# Is it true that for any square row-stochastic matrix one of the eigenvalues is $1$?

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?

• Sep 11 '18 at 12:54

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).

• What do you mean by "algebraic multiplicity"? What does it even mean that an eigenvalue appears with "algebraic multiplicity" for a matrix?
– nbro
May 17 '18 at 8:55
• By saying that an eigenvalue $\lambda$ of a matrix $A$ has algebraic multiplicity $r$, I intend to mean that $(x-\lambda)^{r}$ divides the characteristic polynomial of $A$, but $(x-\lambda)^{r+1}$ does not. With this convention, it is the case that the dimension of the $\lambda$-eigenspace of $A$ is at most $r$ (but in general it can be strictly smaller than $r$). May 17 '18 at 9:11
– nbro
May 17 '18 at 9:21
• 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. May 17 '18 at 9:44

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

• What is non-obvious? The property the OP asks about (that 1 is an eigenvalue) IS obvious once one notices that AU=U for U the column vector of ones... :-)
– Did
Aug 27 '11 at 17:00
• I'm assuming, when he says that the rows sum one, that we are dealing with a "right stochastic matrix", and hence we are speaking of "left eigenvalues" en.wikipedia.org/wiki/Stochastic_matrix If we have a "left stochastic matrix" that additionally has the property of its rows summing one, then it's a double strochastic matrix ( a very particular case) and then you're right, it's obvious. Aug 27 '11 at 17:48
• No particular case. Since AU=U, 1 is an eigenvalue of A (for example in the sense that ker(A-I) is not {0}), hence det(I-A)=0, hence det(I-A^T)=0 (because det(M)=det(M^T) for every matrix M), hence 1 is an eigenvalue of A^T, hence there exists a nonzero vector V such that A^TV=V, hence V^TA=V^T.
– Did
Aug 27 '11 at 18:01
• 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.
– nbro
May 17 '18 at 9:04