A question on the self-adjointness of an operator Given a Hilbert space (separable) $\mathcal{H}$ with an orthonormal basis $\{e_i\}_{i=1}^{\infty}$, define an operator $T$ with domain $\mathcal{D}(T)$ equal to the span of $\{e_i\}$ by $Te_i:=\lambda_ie_i$, for some $λ_i∈R$.
My question: Is $T$ necessarily self-adjoint (i.e. $T=T^*$)? It is easy to see that $T$ is a symmetric operator and therefore $T \subset T^*$. I understand that I can make use of the definition of adjoint $T^*$ and $T$ to show that $T$ and $T^*$ agree on all basis vectors and therefore any finite linear combinations of them? But, how can one show that that they agree on the closure of the span? In other words, how do I show (if it is possible in the first place) that $T^* \subset T$?
Any help will be truly appreciated. Thanks.
 A: No. For example, if the $\lambda_i$ are bounded, we see that $T$ can be extended to an operator $S$ defined everywhere. Then $\text{dom}(T^*)\supseteq \text{dom}(S^*) = \mathcal H$, so $T^*$ is defined everywhere as well.
So can we find the domain of $T^*$ in the general case? Recall that the domain of $T^*$ consists of all $v = \sum_iv_ie_i\in\mathcal H$ for which there is $w\in\mathcal H$ such that for all vectors $u\in\mathcal H$ we have $\langle w,u\rangle = \langle v,Tu\rangle$. Applying this on $u=e_i$ gives $\langle w, e_i \rangle = \lambda_iv_i$, so we get $w = \sum_i\lambda_iv_i$.
Conversely, if $\sum_i\lambda_iv_i \in\mathcal H$ it is easy to check that $w = \sum_i\lambda_iv_i$ satisfies the above condition.
So $v\in\text{dom}(T^*)$ if and only if $\sum_i|\lambda_i|^2|v_i|^2 < \infty$.
There are always such vectors $v$ that are not in the linear span of the basis vectors, for example we can take $|v_i| = \min(\frac1{i^2},\frac1{i^2|\lambda_i|^2})$. So $T$ is never self-adjoint.
