There seem to be two accepted definitions for simple eigenvalues. The definitions involve algebraic multiplicity and geometric multiplicity. When space has a finite dimension, the most used is algebraic multiplicity. As I am interested in the general case (where the dimension of space can be infinite) I will not speak of the characteristic polynomial.

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 $\mathcal{N}_{\lambda}=\cup_{k\in\mathbb{N}}\mathcal{N}((\lambda I-A)^k)$ is $1$ -- algebraic multiplicity $m_a(\lambda)=1$; or

  2. The dimension of $\mathcal{N}(\lambda I - A)$ 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.

As mentioned in this answer, in ergodic theory definition 2 is used. In this case, why is this definition used? Is there a range of useful operators they want to include using this definition?

More precisely, 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).

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 $X$.

  • $\begingroup$ "why is this the definition used"? well, as said in my question, you get clean statements like ergodicity being equivalent to $1$ being a simple eigenvalue of the unitary operator. $\endgroup$ Sep 13, 2021 at 23:47
  • $\begingroup$ @mathworker21 Okay, but I was wondering, how does a map "behave" as the unitary operator with 1 being a simple eigenvalue (in definition 1)? I mean, what kind of maps would we be "missing" using definition 1? $\endgroup$
    – Mrcrg
    Sep 14, 2021 at 0:06
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    $\begingroup$ Consider the two matrices $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$ and $\begin{pmatrix} \mu & \\ & \lambda \end{pmatrix}$ with $\lambda \neq \mu$. Both have a $\lambda$-eigenspace of dimension 1, but the first has a generalised $\lambda$-eigenspace of dimension 2. The nice thing about eigenspaces is that by just knowing the space and the scalar you can determine the action of the matrix; the nasty thing about generalised eigenspaces is that you need to know more: the operator may do something nontrivial on the generalised eigenspace. $\endgroup$
    – Joppy
    Sep 15, 2021 at 0:48


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