Spectral radius in a Banach algebra

Let $A$ be a complex unital Banach algebra and let $a\in A$ . The spectral radius of $a$ is

r_A(a)=\sup\{|\lambda|:\lambda\in\sigma_A(a)\},

where $\sigma_A(a)$ is the spectrum of $a$ in $A$ . Because that spectrum is nonempty and compact, the supremum is a maximum. The spectral-radius formula states that

r_A(a)=\lim_{n\to\infty}\lVert a^n\rVert^{1/n} =\inf_{n\geq1}\lVert a^n\rVert^{1/n}.

For a nonunital Banach algebra, the spectral radius is computed in its unitization. It always satisfies $0\leq r_A(a)\leq\lVert a\rVert$ .

Basic properties

One has $r(a^n)=r(a)^n$ for positive integers $n$ , and more generally polynomial spectral mapping gives $r(p(a))=\max_{\lambda\in\sigma(a)} |p(\lambda)|$ . The spectral radius need not be a norm: nonzero nilpotent elements have radius zero, and subadditivity can fail in noncommutative Banach algebras. The limit formula, including existence of the limit, is a central result of Banach-algebra spectral theory.

The C-star-algebra case

If $a$ is normal in a $C^*$ -algebra, then $r(a)=\lVert a\rVert$ . In particular, this holds for self-adjoint, positive, and unitary elements. For an arbitrary $C^*$ -algebra element, the sharper identity is

\lVert a\rVert^2=r(a^*a).

Thus the norm is spectrally determined on normal elements even though the spectral radius can be strictly smaller than the norm for nonnormal ones.

Examples

For a matrix, the spectral radius is the maximum modulus of its eigenvalues. The matrix $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has spectral radius zero but positive operator norm. For multiplication by a continuous function $f$ on $C(X)$ , the spectral radius is $\max_{x\in X}|f(x)|$ .

References

Gerard J. Murphy, C-Algebras and Operator Theory*, Academic Press, 1990. DOI record. Relevant: §§1.2 and 2.2 on spectra, the spectral-radius formula, and normal elements.