If this is the case then T is the backwards shift operator which for all
Z=(z(1), z(2), ..) in F with max(z(1), z(2), ..) = |Z| < infty:
T(Z) = Z' = (z(2), z(3), .. )
Now if lambda and X are eigen value and eigen vector for T, then:
T(X) = lambda X,
so x(n+1) = lambda x(n), for all n>1.
Hence x(n) = lambda^(n-1) x(1), which is in F iff |lambda| <= 1.
Thus every lambda such that |lambda|<=1 is an eigen value of T and
(1, lambda, lambda^2, .., lambda^n, ..) is an eigen vector corresponding
to the eigen value lambda.
RonL