# If an operator have only Real eigenvalues + symmetric then it's self-adjoint?

I know that if an operator is self-adjoint then has Real eigenvalues but I'm not sure about the converse i.e. if it has only Real eigenvalues and is symmetric then the operator is selfadjoint. Is that true?

----Edit---

The difference between selfadjoint and symmetric being the definition set. Symmetric has an extension which coincide in the original domain, while a Selfadjoint operator has the same domain of definition

I define symmetric as follows. Let be $$\mathcal{A}=\left(A,\mathfrak{D}_{A}\right)$$ an operator densely defined and $$\mathcal{A}^{*}=\left(A^{*},\mathfrak{D}_{A^{*}}\right)$$ the adjoint operator, then $$\mathcal{A}$$ it is called symmetric if $$\begin{eqnarray} & & \mathfrak{D}_{A^{*}}\supseteq\mathfrak{D}_{A}\\ & & A^{*}\psi=A\psi\qquad\forall\psi\in\mathfrak{D}_{A}. \end{eqnarray}$$

While I define Self-adjoint like this Let be $$\mathcal{A}=\left(A,\mathfrak{D}_{A}\right)$$ an operator densely defined and $$\mathcal{A}^{*}=\left(A^{*},\mathfrak{D}_{A^{*}}\right)$$ the adjoint operator, then $$\mathcal{A}$$ it is called selfadjoint if $$\begin{eqnarray} & & \mathfrak{D}_{A^{*}}=\mathfrak{D}_{A}\\ & & A^{*}\psi=A\psi\qquad\forall\psi\in\mathfrak{D}_{A}. \end{eqnarray}$$

• sorry I forgot to add symmetric, now I edited the question Jan 19, 2016 at 8:36
• How do you define symmetric for an arbitrary operator?
– gerw
Jan 19, 2016 at 8:37
• ...what's is the difference between symmetric and self-adjoint? Jan 19, 2016 at 8:38
• Sorry I added the definitions I used... Jan 19, 2016 at 8:45
• Could be indeed interesting! But right now I'm interested in this problem how is it related? Jan 19, 2016 at 8:49

This is far from being true, indeed, every symmetric operator has only real eigenvalues:

If $$\psi\in\ker(T-\lambda),\,\psi\neq 0$$, then $$\lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2,$$ hence $$\lambda=\bar\lambda$$.

Now every symmetric operator that is not self-adjoint yields a counterexample to your conjecture (if you want to be explicit, take $$\Delta$$ on $$C_c^\infty(\mathbb{R}^n)$$ as operator in $$L^2(\mathbb{R}^n)$$).

To get a criterion for self-adjointness, you have to replace the eigenvalues by the spectrum of the operator. Then the following characterization holds:

A symmetric operator $$T$$ is self-adjoint if and only if its spectrum is contained in $$\mathbb{R}$$.

Proof: It suffices to show that $$D(T^\ast)\subset D(T)$$. Let $$z\in\mathbb{C}\setminus\mathbb{R}$$. Since $$\sigma(T)\subset\mathbb{R}$$, the operators $$T-z$$ and $$T-\bar z$$ are invertible.

Let $$\phi\in D(T^\ast)$$ and $$\psi:=(T-z)^{-1}(T^\ast -z)\phi\in D(T)$$. Then we have $$T\psi=T^\ast\psi$$ and $$(T^\ast-z)\phi=(T-z)\psi$$.

It follows that $$(T^\ast-z)(\phi-\psi)=(T-z)\psi-(T-z)\psi=0,$$ that is, $$\phi-\psi\in N(T^\ast-z)=R(T-\bar z)^\perp=\{0\}$$. Hence, $$\phi=\psi\in D(T)$$.

Remark: I used $$R(A)$$ and $$N(A)$$ to denote the range and kernel of $$A$$.

• Thank You Very Much!!! Could you also give me a sketch of the proof of if the spectrum is contained in R then it's self-adjoint? Jan 19, 2016 at 10:46
• Thank you again!!! Why does $\psi$ belong to D(T)? Jan 19, 2016 at 11:38
• Because $(T-z)^{-1}$ maps to $D(T)$ by definition. Jan 19, 2016 at 11:38
• True :D :D :D Perfect! Thank You Jan 19, 2016 at 11:47
• @MaoWao Could you please check whether the last edit to your post is correct? If yes: Could you please explain from where you get $T \psi = T^\ast \psi$, $(T^\ast - z) \psi = (T - z) \psi$ and $(T^\ast - z) (-\psi) = (T - z) \psi$? I can't see how you arrive there.
– Jan
Jun 11, 2020 at 11:49