91Ó°ÊÓ

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

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

For all elements f,g of V and k 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 ${\mathrm{C}}^{∞}$. to ${\mathrm{C}}^{∞}$..

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.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)={\left(\mathrm{f}+\mathrm{g}\right)}^{\text{'}\text{'}}-5{\left(\mathrm{f}+\mathrm{g}\right)}^{\text{'}}+6\left(\mathrm{f}+\mathrm{g}\right) \\ ={\mathrm{f}}^{\text{'}\text{'}}+{\mathrm{g}}^{\text{'}\text{'}}-5{\mathrm{f}}^{\text{'}}-5{\mathrm{g}}^{\text{'}}+6\mathrm{f}+6\mathrm{g} \\ =\left({\mathrm{f}}^{\text{'}\text{'}}-5{\mathrm{f}}^{\text{'}}+6\right)+\left({\mathrm{g}}^{\text{'}\text{'}}-5{\mathrm{g}}^{\text{'}}+6\mathrm{g}\right) \\ \mathrm{T}\left(\mathrm{f}+\mathrm{g}\right)=\mathrm{T}\left(\mathrm{f}\right)+\mathrm{T}\left(\mathrm{g}\right) \end{gathered}$$

Now, to check the second condition as follows.

Let $\mathrm{Î\pm }$be an arbitrary scalar as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{kf}\right)={\left(\mathrm{kf}\right)}^{\text{'}\text{'}}-5{\left(\mathrm{kf}\right)}^{\text{'}}+6\left(\mathrm{kf}\right) \\ ={\mathrm{kf}}^{\text{'}\text{'}}-5{\mathrm{kf}}^{\text{'}}+6\mathrm{kf} \\ =\mathrm{k}\left({\mathrm{f}}^{\text{'}\text{'}}-5{\mathrm{f}}^{\text{'}}+6\mathrm{f}\right) \\ \mathrm{T}\left(\mathrm{kf}\right)=\mathrm{kT}\left(\mathrm{f}\right) \end{gathered}$$

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.

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

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

The set of next equation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=0 \\ {\mathrm{f}}^{\text{'}\text{'}}-5{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right)=0 \end{gathered}$$

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

$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)-5\mathrm{f}\text{'}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right)=0$

The characteristic equation is as follows.

$$\begin{gathered} m^2-5m+6=0 \\ {m}^2-3m-2m-+6=0 \\ m\left(m-3\right)-2\left(m-3\right)=0 \\ \left(m-2\right)\left(m-3\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{m}-2=0 \\ \mathrm{m}=2 \\ \mathrm{m}-3=0 \\ \mathrm{m}=3 \end{gathered}$$

The roots of the equation$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)-5\mathrm{f}\text{'}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right)=0$is${\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^3\mathrm{x}$

The complementary function of the equation $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)-5\mathrm{f}\text{'}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right)=0$as follows.

$f\left(x\right)=c_1{e}^{2x}+c_2{e}^{3}x$

Now, substitute the value$e^{2x}$ for$\mathrm{f}\left(\mathrm{x}\right)$ in$\mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=0$ as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)={\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)-5\mathrm{f}\left(\mathrm{x}\right)+6\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{T}\left({\mathrm{e}}^{2\mathrm{x}}\right)={\left({\mathrm{e}}^{2\mathrm{x}}\right)}^{\text{'}\text{'}}-5{\left({\mathrm{e}}^{2\mathrm{x}}\right)}^{\text{'}}+6\left({\mathrm{e}}^{2\mathrm{x}}\right) \\ =2{\left({\mathrm{e}}^{2\mathrm{x}}\right)}^{\text{'}}-10{\mathrm{e}}^{2\mathrm{x}}+6{\mathrm{e}}^{2\mathrm{x}} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} T\left(e^{2x}\right)=4e^{2x}-10e^{2x}+6e^{2x} \\ =10e^{2x}-10e^{2x} \\ T\left(e^{2x}\right)=0 \end{gathered}$$

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.