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Let $M$ be a doubly stochastic matrix in which every entry is strictly positive.

Prove that for any eigenvalue $\lambda$ we have $\lambda \neq 1 \Longrightarrow |\lambda|< 1$ and the geometric and algebraic multiplicity of the eigenvalue $1$ are the same.

I'm sure this is trivial, but I can't see it! Thanks.

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Hint: try to define an appropriate norm on the matrix so that it is 1 for every doubly stochastic matrix and use the usual equation Av=pv.Now apply norm on this eqn and derive the fact that |p|<1 or you can use the spectral radius formula to directly get the result.1 is the largest eigenvalue so by perron-frobenius theorem,every other eigenvalue has absolute value strictly less than 1

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Got it, thanks. Does that also imply in someway that the multiplicities of $1$ are the same? Oops. Actually I didn't get it. I can only show that $|p| \leq 1$. Can't prove it's strictly smaller than 1. – Linna Jan 3 '13 at 2:21
$\lambda=-1$ is a possibility. – Chris Godsil Jan 3 '13 at 2:40
In order to use the Perron-Frobenius theorem I need $M$ to be irreducible which doesn't necessarily happen, I believe. – Linna Jan 3 '13 at 2:41
perron frobenius holds for positive matrices.. check wiki – Koushik Jan 3 '13 at 2:44
Ok, but why can't $\lambda$ be $-1$ or any root of $1$ for that matter? – Linna Jan 3 '13 at 2:50

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