A stochastic matrix is a square matrix whose entries are non-negative and whose rows all sum to 1. The transition matrix for a finite-state Markov chain is a stochastic matrix, and so they are essential for tackling problems that can be modeled as Markov chains.
One of the many interesting properties of a stochastic matrix is that its largest eigenvalue is 1. I believe this can be proved using the Perron-Frobenius theorem. However, I found a simple, self-contained proof, thanks in part to an answer to a different question of mine on math.SE.
Theorem: The largest eigenvalue of a stochastic matrix is 1.
Proof: First, if A is a stochastic matrix, then A1 = 1, since each row of A sums to 1. This proves that 1 is an eigenvalue of A. Second, suppose there exists λ > 1 and nonzero x such that Ax = λx. Let xi be a largest element of x. Since any scalar multiple of x will also satisfy this equation we can assume, without loss of generality, that xi > 0. Since the rows of A are nonnegative and sum to 1, each entry in λx is a convex combination of the elements of x. Thus no entry in λx can be larger than xi. But since λ > 1, λxi > xi. Contradiction. Therefore, the largest eigenvalue of A is 1.