Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 1 down vote accepted

$\sigma(A)$ 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.