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.

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    $\begingroup$ This was also asked on math.SE, earlier, and with the question actually being more precise: math.stackexchange.com/questions/4333084/… $\endgroup$ Dec 15, 2021 at 8:26
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    $\begingroup$ It's completely fine to ask the same question on both math.SE and MO, but generally people seem to leave the question at math.SE for a few days, to see if there is an answer. I definitely think you should say that your question has been asked elsewhere, to avoid duplicating effort. For example, over at math.SE there is an excellent answer already... $\endgroup$ Dec 15, 2021 at 8:28

1 Answer 1


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.


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