Spectral radius ($\rho (A)$)

The spectral radius is the largest modulus of an operator's or matrix's eigenvalues:

For a finite square matrix, powers tend to zero exactly when every eigenvalue has modulus less than one. In a complex Jordan normal form, the entries of a block power are sums of factors $\left(\genfrac{}{}{0ex}{}{k}{j}\right){\lambda }^{k-j}$, which tend to zero when $|\lambda |<1$; a zero-eigenvalue block becomes zero after finitely many powers. Conversely, an eigenvector gives $G^{k}v={\lambda }^{k}v$, so convergence to zero excludes $|\lambda |\ge 1$. This criterion underlies the convergence theorem for a stationary iterative method for a linear system.

For a holomorphic Banach algebra-valued function, put $M_r(f)={\sup }_{|z|=r}r(f(z))$. Apply the Hadamard three-circle theorem to the holomorphic powers $f^{2^n}$. The continuous functions $\Vert f(z{)}^{2^n}{\Vert }^{1/2^n}$ decrease to the spectral radius. Supremum convergence for decreasing continuous functions passes the boundary suprema to the limit, proving the inequality with the same logarithmic interpolation parameter. In particular, at the geometric-mean radius, the square of the middle spectral radius is at most the product of the two boundary suprema.

If an entire Banach algebra-valued function has bounded spectral radius, that spectral radius is constant. The spectral-radius three-circle inequality on circles about an arbitrary point, with the outer radius tending to infinity, bounds every interior value by the maximum on any smaller circle. Upper semicontinuity of the spectral radius then lets the smaller radius tend to zero. Every value is bounded by the value at the arbitrary center; exchanging centers gives equality. The algebra-valued function itself can be nonconstant, for example $zN$ for a fixed nonzero square-zero matrix $N$.

The joint spectral radius measures the worst exponential growth under arbitrary products from a finite family of matrices. Submultiplicativity gives its existence and invariance under changes of the chosen norm. Every fixed-length product $B$ satisfies $\rho (B{)}^{1/p}\le \hat{\rho }(\mathcal{A})$, while the maximum norm of all length-$p$ products to the power $1/p$ is an upper bound. A repeated product gives the first inequality; grouping an arbitrary long word into blocks proves the second. The spectral radius of each individual matrix does not control growth of all switched products.

For an element $a$ of a complex Banach algebra,

In a complex unital Banach algebra, the spectral radius formula makes the root norms of $(x/t{)}^n$ tend to $r(x)/t$. If $t>r(x)$, those norms are eventually bounded above by a geometric sequence and the powers vanish. Conversely, vanishing powers eventually have norm less than one; the infimum version of the radius formula then forces $r(x)/t<1$. Consequently $r(x)=\inf \{t>0:(x/t{)}^n\to 0\}$, including the case $r(x)=0$.