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,


  • $\begingroup$ Welcome to Math.SE! Do you happen to have any idea yourself that you could share? $\endgroup$
    – Hrodelbert
    Oct 6 '15 at 13:27
  • 2
    $\begingroup$ 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. $\endgroup$
    – 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.

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