Operator with every complex number as an eigenvalue The question :

Let $V$ be the space of all functions $f : \mathbb Z \to \mathbb C$. Find an operator $T : V \to V$ such that every $\lambda \in \mathbb C$ is an eigenvalue i.e. for every $\lambda \in \mathbb C$ , there is some non-zero $f$ such that $Tf = \lambda f$.

This question is surprising, in light of the fact that for a Banach algebra,  the spectrum is a compact subset. So this shows that $V$ cannot be given a Banach algebra structure.
For this question, my idea was to hope that $V$ has an uncountable Hamel basis $e_i : i \in \mathbb C \setminus \{0\}$ . I would then have $T$ such that $Te_1 = 0$,$Te_2 = e_1$, $Te_3 = 2e_2$,..., along with $Te_l = le_l$ for every other $l \neq 1,2,...$. This would work.
I was not able to find such a Hamel basis : I tried using indicator functions of subsets of $\mathbb Z$, but I need a concrete example(i.e. I am aware that a Hamel basis for $V$ exists and must be uncountable, but I need the exact form of the $e_i$), and I am not able to find one.
 A: I guess you require $T$ to be a linear operator.
Define $T$ to be the shift operator $(Tf)(n) = f(n+1)$. For any $\lambda \in \mathbb{C} \backslash \{0\}$, consider $f_\lambda (n) = \lambda^n$, then 
$Tf_\lambda = \lambda f_\lambda.$
A: So, I think this should do the trick (but it is not nice).
For $t>0$, define $$f_t(n)=e^{-nt}$$
Then if $$\sum_{0\leq i\leq k}\lambda_if_{t_i}=0$$
take $i_0$ such that $t_{i_0}=\max\{t_i,\ 0\leq i\leq k \text{ and }\lambda_i\neq0\}$.
Then, $$\sum_{0\leq i\leq k}\lambda_if_{t_i}\sim_{n\to+\infty}\lambda_{i_0}e^{-nt_{i_0}}$$
so that $\lambda_{i_0}=0$, so actually the $(f_t)_t$ are independant.
Now let $\phi$ be a bijection between $\mathbb R_+^*$ and $\mathbb C$, and let
$$g_\lambda:=f_{\phi^{-1}(\lambda)}$$
for $\lambda\in\mathbb C$.
The $(g_\lambda)_\lambda$ are independant so you can complete this family to form a basis $\mathcal B$.
Define $T$ for each element of the basis by $T(g_\lambda)=\lambda g_\lambda$ and $T(b)=0$ for $b\in\mathcal B\setminus\{g_\lambda\}_\lambda$.
This yields a well defined operator $T$ on $V$ which has every complex number as eigenvalue, so that $\text{Sp}\ T=\mathbb C$.
