3
$\begingroup$

I have a stochastic matrix $A \in R^{n \times n}$ whose sum of the entries in each row is $1$. When I found out the eigenvalues and eigenvectors for this stochastic matrix, it always happens that one of the eigenvalues is $1$.

Is it true that for any square row- or right-stochastic matrix (i.e. each row sums up to $1$) one of the eigenvalues is $1$? If so, how do we prove it?

  • 0
    See also: [Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.](https://math.stackexchange.com/q/347408)2018-09-11

2 Answers 2

0

That is a basic and important property of stochastic matrices. It's also non-obvious, unless you are aware of the Perron-Frobenius theorem.

  • 0
    I don't see any proof of this important property in your answer. So I downvoted. If you want me to retract my downvote, add a step-by-step proof to your answer.2018-05-17
3

The column vector with every entry $1$ is an eigenvector with eigenvalue $1$ for your matrix. It is not necessarily true that the eigenvalue $1$ occurs with algebraic multiplicity $1$ as an eigenvalue for your matrix $A$. By the Frobenius Perron-Theorem, that is the case if the entries of $A$ (or even some power of $A$) are all positive. What is true is that $x-1$ is not a repeated factor of the minimum polynomial of $A$ (using the Frobenius Perron theorem on each block).

  • 0
    Just because it seemed to me worthwhile to note that not every eigenvector with eigennvalue 1 of the matrix in the question needs to be a multiple of the all 1's vector in general, which might be of interest to people who know about the Frobenius-Perron theorem.2018-05-17