�}_0^{t}f\left(x\right)dx$and$T\left(g\left(t\right)\right)={鈭�}_0^{t}g\left(x\right)dx$.

Substitute the value${鈭�}_0^{t}f\left(x\right)x$for$T\left(f\left(t\right)\right)$and${鈭�}_0^{t}g\left(x\right)dx$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.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)={鈭�}_0^{t}f\left(x\right)dx+{鈭�}_0^{t}g\left(x\right)dx$

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

$$\begin{gathered} T(f+g(t\left)\right)={鈭�}_0^t\left\{f+g\right\}\left(x\right)dx \\ ={鈭�}_0^{t}f\left(x\right)+g\left(x\right)dx \\ ={鈭�}_0^{t}f\left(x\right)dx+{鈭�}_0^{t}g\left(x\right)dx \\ T(f+g(t\left)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right) \end{gathered}$$

Assume $f鈭�P$and$伪鈭�\mathrm{鈩�}$then $T\left(\left(伪f\right)\left(t\right)\right)={鈭�}_0^t\left(伪f\right)\left(x\right)dx$.

Simplify the equation$T\left(\left(伪f\right)\left(t\right)\right)={鈭�}_0^t\left(伪f\right)\left(x\right)dx$as follows.

$$\begin{gathered} T\left(\left(伪f\right)\left(t\right)\right)={鈭�}_0^t\left(伪f\right)\left(x\right)dx \\ =伪{鈭�}_0^{t}f\left(x\right)dx \\ T\left(\left(伪f\right)\left(t\right)\right)=伪T\left(f\left(t\right)\right) \end{gathered}$$

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 $\mathit{K}\mathit{e}\mathit{r}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\left\{f\left(x\right)鈭�P:T\left\{f\left(x\right)\right\}=0\right\}$.

As the inverse of integral function is differential function means$T\left(f\left(t\right)\right)={鈭�}_0^{t}f\left(x\right)dx$implies $T^{-1}\left(f\left(t\right)\right)=f\text{'}\left(t\right)$.

As the inverse of constant function does not exist means there is no image of constant function.

By the definition of isomorphism, the functionT is not isomorphism.

Hence, the transformation$T\left(f\left(t\right)\right)={鈭�}_0^{t}f\left(x\right)dx$is linear but not isomorphism.