Existence of eigenvalues in a general Banach space Let $(F,||\cdot ||)$ be a Banach space and $T\in\mathcal{L}(F)$. We now ask the following:

In which cases do $T$ admits egienvalues?

Of course, when the dimension of $F$ is finite, conditions of existence are clear. However, when we are dealing with spaces which do not have a finite dimension, things get a little bit more obscure. We give a motivation for this problem.

Let $(X_n)_{n\in\mathbb{N}}$ be a temporally homogenous Markov chain. Then, the transition matrix $P$ acts as a linear operator, i.e. $P\in\mathcal{L}(\ell^1)$. We call $\pi\in\ell^1$ a probability vector if $\sum \pi_k=1$. A common question to ask is whether or not there exists a probability vector $\pi$ associated with $P$ such that $\pi$ is an eigenvector with egienvalue $1$. We call such vectors the stationnary states of $(X_n)_{n\in\mathbb{N}}$. The ergodic theorem tells us about the existence of stationnary states:

Ergodic theorem: If $(X_n)_{n\in\mathbb{N}}$ is irreducible, then the Markov chain admits a stationnary state if and only if it is positive reccurent. Moreover, coordinates of $\pi$ are given by
  $$\pi_k=1/\mathbb{E}_k[T_k].$$


As the problem shows, it is possible to give some conditions for the existence of eigenvalues for even more general spaces than finitely dimensional vector spaces. We clarify that we do not only talk about Hilbert spaces, as shown by our example ($\ell^1$ is not Hilbert). Any ideas on how to proceed for more general operators than transition matrices?
Thank you very much.
 A: There is of course the generalisation of eigenvalues to the notion of spectrum. The spectrum of a bounded operator $A$ is given by
$$
\sigma(A) = \{ \lambda \in \mathbb{C} \mid (\lambda - A) \text{ isn't invertible} \}.
$$
In any text on operator theory it will be proven that $\sigma(A)$ is non-empty. Eigenvalues of $A$ are contained in $\sigma(A)$. But this doesn't mean $A$ necessarily has an eigenvalue.
In fact it's really hard to put a condition on the operator such that it will always have eigenvalues. The only one I'm familiar with (but is too restrictive for your case), is the spectral theorem for self-adjoint compact operators on a Hilbert space.
To illustrate this point, consider the Volterra operator
$$
T:L^2([0,1]) \to L^2([0,1]): f \mapsto (Tf)(x) = \int_0^x f(y) dy.
$$
This is a rather nice compact operator (even on a Hilbert space!), but its set of eigenvalues is empty. In fact its spectrum is simply $\{0\}$. 
Going back to Banach spaces, the non-zero part of the spectrum a compact operator are always eigenvalues. But as the above example demonstrates, there doesn't have to be a non-zero part. So I'm afraid my answer is going to be a negative one, there probably doesn't exist such a condition for general operators. Of course, I'm happy to be proven wrong by any other answers.
