## A Simple Proof that the Largest Eigenvalue of a Stochastic Matrix is 1

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 A11, 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 A= 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.

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### 11 Responses to A Simple Proof that the Largest Eigenvalue of a Stochastic Matrix is 1

What if $\lambda \in \mathbb{C}$ and $\lambda\ge 1$?

2. mzspivey says:

Good question. If $\lambda \in \mathbb{C}$ and $|\lambda| > 1$, the proof still goes through, replacing “largest element” with “element with largest modulus” and applying the triangle inequality in one place. Each entry in $\lambda {\bf x}$ is still a convex combination of elements of x, as that claim has to do with the real entries in A. Thus the modulus of each entry in $\lambda {\bf x}$ is, by the triangle inequality, no larger than the convex combination of the moduli of the entries in x. Thus none of the moduli of the entries in $\lambda {\bf x}$ can be larger than $|x_i|$. But we also have $|\lambda x_i| > |x_i|$ for $\lambda > 1$, by properties of complex numbers. Contradiction.

3. shi says:

I can not understand why, because actually, every entry of in λx is(assume a 3*3 markov matrix A) x1*A11+x2*A21+x3*A31, but what we have is A11+A12+A13 = 1(every row).

• mzspivey says:

Your example $x_1 A_{11} + x_2 A_{21} + x_3 A_{31}$ is left-multiplying A by a row vector x. In my argument I’m right-multiplying A by a column vector x.

4. Rob Kusner says:

Is there a good way to decide how many eigenvalues have modulus < 1?

5. Alex says:

hey i dont get the point : Thus no entry in λx can be larger than x_i
but why not?

• mzspivey says:

A convex combination is a weighted average. If you take a weighted average of a set of numbers (where the weights are nonnegative and sum to 1), then that weighted average can’t be larger than the largest number in the set. For example, suppose you have the numbers 1, 3, 6, and 10. There’s no way you can take a weighted average of those four numbers and get something larger than 10.

6. chundi says:

This is for right stochastic matrix, can u give proof for left stochastic matrix also ?

7. raphael says:

You need that $x_i>0$…

• mzspivey says:

That’s true. It looks like I need to edit my proof. Nice catch!