Eigenvalues of an operator I am stuck on this:

Deduce the eigenvalues of the operator $A^{\dagger} A$ are positive, where
$$A^{\dagger} = -\frac{\text{d}}{\text{d}x} + \tanh(x)$$
$$A = \frac{\text{d}}{\text{d}x} + \tanh(x)$$

The point is that those are not matrices, in the usual sense I understand them, so how shall I proceed?
I shall basically solve
$$(A^{\dagger}A)|\phi\rangle = a |\phi\rangle$$
That is
$$\text{det}(A^{\dagger}A - a\mathbb{1}) = 0$$
And here I somehow stopped.
I also managed to compute the commutator, and I found
$$[A^{\dagger}, A] = -2 \left[\frac{\text{d}}{\text{d}x}, \tanh(x)\right]$$
 A: You don't actually need to solve for either eigenvectors or eigenvalues, and in fact you don't need to know what $A$ is at all.
To prove the positivity, simply consider any eigenvector $\psi$ of $A^\dagger A$ with eigenvalue $\lambda$, 
$$
A^\dagger A \psi = \lambda \psi,
$$
take the inner product of that equation with $\psi$,
$$
⟨\psi,A^\dagger A \psi⟩ = ⟨\psi,\lambda \psi⟩,
$$
and use the defining property of the adjoint,
$$
⟨\psi,A^\dagger A \psi⟩ = ⟨A\psi,A \psi⟩,
$$
to obtain a relationship that involves $\lambda$ and quantities (specifically, norms) which must always be non-negative.

If you want to prove that the eigenvalues are strictly positive, on the other hand, then you will need to work a good deal harder to rule out the possibility of a zero eigenvalue. With the information you've provided it's impossible to tell (as there are not enough boundary conditions), but you can start by showing that $\lambda=0$ requires that
$$
⟨A\psi,A \psi⟩=0
$$
and therefore that $A\psi$ itself be zero; that then gives you a solvable ODE that you can couple with the boundary conditions of the problem (you do have them, right?) to tell whether that $\psi$ is a reasonable state or not.
A: 
The point is that those are not matrices, in the usual sense I
  understand them, so how shall I proceed?

You might proceed by looking for eigenfunctions $\phi(x)$ such that
$$A^\dagger A\,\phi(x) = c\,\phi(x)$$
and show that all the $c$ are real and positive.
For example, consider the simpler case of $A = \frac{d}{dx}$ so that
$$A^\dagger A = -\frac{d^2}{dx^2}$$
and then the eigenvalue equation is
$$-\frac{d^2}{dx^2}\,\phi(x) = c\, \phi(x)$$
The eigenfunctions are well known to be
$$\phi_k(x) = A\sin(kx) + B\cos(kx)$$
with eigenvalues
$$c_k = k^2$$
