Editing Perron–Frobenius theorem (section)

===Inequalities for Perron–Frobenius eigenvalue===
For any non-negative matrix ''A'' its Perron–Frobenius eigenvalue ''r'' satisfies the inequality:

:<math> r \; \le \; \max_i \sum_j a_{ij}.</math>

This is not specific to non-negative matrices: for any matrix ''A'' with an eigenvalue <math>\scriptstyle\lambda</math> it is true
that <math>\scriptstyle |\lambda| \; \le \; \max_i \sum_j |a_{ij}|</math>. This is an immediate corollary of the
[[Gershgorin circle theorem]]. However another proof is more direct:

Any [[Matrix norm#Induced norm|matrix induced norm]] satisfies the inequality <math>\scriptstyle\|A\| \ge |\lambda|</math> for any eigenvalue <math>\scriptstyle\lambda</math> because, if <math>\scriptstyle x</math> is a corresponding eigenvector, <math>\scriptstyle\|A\| \ge |Ax|/|x| = |\lambda x|/|x| = |\lambda|</math>. The [[Matrix norm#Induced norm|infinity norm]] of a matrix is the maximum of row sums: <math>\scriptstyle \left \| A \right \| _\infty = \max \limits _{1 \leq i \leq m} \sum _{j=1} ^n | a_{ij} |. </math> Hence the desired inequality is exactly  <math>\scriptstyle\|A\|_\infty \ge |\lambda|</math> applied to the non-negative matrix ''A''.

Another inequality is:

:<math>\min_i \sum_j a_{ij} \; \le \; r .</math>

This fact is specific to non-negative matrices; for general matrices there is nothing similar. Given  that ''A'' is positive (not just non-negative), then there exists a positive eigenvector ''w'' such that ''Aw'' = ''rw'' and the smallest component of ''w'' (say ''w<sub>i</sub>'') is 1. Then ''r'' = (''Aw'')<sub>''i''</sub> ≥ the sum of the numbers in row ''i'' of ''A''. Thus the minimum row sum gives a lower bound for ''r'' and this observation can be extended to all non-negative matrices by continuity.

Another way to argue it is via the [[Lothar Collatz|Collatz]]-Wielandt formula. One takes the vector ''x''&nbsp;=&nbsp;(1,&nbsp;1,&nbsp;...,&nbsp;1) and immediately obtains the inequality.