91影视

Let us define the${\mathrm{C}}^{鈭�}$ sets as follows.

${\mathrm{C}}^{鈭�}=\left\{\mathrm{f}\left(\mathrm{x}\right)\left|\mathrm{f}\right(\mathrm{x})isasmoothfunction\right\}$

Now, let us define the transformation as follows.

$\mathrm{T}:{\mathrm{C}}^{鈭�}鈫�{\mathrm{C}}^{鈭�}$with $\mathrm{T}(\mathrm{f}\left(\mathrm{x}\right))=\mathrm{f}(\mathrm{x})+\mathrm{f}"(\mathrm{x})$.

Consider two linear spaces V and W. A function T is said to be linear transformation if the following holds.

$$\begin{gathered} \mathbf{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{+}\mathbf{T}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}\mathbf{=}\mathbf{kT}\mathbf{\left(}\mathbf{f}\mathbf{\right)} \end{gathered}$$

For all elements f,g of V and k is scalar.

A linear transformation$\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$is said to be an isomorphism if and only if $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and$\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{W}$or $\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{W}\mathbf{\right)}$.

To check the first condition as follows.

Consider an arbitrary smooth function f(x) and g(x) from${\mathrm{C}}^{鈭�}$and as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\right(\mathrm{x})+\mathrm{g}(\mathrm{x}\left)\right)=\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)\right)+\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)\right)" \\ =\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}"\left(\mathrm{x}\right)+\mathrm{g}"\left(\mathrm{x}\right) \\ =\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}"\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)+\mathrm{g}"\left(\mathrm{x}\right) \\ \mathrm{T}\left(\mathrm{f}\right(\mathrm{x})+\mathrm{g}(\mathrm{x}\left)\right)=\mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)+\mathrm{T}\left(\mathrm{g}\right(\mathrm{x}\left)\right) \end{gathered}$$

Now, to check the second condition as follows.

Let$伪$ be an arbitrary scalar, and $f\left(x\right)鈭�\mathrm{P}$.

$$\begin{gathered} \mathrm{T}\left(\mathrm{伪蹿}\left(\mathrm{x}\right)\right)=\mathrm{伪蹿}\left(x\right)+\left(\mathrm{伪蹿}\left(x\right)\right)" \\ =\mathrm{伪蹿}\left(x\right)+\mathrm{伪蹿}"\left(x\right) \\ =\mathrm{伪}\left(\mathrm{f}\left(x\right)+\mathrm{f}"\left(x\right)\right) \\ \mathrm{T}\left(\mathrm{伪蹿}\left(\mathrm{x}\right)\right)=\mathrm{伪罢}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \end{gathered}$$

Thus, T is a linear transformation.

A linear transformation $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$ is isomorphism if and only if$\mathbf{ker}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$ and$\mathbf{lm}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{W}$

Now, check if $\ker \left(\mathrm{t}\right)=\left\{0\right\}$.

The domain is the space of all smooth functions${\mathrm{C}}^{鈭�}$and hence the zero in the domain will be zero polynomial as follows.

p(x) = 0

And the codomain is the same space, hence the zero will be again $\mathrm{p}\left(\mathrm{x}\right)=0$.

The kernel as follows.

$\ker \left(\mathrm{T}\right)=\left\{\mathrm{x}鈭�\mathrm{P}|\mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=0\right\}$

Let f(x) be an arbitrary smooth function from ${\mathrm{C}}^{鈭�}$.

The set of next equation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=0 \\ \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \end{gathered}$$

The last equation is the first order linear differential equation as follows.

$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}"\left(\mathrm{x}\right)=0$

The characteristic equation is as follows.

$$\begin{gathered} {\mathrm{m}}^2+1=0 \\ \mathrm{m}=卤\mathrm{i} \end{gathered}$$

The roots of the equation$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$is${\mathrm{c}}_1\cos \left(\mathrm{x}\right)+\mathrm{c}{}_{2}sin\left(\mathrm{x}\right)$

The complementary function of the equation$f\left(x\right)+f"\left(x\right)=0$as follows.

$f\left(x\right)=c_1\cos \left(x\right)+c_2\sin \left(x\right)$

Now, substitute the value${\mathrm{c}}_1\cos \left(\mathrm{x}\right)+{\mathrm{c}}_{2}sin\left(\mathrm{x}\right)$for f(x) in$\mathrm{T}(\left(\mathrm{f}\right(\mathrm{x}\left)\right)=0$as follows.

$$\begin{gathered} T\left(f\right(x\left)\right)=f\left(x\right)+f"\left(x\right) \\ T\left(c_1\cos \left(x\right)+c_2\cos \left(x\right)\right)=c_1\cos \left(x\right)+c_2\sin \left(x\right)+\left(-c_1\cos \left(x\right)+\cos c_2\left(x\right)\right)\text{'} \\ =c_1\cos \left(x\right)+c_2\sin \left(x\right)-c_1\cos \left(x\right)-c_2\cos \left(x\right) \\ =0 \end{gathered}$$

This implies that as follows.

$$\begin{gathered} \mathrm{kert}=\left\{\mathrm{f}\left(\mathrm{x}\right)鈭�{\mathrm{C}}^{鈭�}\left|\mathrm{f}\right(\mathrm{x})=-\mathrm{f}"(\mathrm{x})\right\} \\ \mathrm{kert}鈮�\left\{0\right\} \end{gathered}$$

Since,$\mathrm{kert}鈮�\left\{0\right\}$and implies that T is not an isomorphism.

Thus, T is a linear transformation and$\mathrm{kert}鈮�\left\{0\right\}$which implies that T is not an isomorphism.