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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 $\lambda \neq 1 \implies |\lambda|< 1$ 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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  • $\begingroup$ 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. $\endgroup$ – Linna Jan 3 '13 at 2:21
  • $\begingroup$ $\lambda=-1$ is a possibility. $\endgroup$ – Chris Godsil Jan 3 '13 at 2:40
  • $\begingroup$ In order to use the Perron-Frobenius theorem I need $M$ to be irreducible which doesn't necessarily happen, I believe. $\endgroup$ – Linna Jan 3 '13 at 2:41
  • $\begingroup$ perron frobenius holds for positive matrices.. check wiki en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem $\endgroup$ – Koushik Jan 3 '13 at 2:44
  • $\begingroup$ Ok, but why can't $\lambda$ be $-1$ or any root of $1$ for that matter? $\endgroup$ – Linna Jan 3 '13 at 2:50
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This is a consequence of the following two facts:

  1. The Perron-Frobenius theorem for positive matrices.
  2. If $A\ge 0$ (entrywise), then $A$ has row-sums equal to one if and only if $Ae = e$, in which $e$ denotes the all-ones vector, i.e., the spectral radius of $A$ is one.
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