It is a fact that a measure-preserving system $(X,\mathcal{B},\mu,T)$ is ergodic if and only if $1$ is a simple eigenvalue for the associated unitary operator $U_T: L^2(\mu) \to L^2(\mu)$ given by $U_Tf = f\circ T$. I'm curious as to what a simple eigenvalue means here. I know $T$ is ergodic if and only if the only $T$-invariant $L^2$ functions are the constants. So it is clear to me that $T$ is ergodic if and only if the eigenspace associated with the eigenvalue $1$ has dimension $1$.
However, for finite-dimensional transformations, a simple eigenvalue is an eigenvalue $\lambda$ for which if we completely factor the characteristic polynomial, then the exponent of the term with $\lambda$ is $1$. Wikipedia seems to indicate that this is not necessarily the same as the dimension of the eigenspace corresponding to $\lambda$.
So what is a simple eigenvalue for a linear operator on infinite-dimensional space, and why does it coincide with the dimension for the eigenspace corresponding to the eigenvalue?
In ergodic theory it is common to define a simple eigenvalue as one for which the eigenspace has dimension $1$, although as you say this is not the usual definition in some areas. See for example the book Ergodic Theory by Peter Walters; it is not made precise anywhere, as far as I can tell, but that's what the proofs use, see for example the proof of Theorem $1.19$.
Actually, the common definition in dynamical systems is the same as in the particular case of ergodic theory. See for example page 87 of Dynamical Systems: An International Symposium (edited by L. Cesari, J. Hale and J. LaSalle).