Editing Perron–Frobenius theorem (section)

===Perron root is strictly maximal eigenvalue for positive (and primitive) matrices===
If ''A'' is a positive (or more generally primitive) matrix, then there exists a real positive eigenvalue ''r'' (Perron–Frobenius eigenvalue or Perron root), which is strictly greater in absolute value than all other eigenvalues, hence ''r'' is the [[spectral radius]] of ''A''.

This statement does not hold for general non-negative irreducible matrices, which have ''h'' eigenvalues with the same absolute eigenvalue as ''r'', where ''h'' is the period of ''A''.

====Proof for positive matrices====
Let ''A'' be a positive matrix, assume that its spectral radius ρ(''A'') = 1 (otherwise consider ''A/ρ(A)''). Hence, there exists an eigenvalue λ on the unit circle, and all the other eigenvalues are less or equal 1 in absolute value. Suppose that another eigenvalue λ ≠ 1 also falls on the unit circle. Then there exists a positive integer ''m'' such that ''A<sup>m</sup>'' is a positive matrix and the real part of λ''<sup>m</sup>'' is negative. Let ε be half the smallest diagonal entry of ''A<sup>m</sup>'' and set ''T'' = ''A<sup>m</sup>''&nbsp;−&nbsp;''εI'' which is yet another positive matrix. Moreover, if ''Ax'' = ''λx'' then ''A<sup>m</sup>x'' = ''λ<sup>m</sup>x'' thus ''λ''<sup>''m''</sup>&nbsp;−&nbsp;''ε'' is an eigenvalue of ''T''. Because of the choice of ''m'' this point lies outside the unit disk consequently ''ρ''(''T'') > 1. On the other hand, all the entries in ''T'' are positive and less than or equal to those in ''A<sup>m</sup>'' so by [[spectral radius|Gelfand's formula]] ''ρ''(''T'') ≤ ''ρ''(''A<sup>m</sup>'') ≤ ''ρ''(''A'')<sup>''m''</sup> = 1. This contradiction means that λ=1 and there can be no other eigenvalues on the unit circle.

Absolutely the same arguments can be applied to the case of primitive matrices; we just need to mention the following simple lemma, which clarifies the properties of primitive matrices.

====Lemma====
Given a non-negative ''A'', assume there exists ''m'', such that ''A<sup>m</sup>'' is positive, then ''A''<sup>''m''+1</sup>, ''A''<sup>''m''+2</sup>, ''A''<sup>''m''+3</sup>,... are all positive.

''A''<sup>''m''+1</sup> = ''AA''<sup>''m''</sup>, so it can have zero element only if some row of ''A'' is entirely zero, but in this case the same row of ''A<sup>m</sup>'' will be zero.

Applying the same arguments as above for primitive matrices, prove the main claim.