# Spectrum of doubly stochastic matrices

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.

## 2 Answers

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

• 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 en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem – 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

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.