Gelfand's formula

Remarks

For a submultiplicative norm, $\rho(A)^k = \rho(A^k) \le \norm{A^k}$, which gives $\rho(A) \le \norm{A^k}^{1/k}$ for every $k$. For $\varepsilon > 0$, the matrix $A / (\rho(A) + \varepsilon)$ has spectral radius less than $1$, so its powers tend to $0$ by the Jordan normal form; this gives $\norm{A^k} \le (\rho(A) + \varepsilon)^k$ for large $k$. Any two norms are equivalent, and $c^{1/k} \to 1$ for $c > 0$, so the limit does not depend on the norm. Example: $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$, and with the maximum row sum norm $\norm{A^k}^{1/k} = (1 + k)^{1/k} \to 1 = \rho(A)$, although $\norm{A} = 2$. The same formula holds for elements of a complex Banach algebra; see the spectral radius formula.

Going deeper

For a student

The formula explains why $A^k \to 0$ if and only if $\rho(A) < 1$, which is the convergence criterion for iterative methods such as Jacobi and Gauss–Seidel. It also shows that $\rho(A) \le \norm{A}$ for every submultiplicative norm, and that for every $\varepsilon > 0$ there is an operator norm with $\norm{A} \le \rho(A) + \varepsilon$. In the Banach algebra setting, the same statement is proved with the resolvent and complex analysis instead of the Jordan form.