# Perron-Frobenius theorem

(Redirected from Perron–Frobenius theorem)

Let a real square $( n \times n)$- matrix $A$ be considered as an operator on $\mathbf R ^ {n}$, let it be without invariant coordinate subspaces (such a matrix is called indecomposable) and let it be non-negative (i.e. all its elements are non-negative). Also, let $\lambda _ {1} \dots \lambda _ {n}$ be its eigen values, enumerated such that

$$| \lambda _ {1} | = \dots = | \lambda _ {h} | > | \lambda _ {h+} 1 | \geq \dots \geq | \lambda _ {n} | ,\ \ 1 \leq h \leq n.$$

Then,

1) the number $r = | \lambda _ {1} |$ is a simple positive root of the characteristic polynomial of $A$;

2) there exists an eigen vector of $A$ with positive coordinates corresponding to $r$;

3) the numbers $\lambda _ {1} \dots \lambda _ {h}$ coincide, apart from their numbering, with the numbers $r, \omega r \dots \omega ^ {h-} 1 r$, where $\omega = e ^ {2 \pi i/h }$;

4) the product of any eigen value of $A$ by $\omega$ is an eigen value of $A$;

5) for $h > 1$ one can find a permutation of the rows and columns that reduces $A$ to the form

$$\left \| \begin{array}{ccccc} 0 &A _ {1} & 0 &\dots & 0 \\ 0 & 0 &A _ {2} &\dots & 0 \\ \dots &\dots &\dots &\dots &\dots \\ 0 & 0 & 0 &\dots &A _ {h-} 1 \\ A _ {h} & 0 & 0 &\dots & 0 \\ \end{array} \right \| ,$$

where $A _ {j}$ is a matrix of order $nh ^ {-} 1$.

O. Perron proved the assertions 1) and 2) for positive matrices in , while G. Frobenius  gave the full form of the theorem.

How to Cite This Entry:
Perron–Frobenius theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Perron%E2%80%93Frobenius_theorem&oldid=22894