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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

It is well know that if $A$ is normal operator, then $r(A)=\|A\|$.

I want to find an example of operator $A$ which is not normal but $$r(A)=\|A\|,$$ where $r(A)$ and $\|A\|$ denotes respectively the spectral radius of $A$ and the norm of $A$.

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Let $F= \ell_2$ and $A \in \mathcal{B}(F)$ be given by

$A(x_1,x_2,x_3,....)=(0,x_1,x_2,x_3,...)$.

Then we have $||A||=1$ and $ \sigma(A)=\{z \in \mathbb K:|z| \le 1\}$, hence $r(A)=1$.

Furthermore: $A^*A=I \ne AA^*$.

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    $\begingroup$ Very nice - this construction provides a counterexample even in $\mathbb{C}^3$ if one wants a finite-dimensional version. $\endgroup$
    – max_zorn
    May 9 '18 at 5:29
  • $\begingroup$ Is the numerical radius of $A$ is also equals to 1? $\endgroup$
    – Schüler
    May 9 '18 at 10:11
  • $\begingroup$ Yes. If we denote the numerical radius of $A$ by $n(A)$, then it is well known that $r(A) \le n(A) \le ||A||$. $\endgroup$
    – Fred
    May 9 '18 at 10:45

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