# Spectrum of self-adjoint operator on Hilbert space real

My book says that a self-adjoint bounded linear operator $A:H\to H$ on a complex Hilbert (not sure if separability is needed) space has a real spectrum.

I guess that the key is in the fact that any $f\in H^{\ast}$ can be represented by a functional of the form $\langle-,x_0\rangle$ for some $x_0\in H$, but I am not able to use that fact. I also see that $\forall x,y\in H\quad\langle(A-\lambda I)x,y\rangle=\langle x,(A-\bar{\lambda} I)y\rangle$ but I'm not sure that can be useful to prove the statement...

Thank you very much for any help!!!

• I think you forgot to write self-adjoint in your first sentence, even though you write self-adjoint in the title. Commented Sep 9, 2014 at 18:43
• Thank you so much: yes, I did. Edited. Commented Sep 9, 2014 at 19:23

If $$\Im\lambda \ne 0$$, and $$x \in X$$, then $$\Im\lambda \|x\|^{2} = -\Im((A-\lambda I)x,x),\\ |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|.$$ So $$A-\lambda I$$ is injective for all $$\lambda\notin\mathbb{R}$$. The above inequality can be used to show that the range $$\mathcal{R}(A-\lambda I)$$ is closed for $$\lambda\notin\mathbb{R}$$. So $$A-\lambda I$$ is surjective for $$\lambda\notin\mathbb{R}$$ because \begin{align} \mathcal{R}(A-\lambda I)& =\overline{\mathcal{R}(A-\lambda I)} \\ & =\mathcal{N}(A^{\star}-\overline{\lambda}I)^{\perp} \\ & = \mathcal{N}(A-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H. \end{align} Therefore, $$A-\lambda I$$ is injective and surjective for $$\lambda\notin\mathbb{R}$$, which leaves $$\sigma(A)\subseteq\mathbb{R}$$.

• Thank you so much! My text proposes lemmas to be proven as exercises of a level often much higher than the preparation given in theory part, so I have found material on line to understand why $\overline{\mathcal{R}(A-\lambda I)}=\mathcal{N}(A^{\ast}-\bar{\lambda}I)^{\perp}$, but I still don't understand how to realise that such a range is closed. Moreover why $\Im\lambda\|x\|^{2}=\Im((A-\lambda I)x,x)$ (I only see that $((A-\lambda I)x,x)=(Ax,x)-\lambda\|x\|^2$) and why $\mathcal{N}(A^{\ast}-\bar{\lambda}I)^{\perp}=\mathcal{N}(A-\bar{\lambda}I)^{\perp}$? $\infty$ thanks!!! Commented Sep 9, 2014 at 23:52
• @DavideZena : $A^{\star}=A$ which is why $\mathcal{N}(A^{\star}-\overline{\lambda}I)=\mathcal{N}(A-\overline{\lambda}I)$. Because $A=A^{\star}$, you also know that $(Ax,x)$ is real because $(Ax,x)^{\star}=(x,Ax)=(A^{\star}x,x)=(Ax,x)$. Finally, the inequality shows the range is closed because $(A-\lambda I)x_{n}\rightarrow y$ implies $\Im\lambda\|x_{n}-x_{m}\| \le \|(A-\lambda I)(x_{n}-x_{m})\|\rightarrow 0$ as $n,m\rightarrow \infty$; that then gives $\lim_{n}x_{n}=x$ for some $x$ and, so, $(A-\lambda I)x_{n}\rightarrow (A-\lambda)x$ which gives $y=(A-\lambda I)x$ is in the range. Commented Sep 10, 2014 at 0:18
• As to not seeing that $A^{\ast}=A$, I'm quite embarassed :-s. $\aleph_1$ thanks! Commented Sep 10, 2014 at 8:27
• @DavideZena : You're welcome. Glad I could help. Commented Sep 10, 2014 at 13:03
• @Balai_Indah : $|\Im\lambda|\|x\|\le \|(A-\lambda I)x\|$ implies $\|(A-\lambda I)^{-1}y\|\le \frac{1}{|\Im\lambda|}\|y\|$ for all $y$ in the range of $A-\lambda I$, which is a dense subspace. So the inverse has a unique continuous extension. Commented Mar 18, 2021 at 23:46

The fact that the spectrum of $$A$$ is real when $$A = A^*$$ is a general fact about commutative $$C^*$$ algebras; the discussion of $$A$$ as an operator on Hilbert space is not needed. The main point is that in such a $$C^*$$ algebra, we have $$||xx^*|| = ||x||^2$$ for all $$x$$, and the spectral radius $$\rho(x) = ||x||$$ for all $$x$$. Thus if $$Y$$ is a large real number (large multiple of the identity), then $$\rho(A+iY)^2 = ||A+iY||^2 = ||A^2+Y^2|| = Y^2 + O(1).$$ Now if $$x+iy$$ is in the spectrum of $$A$$, say with $$y>0$$, then $$\rho(A+iY)^2 > |x+iy+iY|^2 > (y+Y)^2$$ and hence $$y=0$$. Thus the spectrum of $$A$$ is real.

The representation of the elements of $H^*$ is not needed.

Let $\lambda$ be in the point spectrum of $A$. Then there is $0\neq x\in H=(H,(\cdot,\cdot))$ such that $Ax=\lambda x$ and by the self-adjointness of $A$: $\lambda(x,x)=(\lambda x,x)=(Ax,x)=(x,Ax)=(x,\lambda x)=\overline{\lambda}(x,x)$. Hence $\lambda=\overline{\lambda}$, thus $\lambda\in\mathbb{R}$.

See e.g. http://en.wikipedia.org/wiki/Spectral_theorem for related results in greater generality.

• Forgive me if I don't understand: why $A-\lambda I=0$? I only know that, if $\lambda$ is in the (continuous or punctual) spectrum, $A-\lambda I$ isn't bijective... Thank you so much!!! Commented Sep 9, 2014 at 18:44
• Sorry, that was a mistake. I've edited the answer, but the case where $\lambda\in\sigma\setminus\sigma_p$ still needs to be done.
– Bati
Commented Sep 9, 2014 at 19:02
• No problem: thank you again! Commented Sep 9, 2014 at 23:41