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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?

Rodrigo de Azevedo
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2 Answers2

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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).

Geoff Robinson
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  • What do you mean by "algebraic multiplicity"? What does it even mean that an eigenvalue appears with "algebraic multiplicity" for a matrix? –  May 17 '18 at 08:55
  • 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 $r$). – Geoff Robinson May 17 '18 at 09:11
  • But why do you say it in your answer? –  May 17 '18 at 09:21
  • 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. – Geoff Robinson May 17 '18 at 09:44
  • This answer is a bit misleading. The algebraic and geometric multiplicity of the eigenvalue 1 of a stochastic matrix is the same in all cases. Otherwise, the corresponding Jordan chain in its Jordan normal form would have length $> 1$, which would mean the set of powers of the matrix have unbounded operator norm. However, for a stochastic matrix, its powers are a stochastic matrix too and the operator norm is bounded by the Frobenius norm which in turn is bounded by $n$. – heiner Feb 21 '22 at 12:42
  • @heine: Please read my first comment in answer to user168764. By algebraic multiplicity, of 1, I mean the multiplicity of (x-1) as a factor of the CHARACTERISTIC polynomial. You seem to be discussing the multiplicity of (X-1) as a factor of the MINIMUM polynomial. I agree that for stochastic matrices, the algebraic multiplicity (in the sense I used it) and the geometric multiplicity of 1 as an eigenvalue are the same.. – Geoff Robinson Feb 21 '22 at 13:27
  • Thanks for your reply all these years later. Perhaps I misunderstand, but I think this part is still wrong: "With this convention, it is the case that the dimension of the -eigenspace of is at most (but in general it can be strictly smaller than )". I don't think there's an example of a (row- or column-) stochastic matrix where the dimension of the =1-eigenspace is strictly smaller than , where $(x-1)^r$ is the `maximum' factor of the characteristic polynomial as per your definition. Happy to hear about a counterexample! – heiner Feb 21 '22 at 15:40
  • That part was referring to a general matrix $A$, not a stochastic one. – Geoff Robinson Feb 21 '22 at 16:37
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That is a basic and important property of stochastic matrices. It's also non-obvious, unless you are aware of the Perron-Frobenius theorem.

leonbloy
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    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" http://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. – leonbloy Aug 27 '11 at 17:48
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    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. –  May 17 '18 at 09:04