I was seeing these links in the forum,

Spectrum of symmetric, non-selfadjoint operator on Hilbert space

Distinguishing between symmetric, Hermitian and self-adjoint operators

I know the following definition for "Diagonal Linear Operator on Hilbert Spaces" :

Let $\mathbb K \in $ {$\mathbb R, \mathbb C$} and let $(H,<.,.>_{H},||\cdot||_{H})$ be a $\mathbb K$-Hilbert Space. Then $A$ is a diagonal linear operator on $(H,<.,.>_{H},||\cdot||_{H})$ iff, $\exists \quad \mathbb H \in \mathcal P(H)$ (-the power set of $H$) & a mapping $\lambda$ from $\mathbb H$ to $\mathbb K$ such that:

  1. $\mathbb H$ is an orthonormal basis of $H$,

  2. $A$ is a mapping from {$v \in H : \sum_{h \in \mathbb H}|\lambda_{h}<h,v>_{H}|^{2}$ $\lt +\infty$} to $H$,

  3. $\forall v \in \mathcal D(A)$, $Av = \sum_{h \in \mathbb H}\lambda_{h}<h,v>_{H}h$

Seeing the implications in those links, I am just asking for a concrete example of a Diagonal linear operator $A$, which does NOT satisfy $\forall v, w \in \mathcal D(A)$, $<Av,w>_{H}=<v,Aw>_{H}$.

Intuitively, I think there should exist such an example.

Also, if someone thinks or knows this intuition is wrong, can someone please show me : "Diagonal Linear Operator on Hilbert Spaces is symmetric" ??

Thanks in advance.

EDIT: Right now, I am trying to think of Hilbert-Schmidt Operator with a "non-symmetric" kernel. But, I guess, there should be more easier example.


Symmetry follows directly from the third item in your definition when $\mathbb{K} = \mathbb{R}$. Take $v,w \in \mathcal{D}(A)$. Then $$ \langle Av, w \rangle = \left \langle \sum_{h} \lambda_h \langle h,v \rangle h, w \right \rangle = \sum_h \lambda_h \langle h,v \rangle \langle h,w \rangle \\ =\left \langle v, \sum_{h} \lambda_h \langle h,w \rangle h \right \rangle = \langle v, A w \rangle. $$ The convergence of the series and the manipulation of the the terms in this way is justified by the inclusion $v,w \in \mathcal{D}(A)$ and the fact that you're working over an ON basis.

EDIT after field clarification: Is the operator $A = i I$ not to your liking in terms of concreteness? In this case $\lambda_h = i$ for every $h$ in your basis.

  • $\begingroup$ Sorry, but the trouble is causing by an exercise, given later: "Let $\mathbb K \in$ {$\mathbb R , \mathbb C$}, let $(H, <.,.>_{H}, ||\cdot||_{H})$-be a $\mathbb K$-Hilbert space and let $A: D(A)(\subset H) \to H$ be a diagonal linear operator. Prove that: $A$ is symmetric if and only if $\sigma_{P}(A) \subset \mathbb R$". So, at some point the difference will be created by $H$ being a $\mathbb R$-Hilbert Space, instead of general $\mathbb K$-Hilbert Space. $\endgroup$
    – user92360
    Mar 16 '17 at 13:21
  • $\begingroup$ I mean instead of $\left \langle v, \sum_{h} \lambda_h \langle h,w \rangle h \right \rangle$ in the L.H.S. of last equality it should have been $\left \langle v, \sum_{h} \overline{\lambda_h} \langle h,w \rangle h \right \rangle$. Since $H$ is a $\mathbb K$-Hilbert Space. And $\mathbb K$ can be $\mathbb C$ also. $\endgroup$
    – user92360
    Mar 16 '17 at 14:51
  • 1
    $\begingroup$ Oops... my eyes zipped right past your $\mathbb{K}$ definition, and I thought you were working over the reals. You are right: my argument only works in the real case. $\endgroup$
    – Glitch
    Mar 16 '17 at 17:12

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