Spectrum of self-adjoint operator

Question

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}\}$.

Answer 1

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 $A = 0$.