If $r(A)=\omega(A)$, is $r(A)=\|A\|$?

Question

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

It is well know that
$$r(A)\leq\omega(A)\leq\|A\|,$$ for every $A\in\mathcal{B}(F)$, where $r(A)$, $\omega(A)$ and $\|A\|$ denote respectively the spectral radius, the numerical radius and the norm of $A$.

So clearly if $r(A)=\|A\|$ then $r(A)=\omega(A)$.

If $r(A)=\omega(A)$, is $r(A)=\|A\|$?

Answer 1

No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=B\oplus C\in M_3(\mathbb C)$, where $B=1$ and $C=2E_{12}$. That is, $$ A=\begin{bmatrix} 1&0&0\\0&0&2\\0&0&0\end{bmatrix}. $$ Then

• $\sigma(A)=\{0,1\}$ and thus $r(A)=1$

• the numerical range of $A$ is the convex hull of the ranges of $B$ ($=\{1\})$ and of $C$ (disk of radius $1$ centered at the origin). So $\omega(A)=1$.

• the norm of $A$ is the maximum of the two norms, so $\|A\|=2$.