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?


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}

  • $\begingroup$ sorry I forgot to add symmetric, now I edited the question $\endgroup$
    – Picard
    Jan 19 '16 at 8:36
  • 1
    $\begingroup$ How do you define symmetric for an arbitrary operator? $\endgroup$
    – gerw
    Jan 19 '16 at 8:37
  • 2
    $\begingroup$ ...what's is the difference between symmetric and self-adjoint? $\endgroup$ Jan 19 '16 at 8:38
  • 1
    $\begingroup$ Sorry I added the definitions I used... $\endgroup$
    – Picard
    Jan 19 '16 at 8:45
  • $\begingroup$ Could be indeed interesting! But right now I'm interested in this problem how is it related? $\endgroup$
    – Picard
    Jan 19 '16 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 $(T^\ast-z)\psi=(T-z)\psi$.

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 $A$.

  • $\begingroup$ 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? $\endgroup$
    – Picard
    Jan 19 '16 at 10:46
  • $\begingroup$ Thank you again!!! Why does $\psi$ belong to D(T)? $\endgroup$
    – Picard
    Jan 19 '16 at 11:38
  • $\begingroup$ Because $(T-z)^{-1}$ maps to $D(T)$ by definition. $\endgroup$
    – MaoWao
    Jan 19 '16 at 11:38
  • $\begingroup$ True :D :D :D Perfect! Thank You $\endgroup$
    – Picard
    Jan 19 '16 at 11:47
  • $\begingroup$ @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. $\endgroup$
    – Jan
    Jun 11 '20 at 11:49

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