Perron–Frobenius theorem

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, the theorem gives a unique stationary distribution.

Going deeper

For a student

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 theory and power iteration for the dominant eigenvalue.