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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.

5 Answers 5

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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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    @Bach: Right; this does not show that 1 is a simple eigenvalue.2016-03-10
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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 (since $k$ 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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If we can show that $A$ doesn't increase the 1-norm, i.e., $\|Ax\|_1\leq\|x\|_1$ Then $\|Ax\|_1=\|\lambda x\|_1=|\lambda|\|x\|_1\leq\|x\|_1$ which is $|\lambda|\leq 1$, we are done, but how to show above inequality? For convenience, let's set stochastic matrix $A=\begin{pmatrix}a_{11}& a_{12}\\a_{21}& a_{22}\end{pmatrix}$ Then \begin{eqnarray*}\|Ax\|_1&=&|a_{11}x_1+a_{12}x_2|+|a_{21}x_1+a_{22}x_2|\\&\leq& a_{11}|x_1|+a_{12}|x_2|+a_{21}|x_1|+a_{22}|x_2|\\&=&|x_1|+|x_2|\\&=&\|x\|_1\end{eqnarray*} For n-dimensional matrix, it can be shown in same manner.

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Call A a $n \times n$ stochastic matrix and denote with $(\lambda,\textbf{x})$ one of its eigenpair.

Obviously $1$ is an eigenvalue for $A$, indeed follow directly from the definition of row-stochastic [column-stochastic] matrix that $\textbf e$ $=(1\dots1)^{T}$ is a right [left] eigenvector associated with $1$.

Using induced matrix norm $\parallel\parallel_{1}$ or $\parallel\parallel_{\infty}$ , it's easy to prove that the spectral radius $\rho(A)\leq 1$ :

$ |\lambda|= \frac{||A x||}{||x||} \leq max_{||x||=1} ||Ax||= ||A|| $ Now, since $||A||_{1}$ [respectively $||A||_{\infty}$ ] is the maximum absolute column [row] sum of the matrix, we have

$||A||_{1}=1$ if $A$ is a column-stochastic matrix and

$||A||_{\infty}=1$ if $A$ is a row-stochastic matrix,

and then in any case $|\lambda|\leq 1$.