# Looking for a NON-Symmetric Diagonal Linear Operator!!

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 {$$v \in H : \sum_{h \in \mathbb H}|\lambda_{h}_{H}|^{2}$$ $$\lt +\infty$$} to $$H$$,

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

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

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" ??

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.
• 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. – user92360 Mar 16 '17 at 13:21
• 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. – user92360 Mar 16 '17 at 14:51
• 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. – Glitch Mar 16 '17 at 17:12