---
tag: 62NS7PV1M9
kind: statement
role: theorem
name: Perron–Frobenius theorem
msc: [15B48, 15A18]
uses: [JPQCCFCY8X, 7Q60BB01YG, 9V7YQA85PC, 0BQAA2BK3Q, 0N6960W0BV, 0VSSPAFM2C]
about: [JPQCCFCY8X, 7Q60BB01YG]
sources:
  - url: https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem
same_as:
  wikidata: Q1564541
  wikipedia: Perron–Frobenius theorem
---

## Statement

Let $n \ge 1$, let $A$ be a real $n \times n$ [[nonnegative matrix|JPQCCFCY8X]], and let $r = \rho(A)$ be its [[spectral radius|9V7YQA85PC]].

1. The number $r$ is an [[eigenvalue|0BQAA2BK3Q]] of $A$, and it has an [[eigenvector|0N6960W0BV]] $v \ne 0$ with all entries nonnegative.
2. If $n \ge 2$ and $A$ is [[irreducible|7Q60BB01YG]], then $r > 0$ and $r$ is a simple root of the [[characteristic polynomial|0VSSPAFM2C]] of $A$.
3. If $n \ge 2$ and $A$ is irreducible, then $r$ has an eigenvector $v$ with all entries positive.
4. If $n \ge 2$ and $A$ is irreducible, then every nonnegative eigenvector of $A$ is a positive multiple of $v$.
5. If $n \ge 2$, $A$ is irreducible, and $h$ is the number of eigenvalues of $A$ of absolute value $r$, then these eigenvalues are $r e^{2 \pi i k / h}$ for $k = 0, 1, \ldots, h - 1$.
6. If all entries of $A$ are positive, then the conclusions of 2, 3 and 4 hold, and $\abs{\lambda} < r$ for every eigenvalue $\lambda \ne r$ of $A$.

## Remarks

One proof of part 6 looks at the map $x \mapsto Ax / \sum_i (Ax)_i$ on the simplex of nonnegative vectors with entry sum $1$; Brouwer's fixed point theorem gives a positive eigenvector.
For an irreducible matrix, another proof uses the Collatz–Wielandt formula $r = \max_{x \ge 0,\, x \ne 0} \min_{x_i > 0} (Ax)_i / x_i$.
Part 1 follows from part 6 by continuity, applied to $A + \varepsilon J$ where $J$ has all entries $1$ and $\varepsilon \to 0$.
Example: $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is irreducible with $r = 1$, eigenvector $(1, 1)$, and eigenvalues $1$ and $-1$, so $h = 2$; the strict inequality of part 6 fails because $A$ has zero entries.
Without irreducibility, simplicity can fail: the identity matrix $I_2$ has $r = 1$ as a double eigenvalue.
For the transition matrix of a finite irreducible [[Markov chain|0G48PYM96Q]], the theorem gives a unique stationary distribution.

## Explanations

### L1 human

A matrix with nonnegative entries has a special eigenvalue: its spectral radius, the largest absolute value of an eigenvalue, is itself an eigenvalue, a real number $r \ge 0$.
Its eigenvector can be chosen with nonnegative entries, so it makes sense as a list of sizes, shares or probabilities.
When all entries are positive, this eigenvalue strictly beats all the others, and repeated multiplication by $A$ pushes every positive vector towards the direction of that eigenvector.
This is why rankings such as PageRank and long-run behaviour of population models are computed from it.

### L2 human

The theorem has three layers: nonnegative matrices give a nonnegative eigenvector for $\rho(A)$; irreducible ones give simplicity and positivity; positive ones give a spectral gap.
The number $h$ in part 5 is the period of the directed graph of $A$, and $A$ is called primitive when $h = 1$.
For a primitive matrix, $A^k / r^k$ converges to the rank one matrix $v w^T / (w^T v)$, where $w$ is the positive left eigenvector.
The theorem underlies the convergence of finite Markov chains, the [[M-matrix|DNKXHMGQY1]] theory and power iteration for the dominant eigenvalue.