rad(T)=||T|| for non-normal T It is well-known that for normal bounded operators $T$ on a Hilbert space one has $\mathrm{rad}(T)=\|T\|$ (where rad is the spectral radius).
Are there any sufficient conditions under which a non-normal operator satisfies $\mathrm{rad}(T)=\|T\|$ ?
Thanks in advance,
Mark
 A: Here are some thoughts - not sure if there is a clear-cut answer to your question even for matrices (I use $r(T)$ to denote $\text{rad}(T)$) :


*

*$T$ has this property iff $r(T) \geq \|T\|$, which happens iff
$$
\|T(x)\| \leq r(T)\|x\| \quad\forall x
$$
$$
\Leftrightarrow \langle Tx,Tx\rangle \leq r(T)^2 \langle x,x\rangle
$$
$$
\Leftrightarrow r(T)^2I - T^{\ast}T \text{ is a positive operator}
$$

*If $T$ is unitarily equivalent to an operator with this property, then $T$ has this property.

*If $T$ is a $2\times 2$ matrix with this property, then it is normal.
Proof: We may assume (as above) that $T$ has the form
$$
T = \begin{pmatrix}
\lambda & 0 \\
a & \mu
\end{pmatrix}
$$
where $|\lambda| \geq |\mu|$. Then if $e_1 = (1,0)$,
$$
|\lambda| = r(T)  \geq \|T\| \geq \|T(e_1)\| = (|\lambda|^2 + |a|^2)^{1/2}
$$
and so $a=0$, whence $T$ is normal.


*If $N$ is the standard nilpotent matrix with all zeroes except under the diagonal, then the block matrix
$$
T = \begin{pmatrix} 
I & 0 \\
0 & N
\end{pmatrix}
$$
has the property that $r(T) = \|T\| = 1$, but it is not unitarily equivalent to a normal matrix. In particular, there is a $3\times 3$ matrix with this property that is not normal.

