# The spectral radius of normal operator

Let $H$ be a Hilbert space and $T$ be linear bounded operator in $H$. Prove that if $T$ is normal then the spectral radius of $T$,

$$r(T)=\|T\|.$$

Is this TRUE?

• Yes. ${}{}{}{}{}$ Dec 5 '14 at 6:48
• @copper.hat Could you give me hints to prove this? Dec 5 '14 at 6:51
• Look at Corollary 7.2 in math.ksu.edu/~nagy/real-an/2-07-op-th.pdf Dec 5 '14 at 7:54

Recall Gelfand's formula for the spectral radius of a bounded operator $T$: $$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n}.$$ If $T$ is a self-adjoint operator and $\|f\|=1$ then $$\|Tf\|^2 = \langle Tf, Tf\rangle = \langle T^2f, f\rangle\leqslant\|T^2f\|\|f\|=\|T^2f\|$$ which implies $\|T^2\|=\|T\|^2$. By induction it follows that $\|T^{2^n}\|=\|T\|^{2^n}$ for all $n$ and hence $$r(T) = \lim_{n\to\infty} \|T^{2^n}\|^{\frac1{2^n}} = \lim_{n\to\infty}\|T\|=\|T\|.$$ If $T$ is normal, then by induction $\|(T^*T)^n\|=\|T^n\|^2$, and as $T^*T$ is self-adjoint, $$r(T^*T)=r(T)^2=\|T\|^2,$$ from which we conclude that the spectral radius of a normal operator is equal to its operator norm.
• I am confusing a little bit in your prove. How can we claim that $r \left( T^*T \right) = r (T)^2?$ May 2 '18 at 7:33
• @TrầnLinh: you have $$r(t)^2=\lim_n\|T^n\|^{2/n}=\lim_n\|T^{*n}T^n\|^{1/n}=\lim_n\|(T^*T)^n\|^{1/n}=r(T^*T).$$ Nov 29 '18 at 1:21
Let $E$ be the unique spectral decomposition of $T$. Then, there exists a isometric isomorphism $\phi :L^{\infty}(E)\mapsto B(H)$ such that $\phi$ maps the identity function to $T$ where $B(H)$ is the dual of $H$ ($L^{\infty}(E)$ is a Banach algebra of bounded Borel functions on $\sigma(T))$ with the sup norm). Let $I:\sigma(T)\mapsto \mathbb{C}$ be the identity map on the spectrum of $T$. Then, since $I\in L^{\infty}(E)$ (Since $\sigma(T)$ is compact in $\mathbb{C}$), it follows that \begin{align*} \rho(T):=sup\left\{\lambda\in \mathbb{C}|\lambda\in \sigma(T)\right\}=&\rVert I\rVert_{L^{\infty}(E)}\\ =&\rVert \phi(I)\rVert_{B(H)} \quad \text{ $\because$ $\phi$ is an isometry}\\ =&\rVert T\rVert_{B(H)} \quad \text{ $\because$ $\phi$ maps identity to $T$} \end{align*}