Editing Perron–Frobenius theorem (section)

===Further proofs===

====Perron projection====
The proof now proceeds using [[Eigendecomposition of a matrix|spectral decomposition]]. The trick here is to split the Perron root from the other eigenvalues. The spectral projection associated with the Perron root is called the Perron projection and it enjoys the following property:

The Perron projection of an irreducible non-negative square matrix is a positive matrix.

Perron's findings and also (1)–(5) of the theorem are corollaries of this result. The key point is that a positive projection always has rank one. This means that if ''A'' is an irreducible non-negative square matrix then the algebraic and geometric multiplicities of its Perron root are both one. Also if ''P'' is its Perron projection then ''AP'' = ''PA'' = ρ(''A'')''P'' so every column of ''P'' is a positive right eigenvector of ''A'' and every row is a positive left eigenvector. Moreover, if ''Ax'' = λ''x'' then ''PAx'' = λ''Px'' = ρ(''A'')''Px'' which means ''Px'' = 0 if λ ≠ ρ(''A''). Thus the only positive eigenvectors are those associated with ρ(''A''). If ''A'' is a primitive matrix with ρ(''A'') = 1 then it can be decomposed as ''P'' ⊕ (1&nbsp;−&nbsp;''P'')''A'' so that ''A<sup>n</sup>'' = ''P'' + (1&nbsp;−&nbsp;''P'')''A''<sup>''n''</sup>. As ''n'' increases the second of these terms decays to zero leaving ''P'' as the limit of ''A<sup>n</sup>'' as ''n''&nbsp;→&nbsp;∞.

The power method is a convenient way to compute the Perron projection of a primitive matrix. If ''v'' and ''w'' are the positive row and column vectors that it generates then the Perron projection is just ''wv''/''vw''. The spectral projections aren't neatly blocked as in the Jordan form. Here they are overlaid and each generally has complex entries extending to all four corners of the square matrix. Nevertheless, they retain their mutual orthogonality which is what facilitates the decomposition.

====Peripheral projection====
The analysis when ''A'' is irreducible and non-negative is broadly similar. The Perron projection is still positive but there may now be other eigenvalues of modulus ρ(''A'') that negate use of the power method and prevent the powers of (1&nbsp;−&nbsp;''P'')''A'' decaying as in the primitive case whenever ρ(''A'') = 1. So we consider the '''peripheral projection''', which is the spectral projection of ''A'' corresponding to all the eigenvalues that have modulus ''ρ''(''A''). It may then be shown that the peripheral projection of an irreducible non-negative square matrix is a non-negative matrix with a positive diagonal.

====Cyclicity====
Suppose in addition that ρ(''A'') = 1 and ''A'' has ''h'' eigenvalues on the unit circle. If ''P'' is the peripheral projection then the matrix ''R'' = ''AP'' = ''PA'' is non-negative and irreducible, ''R<sup>h</sup>'' = ''P'', and the cyclic group ''P'', ''R'', ''R''<sup>2</sup>, ...., ''R''<sup>''h''−1</sup> represents the harmonics of ''A''. The spectral projection of ''A'' at the eigenvalue λ on the unit circle is given by the formula <math>\scriptstyle h^{-1}\sum^h_1\lambda^{-k}R^k</math>. All of these projections (including the Perron projection) have the same positive diagonal, moreover choosing any one of them and then taking the modulus of every entry invariably yields the Perron projection. Some donkey work is still needed in order to establish the cyclic properties (6)–(8) but it's essentially just a matter of turning the handle. The spectral decomposition of ''A'' is given by ''A''&nbsp;=&nbsp;''R''&nbsp;⊕&nbsp;(1&nbsp;−&nbsp;''P'')''A'' so the difference between ''A<sup>n</sup>'' and ''R<sup>n</sup>'' is ''A<sup>n</sup>''&nbsp;−&nbsp;''R<sup>n</sup>'' =&nbsp;(1&nbsp;−&nbsp;''P'')''A''<sup>''n''</sup> representing the transients of ''A<sup>n</sup>'' which eventually decay to zero. ''P'' may be computed as the limit of ''A<sup>nh</sup>'' as ''n''&nbsp;→&nbsp;∞.