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# Perron–Frobenius theorem

### From Intelligent Perception

**Perron–Frobenius theorem**.

Suppose $A = (a_{ij})$ is an $n × n$ positive matrix: $a_{ij} > 0$ for $1 ≤ i, j ≤ n$. Then there is a real eigenvalue $r>0$ of A such that

- $r$ is a simple root of the characteristic polynomial of A and its eigenspace is 1-dimensional (the geometric multiplicity of $r$ is $1$).
- there exists an eigenvector $v = (v_1,…,v_n)$ corresponding to $r$ such that all its components are positive: $v_i > 0$ for $ 1 ≤ i ≤ n$.
- any other eigenvalue $λ$ satisfies $|λ| < r$.

Then $r$ is called the *Perron root* or the Perron–Frobenius eigenvalue. It is the spectral radius of $A$.

Suppose $A = (a_{ij})$ is an $n × n$ irreducible non-negative matrix: $a_{ij} \geq 0$ for $1 ≤ i, j ≤ n$. Then there is a positive real eigenvalue $r>0$ such that

- r is a simple root of the characteristic polynomial of A and its eigenspace is 1-dimensional (the geometric multiplicity of $r$ is $1$).
- there exists an eigenvector $v = (v_1,…,v_n)$ of $r$ such that all its components are positive: $v_i > 0$ for $ 1 ≤ i ≤ n$.
- there are "maximal" eigenvalues: $\mu = e^{2πik/m}r,k = 0, 1, ..., m − 1$, where $m$ is some integer, $r>0$ is a real eigenvalue (all such eigenvalues are simple roots of the characteristic polynomial), while any other eigenvalue $λ$ satisfies $|λ| < r$.

The *Perron root* satisfies:
$$\min_i \sum_{j} a_{ij} \le r \le \max_i \sum_{j} a_{ij}.$$
Therefore, for a row stochastic matrix,
$$r=1.$$