91Ó°ÊÓ

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

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

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

The given transformation as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}-5\mathrm{f}\text{'}+6\mathrm{f}$, from $C^{∞}$. to $C^{∞}$..

By using the definition of linear transformation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \\ \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kT}\left(\mathrm{A}\right) \end{gathered}$$

Now, to check the first condition as follows.

Consider the function as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)={\left(\mathrm{f}+\mathrm{g}\right)}^{\text{'}\text{'}}+2{\left(\mathrm{f}+\mathrm{g}\right)}^{\text{'}}+\left(\mathrm{f}+\mathrm{g}\right) \\ ={\mathrm{f}}^{\text{'}\text{'}}+{\mathrm{g}}^{\text{'}\text{'}}+2{\mathrm{f}}^{\text{'}}+2{\mathrm{g}}^{\text{'}}+\mathrm{f}+\mathrm{g} \\ =\left({\mathrm{f}}^{\text{'}\text{'}}+2{\mathrm{f}}^{\text{'}}+f\right)+\left({\mathrm{g}}^{\text{'}\text{'}}+2{\mathrm{g}}^{\text{'}}+\mathrm{g}\right) \\ \mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)=\mathrm{T}\left(\mathrm{f}\right)+\mathrm{T}\left(\mathrm{g}\right) \end{gathered}$$

Now, to check the second condition as follows.

Let role="math" localid="1659755819661" $\mathrm{Î\pm }$be an arbitrary scalar as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{kf}\right)={\left(\mathrm{kf}\right)}^{\text{'}\text{'}}+2{\left(\mathrm{kf}\right)}^{\text{'}}+\left(\mathrm{kf}\right) \\ ={\mathrm{kf}}^{\text{'}\text{'}}+2{\mathrm{kf}}^{\text{'}}+\mathrm{kf} \\ =\mathrm{k}\left({\mathrm{f}}^{\text{'}\text{'}}-2{\mathrm{f}}^{\text{'}}+\mathrm{f}\right) \\ \mathrm{T}\left(\mathrm{kf}\right)=\mathrm{kT}\left(\mathrm{f}\right) \end{gathered}$$

Thus, T is a linear transformation.

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

The kernel as follows.

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

Let$\mathrm{f}\left(\mathrm{x}\right)$be an arbitrary smooth function from ${\mathrm{C}}^{∞}$.

The set of next equation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=0 \\ {\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)+2{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)+2\mathrm{f}\left(\mathrm{x}\right)=0 \end{gathered}$$

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

$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)+2\mathrm{f}\text{'}\left(\mathrm{x}\right)+2\mathrm{f}\left(\mathrm{x}\right)=0$

The characteristic equation is as follows.

$$\begin{gathered} {\mathrm{m}}^2+2\mathrm{m}+1=0 \\ {\mathrm{m}}^2+\mathrm{m}+\mathrm{m}+6=0 \\ \mathrm{m}\left(\mathrm{m}+1\right)+1\left(\mathrm{m}+1\right)=0 \\ \left(\mathrm{m}+1\right)\left(\mathrm{m}+1\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{m}+1=0 \\ \mathrm{m}=1 \\ \mathrm{m}+1=0 \\ \mathrm{m}=1 \end{gathered}$$

The roots of the equation $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)+2\mathrm{f}\text{'}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=0$is ${\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^x$

The complementary function of the equation$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)+2\mathrm{f}\text{'}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=0$ as follows.

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}$

Now, substitute the value$e^{-x}$ for$f\left(x\right)$ in$T\left(f\right(x\left)\right)=0$ as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)={\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)+2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{T}\left({\mathrm{e}}^{-\mathrm{x}}\right)={\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{\text{'}\text{'}}+2{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{\text{'}}+\left({\mathrm{e}}^{-\mathrm{x}}\right) \\ =-{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{\text{'}}-2{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}} \\ ={\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} T\left(e^{-x}\right)=2e^{-x}-2e^{-x} \\ T\left(e^{-x}\right)=0 \end{gathered}$$

This implies that as follows.

$\mathrm{kert}≠\left\{0\right\}$

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

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