Question

Find all the eigenvalues and “eigenvectors” of the linear transformations.

$\mathbf{T}\left(f\right)\mathbf{=}\mathbf{f}\mathbf{\text{'}}\mathbf{‐}\mathbf{f}\mathbf{}\mathbf{from}\mathbf{}\mathbf{C}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{}\mathbf{to}\mathbf{}\mathbf{C}\mathbf{\text{'}}\mathbf{\text{'}}$

Short answer

The eigenvalues and eigenvectors of the linear transformation is $\mathrm{\lambda }\in {\mathrm{R}}_{,}{\mathrm{E}}_{\mathrm{\lambda }}=\mathrm{span}\left({\mathrm{e}}^{\left(\mathrm{\lambda }+1\right)\mathrm{t}}\right)$.

Step-by-step solution

Define eigenvalues

The scalar values that are associated with the vectors of the linear equations in the matrix are called eigenvalues.

$\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{=}\mathbf{\lambda }\overrightarrow{\mathbf{x}}$, here $\overrightarrow{\mathbf{x}}$is eigenvector and $\mathbf{\lambda }$is the eigenvalue.

Use the formula and find the eigenvalues and eigenvectors

Consider the given equation,$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}-\mathrm{f}$

Solve,

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{\lambda f}$

Substitute the value of$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{\lambda f}$

$$\begin{gathered} \mathrm{f}\text{'}-\mathrm{f}=\mathrm{\lambda f} \\ \mathrm{f}\text{'}=\left(\mathrm{\lambda }+1\right)\mathrm{f} \\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{C}·{\mathrm{e}}^{\left(\mathrm{\lambda }+1\right)\mathrm{x}},\mathrm{C}\in \mathrm{R} \end{gathered}$$

Hence, for the eigenvalue $\mathrm{\lambda }\in \mathrm{R}$, the eigenvector space is${\mathrm{E}}_{\mathrm{\lambda }}=\mathrm{span}\left({\mathrm{e}}^{\left(\mathrm{\lambda }+1\right)\mathrm{x}}\right)$

Thus,$\mathrm{\lambda }\in \mathrm{R},{\mathrm{E}}_{\mathrm{\lambda }}=\mathrm{span}\left({\mathrm{e}}^{\left(\mathrm{\lambda }+1\right)\mathrm{t}}\right)$