proof of Gelfand spectral radius theorem
For any $\u03f5>0$, consider the matrix

$$\stackrel{~}{A}={(\rho (A)+\u03f5)}^{1}A$$ 

Then, obviously,
and, by a wellknown result on convergence of matrix powers,

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

That means, by sequence limit definition, a natural number^{} ${N}_{1}\in \mathbf{N}$ exists such that
which in turn means:
or
Let’s now consider the matrix

$$\stackrel{\u02c7}{A}={(\rho (A)\u03f5)}^{1}A$$ 

Then, obviously,

$$\rho (\stackrel{\u02c7}{A})=\frac{\rho (A)}{\rho (A)\u03f5}>1$$ 

and so, by the same convergence theorem^{},$\parallel {\stackrel{\u02c7}{A}}^{k}\parallel $ is not bounded.
This means a natural number ${N}_{2}\in \mathbf{N}$ exists such that

$$\forall k\ge {N}_{2}\Rightarrow \parallel {\stackrel{\u02c7}{A}}^{k}\parallel >1$$ 

which in turn means:

$$\forall k\ge {N}_{2}\Rightarrow \parallel {A}^{k}\parallel >{(\rho (A)\u03f5)}^{k}$$ 

or

$$\forall k\ge {N}_{2}\Rightarrow {\parallel {A}^{k}\parallel}^{1/k}>(\rho (A)\u03f5).$$ 

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 selfconsistent (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

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