proof of Gelfand spectral radius theorem

For any $ϵ>0$, consider the matrix

$$\tilde{A}={(\rho (A)+ϵ)}^{-1}A$$

Then, obviously,

and, by a well-known result on convergence of matrix powers,

$$\underset{k\to \mathrm{\infty }}{lim}{\tilde{A}}^k=0.$$

That means, by sequence limit definition, a natural number $N_1\in 𝐍$ exists such that

which in turn means:

or

Let’s now consider the matrix

$$\check{A}={(\rho (A)-ϵ)}^{-1}A$$

Then, obviously,

$$\rho (\check{A})=\frac{\rho (A)}{\rho (A)-ϵ}>1$$

and so, by the same convergence theorem,$\parallel {\check{A}}^k\parallel$ is not bounded. This means a natural number $N_2\in 𝐍$ exists such that

$$\forall k\ge N_2\Rightarrow \parallel {\check{A}}^k\parallel >1$$

which in turn means:

$$\forall k\ge N_2\Rightarrow \parallel A^k\parallel >{(\rho (A)-ϵ)}^k$$

or

$$\forall k\ge N_2\Rightarrow {\parallel A^k\parallel }^{1/k}>(\rho (A)-ϵ).$$

Taking $N:=max(N_1,N_2)$ and putting it all together, we obtain:

which, by definition, is

$$\underset{k\to \mathrm{\infty }}{lim}{\parallel A^k\parallel }^{1/k}=\rho (A).\mathrm{\square }$$

Actually, in case the norm is self-consistent (http://planetmath.org/SelfConsistentMatrixNorm), the proof shows more than the thesis; in fact, using the fact that $|\lambda |\le \rho (A)$, we can replace in the limit definition the left lower bound with the spectral radius itself and write more precisely:

which, by definition, is

$$\underset{k\to \mathrm{\infty }}{lim}{\parallel A^k\parallel }^{1/k}=\rho {(A)}^{+}.$$

Title	proof of Gelfand spectral radius theorem
Canonical name	ProofOfGelfandSpectralRadiusTheorem
Date of creation	2013-03-22 15:33:55
Last modified on	2013-03-22 15:33:55
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	7
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 34L05