ProofOfGelfandSpectralRadiusTheorem 2005-11-06 08:09:41 2006-09-11 14:19:15 Proof Gelfand spectral radius theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here For any $\epsilon>0$, consider the matrix \[ \tilde{A}=(\rho(A)+\epsilon)^{-1}A \] Then, obviously, \[ \rho(\tilde{A}) = \frac{\rho(A)}{\rho(A)+\epsilon}<1 \] and, by a well-known result on convergence of matrix powers, \[ \lim_{k \to \infty}\tilde{A}^k=0. \] That means, by sequence limit definition, a natural number $N_1\in \mathbf{N}$ exists such that \[ \forall k\geq N_1 \Rightarrow \|\tilde{A}^k\| < 1 \] which in turn means: \[ \forall k\geq N_1 \Rightarrow \|A^k\| < (\rho(A)+\epsilon)^k \] or \[ \forall k\geq N_1 \Rightarrow \|A^k\|^{1/k} < (\rho(A)+\epsilon). \] Let's now consider the matrix \[ \check{A}=(\rho(A)-\epsilon)^{-1}A \] Then, obviously, \[ \rho(\check{A}) = \frac{\rho(A)}{\rho(A)-\epsilon}>1 \] and so, by the same convergence theorem,$\|\check{A}^k\|$ is not bounded. This means a natural number $N_2\in \mathbf{N}$ exists such that \[ \forall k\geq N_2 \Rightarrow \|\check{A}^k\|>1 \] which in turn means: \[ \forall k\geq N_2 \Rightarrow \|A^k\| > (\rho(A)-\epsilon)^k \] or \[ \forall k\geq N_2 \Rightarrow \|A^k\|^{1/k} > (\rho(A)-\epsilon). \] Taking $N:=max(N_1,N_2)$ and putting it all together, we obtain: \[ \forall \epsilon>0, \exists N\in\mathbb{N}: \forall k\geq N \Rightarrow \rho(A)-\epsilon < \|A^k\|^{1/k} < \rho(A)+\epsilon \] which, by definition, is \[ \lim_{k \to \infty}\|A^k\|^{1/k} = \rho(A).\,\,\square \] Actually, in case the norm is \PMlinkname{self-consistent}{SelfConsistentMatrixNorm}, the proof shows more than the thesis; in fact, using the fact that $|\lambda|\leq\rho(A)$, we can replace in the limit definition the left lower bound with the spectral radius itself and write more precisely: \[ \forall \epsilon>0, \exists N\in\mathbb{N}: \forall k\geq N \Rightarrow \rho(A) \leq \|A^k\|^{1/k} < \rho(A)+\epsilon \] which, by definition, is \[\lim_{k \to \infty}\|A^k\|^{1/k} = \rho(A)^+. \]