Editing Perron–Frobenius theorem (section)

===Multiplicity one===
This section proves that the Perron–Frobenius eigenvalue is a simple root of the characteristic polynomial of the matrix. Hence the eigenspace associated to Perron–Frobenius eigenvalue  ''r'' is one-dimensional. The arguments here are close to those in Meyer.<ref name="Meyer">{{harvnb|Meyer|2000|pp=[http://www.matrixanalysis.com/Chapter8.pdf chapter 8 page 665] {{cite web |url=http://www.matrixanalysis.com/Chapter8.pdf |title=Archived copy |access-date=2010-03-07 |url-status=dead |archive-url=https://web.archive.org/web/20100307021652/http://www.matrixanalysis.com/Chapter8.pdf |archive-date=March 7, 2010 }}}}</ref>

Given a strictly positive eigenvector ''v'' corresponding to ''r'' and another eigenvector ''w'' with the same eigenvalue. (The vectors ''v'' and ''w'' can be chosen to be real, because ''A'' and ''r'' are both real, so the null space of ''A-r'' has a basis consisting of real vectors.) Assuming at least one of the components of ''w'' is positive (otherwise multiply ''w'' by −1). Given  maximal possible ''α'' such that ''u=v- α w'' is non-negative, then one of the components of ''u'' is zero, otherwise ''α'' is not maximum. Vector ''u'' is an eigenvector. It is non-negative, hence by the lemma described in the [[#Power method and the positive eigenpair|previous section]] non-negativity implies strict positivity for any eigenvector. On the other hand, as above at least one component of ''u'' is zero. The contradiction implies that ''w'' does not exist.

Case: There are no Jordan blocks corresponding to the Perron–Frobenius eigenvalue ''r'' and all other eigenvalues which have the same absolute value.

If there is a Jordan block, then the [[Matrix norm#Induced norm|infinity norm]]
(A/r)<sup>k</sup><sub>∞</sub> tends to infinity for ''k → ∞ '',
but that contradicts the existence of the positive eigenvector.

Given ''r'' = 1, or ''A/r''. Letting ''v'' be a Perron–Frobenius strictly positive eigenvector, so ''Av=v'', then:

<math> \|v\|_{\infty}= \|A^k v\|_{\infty} \ge \|A^k\|_{\infty} \min_i (v_i), ~~\Rightarrow~~ \|A^k\|_{\infty} \le \|v\|/\min_i (v_i) </math>
So ‖''A<sup>k</sup>''‖<sub>∞</sub> is bounded for all ''k''. This gives another proof that there are no eigenvalues which have greater absolute value than Perron–Frobenius one. It also contradicts the existence of the Jordan block for any eigenvalue which has absolute value equal to 1 (in particular for the Perron–Frobenius one), because existence of the Jordan block implies that ‖''A<sup>k</sup>''‖<sub>∞</sub> is unbounded. For a two by two matrix:
: <math>
J^k= \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} ^k
=
\begin{pmatrix} \lambda^k & k\lambda^{k-1} \\ 0 & \lambda^k \end{pmatrix},
</math>
hence ‖''J''<sup>''k''</sup>‖<sub>∞</sub> = |''k'' + ''λ''| (for |''λ''| = 1), so it tends to infinity when ''k'' does so. Since ''J<sup>k</sup>'' = ''C''<sup>−1</sup> ''A''<sup>''k''</sup>''C'', then ''A''<sup>''k''</sup> ≥ ''J''<sup>''k''</sup>/ (''C''<sup>−1</sup> ''C'' ), so it also tends to infinity. The resulting contradiction implies that there are no Jordan blocks for the corresponding eigenvalues.

Combining the two claims above reveals that the Perron–Frobenius eigenvalue ''r'' is simple root of the characteristic polynomial. In the case of nonprimitive matrices, there exist other eigenvalues which have the same absolute value as ''r''.  The same claim is true for them, but requires more work.