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Can spectrum $\sigma(A)$ of self-adjoint non-zero operator $A$ ($\langle Ax, y\rangle = \langle x, Ay \rangle $) be equal zero:

$\sigma(A) = \{0\}$?

I know that its real and closed, but i do not know how to show this, any advices pls.

Spectrum is a set $ \sigma(A) = \{\lambda \in \mathbb{C}: (A - \lambda I)^{-1} \text{ does not exist}\}$.

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    $\begingroup$ How about $A = 0$? $\endgroup$ – Mees de Vries May 23 '18 at 12:36
  • $\begingroup$ @MeesdeVries sorry, edited question: non-zero operator $\endgroup$ – Alexander Markov May 23 '18 at 12:38
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    $\begingroup$ The functional calculus implies that in the Hilbert case, if $\sigma(A)=0$ then $A=0$. What do you mean by self-adjoint for Banach spaces? $\endgroup$ – Luiz Cordeiro May 23 '18 at 12:47
  • $\begingroup$ @LuizCordeiro I think its just my mistake... How can I show that only zero operator satisfies the condition in Hilbert space? I know that $A$ has no inverse, but what should i do next? $\endgroup$ – Alexander Markov May 23 '18 at 12:51
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The only possibility is $A = 0$.

Recall that the spectral radius of $A$ is defined as $$\rho(A) = \inf_{n\in\mathbb{N}} \|A^n\|^{1/n}$$

It is known that $\rho(A) = \max\{|\lambda| : \lambda \in \sigma(A)\}$.

$A$ is self-adjoint, so in particular $A$ is normal. This implies $\|A^n\| = \|A\|^n, \forall n \in \mathbb{N}$ so $\rho(A) = \|A\|$.

Therefore $\sigma(A) = \{0\}$ implies $\|A\| = \rho(A) = 0$ so $A = 0$.

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