# 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 $\mathrm{rad}(T)=\|T\|$ ?

Mark

• Welcome to Math.SE! Do you happen to have any idea yourself that you could share? Oct 6 '15 at 13:27
• Not really, sorry. I ran into the problem because I want to estimate the norm of a semigroup $e^{-tH}$, where $H$ is non-normal and I know its spectrum.
– Mark
Oct 6 '15 at 13:42

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 $\text{rad}(T)$) :
1. $T$ has this property iff $r(T) \geq \|T\|$, which happens iff $$\|T(x)\| \leq r(T)\|x\| \quad\forall x$$ $$\Leftrightarrow \langle Tx,Tx\rangle \leq r(T)^2 \langle x,x\rangle$$ $$\Leftrightarrow r(T)^2I - T^{\ast}T \text{ is a positive operator}$$
2. If $T$ is unitarily equivalent to an operator with this property, then $T$ has this property.
3. 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.
1. 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.