Im trying to understand Lidskii Theorem which states the following.
If $H$ is a separable Hilbert Space, $T:H \rightarrow H$ a (compact) trace class operator and $\{\gamma _n\}_{n\in \Bbb N}$ are the eigenvalues of $T$, each repeted as many times as its algebraic multiplicity, then
$$Trace(T)= \sum_{n=1}^\infty \gamma_n$$
And I couldn't find anywhere the definiton of algebraic multiplicity of an eigenvalue in the infinite dimensional case. I started believing it has something to do with the following statement. Let $T:H \rightarrow H$
be a compact operator, for every eigenvalue $\gamma$ there exist a $m\in \Bbb N_0$ such that
$$\{0 \} \subsetneq Ker(T-\gamma.Id)\subsetneq Ker(T-\gamma.Id)^2 \subsetneq \dots\subsetneq Ker(T-\gamma.Id)^m=Ker(T-\gamma.Id)^{m+1}=Ker(T-\gamma.Id)^{m+2}=\dots$$
I thought $m$ was the algebraic multiplicity of $\gamma$ but then realiced this does not match with the definition in the finite dimensional case, for example, taking $T=Id_{\Bbb R^2}$ and $\gamma=1$. So my question, again, is "what does algebraic multiplicity means?"
Best Answer
The algebraic multiplicity of an eigenvalue $\gamma$ of $T\in\mathcal L(X)$, where $X$ is a Hilbert space, equals $\dim \cup_{k=1}^\infty \ker(T-\gamma I)^k$, which in the case of a compact operator equals $\dim \ker(T-\gamma I)^m$ for the $m$ in your question.
The latter holds because 1. clearly always $\ker(T-\gamma I)^m\subset \ker(T-\gamma I)^{m+1}$, and 2. for compact operators there exists a maximal $m$ such that the "$\subset$" is "$\subsetneq$".
Corresponding vectors are called generalized eigenvectors (of order $k$, where $k$ is minimal).
References
This and the fact on compact operators is stated on p. 24, section 1.4.2 of Finite Element Methods for Eigenvalue Problems, by Jiguang Sun, Aihui Zhou, CRC Press, 19.8.2016 - 343 pp. (By p. 18, X is a Hilbert space.) Link to page 24: https://books.google.fi/books?id=YC7FDAAAQBAJ&pg=PA24
Theory and Applications of Volterra Operators in Hilbert Space, by Israel Gohberg, M. G. Krein, p. 49, footnote 30 states the same for $\gamma\ne0$: https://books.google.fi/books?id=HUkR9eQhKLYC&pg=PA49
Note: $(T-\gamma I)^m = T^m -\gamma T^{m-1} + \cdots +(-\gamma)^m I=T'+(-\gamma)^m I$, where $T'$ is compact. Therefore, the algebraic multiplicity of a nonzero eigenvalue of a compact operator is always finite (use Rudin F.A., Theorem 4.25a). (That of $0$ may be infinite; take $T=0$.)
BTW, "analytic multiplicity" may be different from the algebraic and geometric ones. "Geometric multiplicity" is $\dim \ker(T-\gamma I)$, hence at most the algebraic one.
Edit: much the same is said here, including a proof for the existence of a maximal $m$ (still assuming a Hilbert space; I haven't checked if it is necessary): https://math.stackexchange.com/a/406371