Boundness of the operator when Hamel basis vectors are eigenvectors Let $X$ be infinite-dimensional Banach space, $\{e_j\}_{j\in J}$ be normalized Hamel basis for it.
Suppose for each $j\in J$ we choose a number $\lambda_j\in\mathbb{R}$. Then we define a linear map
$$A:X\to X$$
$$Ae_j = \lambda_je_j$$
and extend by linearity.
Is it possible to say something about boundness of $A$ knowing the set $\Lambda = \{\lambda_j\: |\: j\in J\}$? Suppose $0\not\in\Lambda$ so $A$ is a bijection.
Trivially if $\Lambda$ is unbounded then $A$ is unbounded too. If $0$ is a limit point of $\Lambda$ then $A$ is a bijection with unbounded inverse, so it also must be unbounded.
What if $\Lambda$ is bounded and is away from zero?
Can anything be done in a general case? Any necessary/sufficient conditions?

No.
Say $Y$ is a subspace of $X$ and $Y$ is not closed. Say $B_1$ is a Hamel basis for $Y$ and let $B$ be a Hamel basis for $X$ with $B_1\subset B$.
Let $$\lambda_j=\begin{cases}1,&(e_j\in B_1),
\\2,&(e_j\notin B_1).\end{cases}$$
Now if $A$ were continuous then $Y=\{x: Ax=x\}$ would be closed.
