If a matrix contains a differential operator, can it have eigenvalues? If yes, how can I determine them? I will assume that the answer is yes and try to solve the following problem.
Consider the matrix $A$ defined as:
$$ A = \begin{bmatrix}0 & {d \over dx} \\1 & 0 \end{bmatrix} \tag 1$$
The eigenvectors can be found by solving a system of equations given by:
$$ A\vec v - \lambda \vec v =0 \tag 2$$
where $\vec v$ is the eigenvector and $\lambda$ is the eigenvalue. Let $f(x)$ and $g(x)$ be components of $\vec v$:
$$ \vec v = \begin{bmatrix}f(x) \\g(x) \end{bmatrix} \tag 3 $$
Now equation $(2)$ becomes:
$$ \begin{bmatrix}0 & {d \over dx} \\1 & 0 \end{bmatrix} \begin{bmatrix}f(x) \\g(x) \end{bmatrix} = \lambda \begin{bmatrix}f(x) \\g(x) \end{bmatrix}\tag 4$$
which is essentially the following system of equations:
$$ {d g(x) \over dx} - \lambda f(x)=0 \tag 5$$ $$ f(x) - \lambda g(x)=0 \tag 6$$
The solutions of this system of equations are:
$$ f(x) = C \lambda e^{\lambda ^2 x} \tag 7$$ $$ g(x) = C e^{\lambda ^2 x} \tag 8$$
The constant $C$ should be determined from the boundary condition. However, I don't know how to actually determine the value of $\lambda$ (assuming that it actually exists). If $A$ was a linear operator, the eigenvalues would be calculated from its determinant. But now $A$ contains a differential operator. So my question is, how can the eigenvalues of $A$ be determined?
Suppose $v=(f,g)$ is an eigenvector for $A$. Then, there exists a $\lambda \in \mathbb R$ satisfying the equations you wrote. But If $g'(x) = \lambda f(x)$, and $f(x) = \lambda g(x)$, then $g'(x)=\lambda^2g(x)$. Therefore there exists $\delta \in \mathbb R$ for which $g(x) = \delta \exp(\lambda^2 x)$. But that completely determines $f$. $f(x)=\lambda g(x) = \lambda \delta \exp(\lambda^2 x)$.
Pick $(\lambda, \delta) \in \mathbb R^2$. Then $(\delta \exp(\lambda^2 c) , \lambda \delta \exp(\lambda^2 x))$ are the set of eigenvectors of $A$. Which value of $\lambda$ works? All of them! $A$ has a spectrum of eigenvalues instead of finite, and in this case, any $\lambda \in \mathbb R$ is an eigenvalue of $A$, even 0, with the functions $(x \mapsto 0, x \mapsto 0)$.