# Norm of normal Operator A

I just found the following equality $||A||=sup_{\lambda\in\sigma(A)} |\lambda|$

My question: What exactly means $\sigma(A)$ and why this is true ?

I always thouht the only way to get the operatornorm is $||A||=sup_{||x||=1}||A(x)||_{Y}$

where $Y$ is some Vectorspac

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$\sigma(A)$ is called the spectrum, and it generalizes the set of eigenvalues. If $\lambda$ is an eigenvalue of $A$, then $(A-\lambda I)v=0$ for some nonzero vector $v$, so $A-\lambda I$ cannot be invertible. ($I$ stands for the identity operator.) The spectrum is defined as: $$\sigma(A):=\{\lambda : \nexists (A-\lambda I)^{-1} \}.$$ The quantity $\sup\{|\lambda| : \lambda\in\sigma(A)$ is also called the spectral radius, and it indeeds equals to the norm in case of normal operators, see the spectral radius formula on the wikipage.