Read-back: eigenvalue_basic

Let $V$ be an arbitrary finite type with decidable equality (the empty type is allowed), and let $P$ be a $V \times V$ matrix of real numbers assumed to be stochastic in the following sense: every entry satisfies $P_{x,y} \ge 0$, and every row sums to one, $\sum_{y} P_{x,y} = 1$ for all $x$. Under this single hypothesis the theorem asserts the conjunction of the following three statements.

(1) For every real number $\lambda$, if $\lambda$ is an eigenvalue of $P$ — meaning there exists a function $f : V \to \mathbb{R}$ that is not identically zero and satisfies the matrix–vector equation $P f = \lambda f$, i.e. $\sum_y P_{x,y} f(y) = \lambda f(x)$ for all $x$ — then $|\lambda| \le 1$. (Only real eigenvalues with a real eigenvector are covered by this eigenvalue notion; complex eigenvalues are not mentioned.)

(2) If $P$ is irreducible — here meaning: for every pair of states $x, y \in V$ there exists a natural number $t$ (the value $t = 0$ is allowed, in which case $P^0$ is the identity matrix) with $(P^t)_{x,y} > 0$ — then every function $f : V \to \mathbb{R}$ satisfying $P f = f$ (that is, $\sum_y P_{x,y} f(y) = f(x)$ for all $x$; no non-vanishing condition is imposed on $f$) is constant: $f(x) = f(y)$ for all $x, y \in V$.

(3) If $P$ is irreducible (in the sense of (2)) and aperiodic, then $-1$ is not an eigenvalue of $P$ in the sense of (1); that is, there is no nonzero real function $f$ with $P f = -f$. Here aperiodic means: for every state $x$, the period of $x$ equals $1$, where the period of $x$ is defined as the supremum (in $\mathbb{N}$, where the supremum of an empty or unbounded set is the junk value $0$) of the set of natural numbers $d$ that divide every element of the return set $\{t \in \mathbb{N} \mid t \ge 1 \text{ and } (P^t)_{x,x} > 0\}$. Note that if the return set of some $x$ is empty, every $d$ vacuously divides all its elements, the divisor set is unbounded, and the period of that $x$ is $0 \ne 1$, so such a chain is not aperiodic under this definition.

Edge cases: when $V$ is empty, part (1) and part (3) hold vacuously (no nonzero $f$ exists), and part (2)'s conclusion quantifies over no states; also, for empty $V$ the row-sum condition in the stochasticity hypothesis is vacuous. The three parts are asserted simultaneously as one conjunction, each with only the hypotheses listed for it.