Let $A$ the vector space of infinitely differentiable functions $\mathbb{R}\rightarrow\mathbb{R}$ and $T$ the linear operator $T(f)=f'+f''$. What are the real eigenvalues and eigenvectors? My approach was solving the equation $f''+f'-\lambda f=0$. I set $u=f,v=f'$ and have the matrix $M=\begin{pmatrix}0 & 1\\\lambda & -1 \end{pmatrix}$ which needs to have real eigenvalues hence $1+4\lambda\geq 0$ so $\lambda\geq -1/4$. Let $t_1=(-1+\sqrt{1+\lambda})/2$ and $t_2=(-1-\sqrt{1+\lambda})/2$. The eigenvectors are $f=c_1e^{t_1x}+c_2e^{t_2x}$.
Is this approach correct?
If $Tf = \lambda f$, then letting $x_1 = f, x_2 = f'$ we obtain the system $x' =M x$, with $x(0) = (f(0),f'(0))^T$ and $M=\begin{bmatrix} 0 & 1 \\ \lambda & -1 \end{bmatrix}$.
The eigenvalues of the matrix $M$ are ${1 \over 2} (-1 \pm \sqrt{ 1+4 \lambda } )$ which are real iff $\lambda \ge -{1 \over 4}$ and the corresponding eigenvector is $f(t) = [e^{Mt} (c_1, c_2)^T]_1$, where $c_1,c_2$ are arbitrary constants.
More detail:
Suppose $\lambda>-{1 \over 4}$, then $M$ has distinct eigenvalues $\mu_1, \mu_2$, and $M$ is diagonalizable, that is there is some $V$ such that $M = V\begin{bmatrix} \mu_1 & 0 \\ 0 & \mu_2 \end{bmatrix} V^{-1}$, and so $e^{Mt} = V\begin{bmatrix} e^{\mu_1 t} & 0 \\ 0 & e^{\mu_2 t} \end{bmatrix} V^{-1}$. It follows that all solutions are of the form $t \mapsto d_1 e^{\mu_1 t} + d_2 e^{\mu_2 t}$.
If $\lambda=-{1 \over 4}$ we see that $\dim \ker (M+{1 \over 2}I) = 1$, hence $M$ is not diagonalizable. Hence using the Jordan form we can write $M = V\begin{bmatrix} -{1 \over 2} & 1 \\ 0 & -{1 \over 2} \end{bmatrix} V^{-1}$, and hence $e^{Mt} = V\begin{bmatrix} e^{-{1 \over 2}t} & t e^{-{1 \over 2}t} \\ 0 & e^{-{1 \over 2}t} \end{bmatrix} V^{-1}$. It follows that all solutions are of the form $t \mapsto d_1 e^{-{1 \over 2}t} + d_2 t e^{-{1 \over 2}t}$.