Consider the set V of all infinite sequences in $\Bbb{F}$. $$V = { \{ (a_0, a_1, a_2, ...)\ |\ a_i \in \Bbb{F}, \text{for } i=0, 1, 2, ... \} }$$ and the function T:V $\rightarrow$ V defined by $$T(a_0, a_1, a_2,a_3, ...) = (a_1, a_2,a_3, a_4, ...)$$ assuming that V is a vector space over $\Bbb{F}$ and T is a linear transformation prove that every $\lambda$ in $\Bbb{F}$ is an eigenvalue of T, and find the corresponding eigenspaces.
Please show me how to prove this using $$T(v) = \lambda v$$
Hint: suppose $T(a_{0}, a_{1}, a_{2}, a_{3}, \ldots) = (a_{1}, a_{2}, a_{3}, a_{4}, \ldots) = \lambda(a_{0}, a_{1}, a_{2}, a_{3}, \ldots)$. Then $a_{0} = \lambda a_{1}$; $a_{2} = \lambda a_{1} = \lambda(\lambda a_{0}) = \lambda^{2} a_{0}$, and so on. By an induction argument, one can see that $a_{n} = \lambda^{n}a_{0}$ for each $n \in \mathbb{N}$. Now use this to classify the possible eigenvalues and the corresponding eigenspaces.