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The largest eigenvalue of a stochastic matrix (i.e. a matrix whose entries are positive and whose rows add up to 1) is 1.

Wikipedia marks this as a special case of the Perron-Frobenius theorem... but I wonder if there is a simpler (more direct) way to demonstrate this result?

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up vote 30 down vote accepted

Here's a really elementary proof (which is a slight modification of Fanfan's answer to a question of mine). As Calle shows, it is easy to see that the eigenvalue $1$ is obtained. Now, suppose $Ax = \lambda x$ for some $\lambda > 1$. Since the rows of $A$ are nonnegative and sum to $1$, each element of vector $Ax$ is a convex combination of the components of $x$, which can be no greater than $x_{max}$, the largest component of $x$. On the other hand, at least one element of $\lambda x$ is greater than $x_{max}$, which proves that $\lambda > 1$ is impossible.

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Thank you, simple and self-contained, that is what I had in mind! – koletenbert May 24 '11 at 6:59
Unlike Perron-Frobenius, however, this does not show that $\lambda=1$ is a simple eigenvalue (that is, having a 1-dimensional eigenspace) in case the matrix in question is irreducible. Right? – Bach Feb 14 at 14:33
@Bach: Right; this does not show that 1 is a simple eigenvalue. – Mike Spivey Mar 10 at 22:56

Say $A$ is a $n \times n$ row stochastic matrix. Now: $$A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} \sum_{i=1}^n a_{1i} \\ \sum_{i=1}^n a_{2i} \\ \vdots \\ \sum_{i=1}^n a_{ni} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} $$ Thus the eigenvalue 1 is attained.

To show that the this is the largest eigenvalue you can use the Gershgorin circle theorem. Take row $k$ in $A$. The diagonal element will be $a_{kk}$ and the radius will be $\sum_{i\neq k} |a_{ki}| = \sum_{i \neq k} a_{ki}$ since all $a_{ki} \geq 0$. This will be a circle with its center in $a_{kk} \in [0,1]$, and a radius of $\sum_{i \neq k} a_{ki} = 1-a_{kk}$. So this circle will have 1 on its perimeter. This is true for all Gershgorin circles for this matrix (sine $k$ was taken arbitrarily). Thus, since all eigenvalues lie in the union of the Gershgorin circles, all eigenvalues $\lambda_i$ satisfy $|\lambda_i| \leq 1$.

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