This question was posted a long time ago on the mathexchange, but I didn't get any answers there, and despite having discussed it with some colleagues, I don't think I have a definitive answer.

I am wondering about simple eigenvalue definitions in the context of ergodic theory. There seem to be two accepted definitions for simple eigenvalues. The definitions involve algebraic multiplicity and geometric multiplicity.

Let $E$ be a Banach space (possible infinite) and $A:E\to E$ a linear operator, then the eigenvalue $\lambda$ is simple if

  1. The dimension of (generalized eigenspace) $\mathcal{N}_{\lambda}=\cup_{k\in\mathbb{N}} \mathcal{N}((\lambda I-A)^k)$ is $1$: algebraic multiplicity $m_a(\lambda)=1$; or

  2. The dimension of (eigenspace) $\mathcal{N}(\lambda I - A)$ is $1$: geometric multiplicity $m_g(\lambda)=1$.

Note that the definitions are not equivalent as we have $m_a \ge m_g$. We have that, if an eigenvalue is simple in definition 1, it will be simple in definition 2, but the opposite is not true.

More specifically, I am interested in the difference when we characterize ergodicity. A system is ergodic iff 1 is a simple eigenvalue of the Koopman operator. See question 4 or this post. Another moment where the simplicity of the eigenvalue appears is in the Perron-Frobenius theorem (finite version).

For example, in Baladi's book Positive Transfer Operators and Decay of correlation, a simple eigenvalue is given by definition 1. But in the book Ergodic Theory by Petersen or Ergodic Theory by Walters, they seem to use definition 2, more precisely, in the demonstrations that use the fact of being simple, they use that eigenspace has dimension 1, and as mentioned these definitions are not equivalent. But in the case of definition 1, if it is shown that the eigenspace has dimension 1, it is not guaranteed to be simple, since the generalized eigenspace has a greater or equal dimension.

In this context of ergodic theory (mainly, speaking of the Koopman operator), I was told that the definitions are equivalent, but I was not convinced by the explanation.

$\mathcal{N}(X)$ is the kernel or nullpace of $X$.


1 Answer 1


If $\lambda$ has modulus $1$, then both definitions are equivalent for power-bounded operators, i.e., for operators $A$ that satisfy $\sup_{n = 0,1,2,\dots} \|A^n\| < \infty$.

Indeed, if $\lvert \lambda \rvert = 1$ and if there exists a vector $x$ that is in the kernel of $(\lambda-A)^2$ but not in the kernel of $\lambda-A$, then a brief computation shows that the sequence $(A^n x)$ is unbounded.

The transfer operator and the Koopman operator that belong to a measure-preserving system on a probability space are always power-bounded (in fact, they have norm $1$) on $L^p$ for every $p \in [1,\infty]$. It seems that this explains why there is no inconsistency in the ergodic theoretic situation that you mentioned.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.