Editing Perron–Frobenius theorem (section)

===Power method and the positive eigenpair===
For a positive (or more generally irreducible non-negative) matrix ''A'' the dominant [[eigenvector]] is real and strictly positive (for non-negative ''A'' respectively non-negative.)

This can be established using the [[power method]], which states that for a sufficiently generic (in the sense below) matrix ''A'' the sequence of vectors  ''b''<sub>''k''+1</sub> = ''Ab''<sub>''k''</sub> / | ''Ab''<sub>''k''</sub> | converges to the [[eigenvector]] with the maximum [[eigenvalue]]. (The initial vector ''b''<sub>0</sub> can be chosen arbitrarily except for some measure zero set). Starting with a non-negative vector ''b''<sub>0</sub> produces the sequence of non-negative vectors ''b<sub>k</sub>''. Hence the limiting vector is also non-negative. By the power method this limiting vector is the dominant eigenvector for ''A'', proving the assertion. The corresponding eigenvalue is non-negative.

The proof requires two additional arguments. First, the power method converges for matrices which do not have several eigenvalues of the same absolute value as the maximal one. The previous section's argument guarantees this.

Second, to ensure strict positivity of all of the components of the eigenvector for the case of irreducible matrices. This follows from the following fact, which is of independent interest:

:Lemma: given a positive (or more generally irreducible non-negative) matrix ''A'' and ''v'' as any non-negative eigenvector for ''A'', then it<!--the eigenvector?--> is necessarily strictly positive and the corresponding eigenvalue is also strictly positive.

Proof. One of the definitions of irreducibility for non-negative matrices is that for all indexes ''i,j'' there exists ''m'', such that (''A''<sup>''m''</sup>)<sub>''ij''</sub> is strictly positive. Given a non-negative eigenvector ''v'', and that at least one of its components say ''i''-th is strictly positive, the corresponding eigenvalue is strictly positive, indeed, given ''n'' such that (''A''<sup>''n''</sup>)<sub>''ii''</sub> >0, hence:  ''r''<sup>''n''</sup>''v''<sub>''i''</sub> =
''A''<sup>''n''</sup>''v''<sub>''i''</sub> ≥
(''A''<sup>''n''</sup>)<sub>''ii''</sub>''v''<sub>''i''</sub>
>0. Hence ''r'' is strictly positive. The eigenvector is strict positivity. Then given ''m'', such that (''A''<sup>''m''</sup>)<sub>''ji''</sub> >0, hence:  ''r''<sup>''m''</sup>''v''<sub>''j''</sub> =
(''A''<sup>''m''</sup>''v'')<sub>''j''</sub> ≥
(''A''<sup>''m''</sup>)<sub>''ji''</sub>''v''<sub>''i''</sub> >0, hence
''v''<sub>''j''</sub> is strictly positive, i.e., the eigenvector is strictly positive.