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Someone should really start cleaning up the math-related articles on wikipedia. This one is very unclear, especially for the people who are most likely to be reading wikipedia. The proof at the beginning should be more detailed. —Preceding unsigned comment added by 66.49.227.118 (talk) 23:21, 2 February 2011 (UTC)Reply

There is also a TeX formatting error near the bottom, I'd have a go at fixing it but I'm not certain what that step is supposed to contain. Slacr (talk) 16:44, 17 February 2014 (UTC)Reply

Much of the content of this article should be moved to a proof page, as per Wikipedia:WikiProject Mathematics/Proofs. See Category:Article proofs for examples of how other articles have done this. linas 15:12, 4 December 2005 (UTC)Reply

I believe there is a mistake in proving the upper bound for gelfand's theorem p(A)<¦A^k¦^{1/k}+e , but it is easily fixed since an upperbound exists from the first lemma.

What is the source for the proof of Gelfand's Fomula? Please cite! —Preceding unsigned comment added by 81.172.140.37 (talk) 09:23, 14 February 2008 (UTC)Reply

I don't think that the given definition of spectral radius of a graph has to be limited to PLANAR graph. Do you? --achab 06:37, 19 April 2007 (UTC)Reply

The follwing statement

Gelfand's formula leads directly to a bound on the spectral radius of a product of finitely many matrices, namely ${\displaystyle \rho (A_{1}A_{2}\ldots A_{n})\leq \rho (A_{1})\rho (A_{2})\ldots \rho (A_{n}).}$

is definitely not valid. Gelfand's formula cannot imply the specified bound on the spectral radius of a product of matrices simply because such a bound is not valid.

Example.

Let

${\displaystyle A_{1}={\begin{bmatrix}0&2\\1/2&0\end{bmatrix}},\qquad A_{2}={\begin{bmatrix}0&1/2\\2&0\end{bmatrix}}.}$

Then

${\displaystyle A_{1}A_{2}={\begin{bmatrix}4&0\\0&1/4\end{bmatrix}}.}$

So, ${\displaystyle \rho (A_{1})=\rho (A_{2})=1}$ while ${\displaystyle \rho (A_{1}A_{2})=4}$. Thus, ${\displaystyle \rho (A_{1}A_{2})>\rho (A_{1})\rho (A_{2}).}$ --79.139.218.53 (talk) 04:43, 15 May 2008 (UTC)Reply

Ah but: ${\displaystyle A_{2}A_{1}={\begin{bmatrix}1/4&0\\0&4\end{bmatrix}}}$ so ${\displaystyle A_{1}}$, ${\displaystyle A_{2}}$ don't commute, which is assumed in the main text. 78.105.183.186 (talk) 12:20, 7 October 2008 (UTC)Reply

if the spectral radius is the supremum of the absolute values of a matrix than the 1 st formula should be changed accordingly (write sup(...) instead of max(....)). regards مبتدئ (talk) 19:28, 31 August 2009 (UTC)Reply

Since it is a supremum over a finite set, the difference is meaningless.--84.161.219.86 (talk) 20:46, 5 January 2013 (UTC)Reply

In the beginning of the proof of Gelfand's formula, ${\displaystyle \|A^{k}\|\sim \rho (A)^{k}\quad k\to \infty }$ is not restating of the theorem's statement. For example, it could happen that ${\displaystyle \|A^{k}\|=\rho (A)^{k}\cdot k^{10}}$ or something else with subexponential growth instead of ${\displaystyle k^{10}}$. Maybe we should delete from "In other words" to "proof"? Andrey Petrov (talk) 19:05, 20 August 2010 (UTC)Reply

I agree with this. I'm getting rid of the statement, which is misleading. Bengski68 (talk) 23:16, 10 June 2015 (UTC)Reply