If your equations look like $x_{i+1} = Ax_i$, where each $x_j$ is an element of a (possibly infinite dimensional) vector space and $A$ is a linear operator thereon, then the general solution is $x_j = \exp(jA)x_0$, with $x_0$ specifying the initial conditions. The properties of $\exp(jA)$ can be deduced from a knowledge of the spectrum and invariant subspaces of $A$. This is precisely spectral theory. So I guess you just have to bite the bullet and start reading up on spectral theory.

The easiest situation is when the $x_j$ are vectors in a Hilbert space and $A$ is a bounded self-adjoint operator. This case is covered in essentially every introductory book on functional analysis. There is a long list of references of introductory and higher level reference books on Wikipedia.

If your $A$ is not of the above form, then you need to be more specific about the structure of your difference equation to get a more relevant suggestion.

If your equations look like $x_{i+1} = Ax_i$, where each $x_j$ is an element of a (possibly infinite dimensional) vector space and $A$ is a linear operator thereon, then the general solution is $x_j = A^j x_0 = \exp(j\log(A))x_0$, with $x_0$ specifying the initial conditions. The properties of $A^j$, or equivalently $\exp(j\log(A))$, can be deduced from a knowledge of the spectrum and invariant subspaces of $A$. This is precisely spectral theory. So I guess you just have to bite the bullet and start reading up on spectral theory.

The easiest situation is when the $x_j$ are vectors in a Hilbert space and $A$ is a bounded self-adjoint operator. This case is covered in essentially every introductory book on functional analysis. There is a long list of references of introductory and higher level reference books on Wikipedia.

If your $A$ is not of the above form, then you need to be more specific about the structure of your difference equation to get a more relevant suggestion.