# What does algebraic multiplicity mean for compact operators

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?"

• First, the definition of algebraic multiplicity is the same as in the finite-dimensional case. Second, you should probably consider the Jordan block decomposition to understand the above definition. Commented Nov 25, 2019 at 5:54
• Well, any linear function function between finite dimensional spaces is compact so $Id_{\Bbb R^2}$ is compact. The definition I have for the finite dimensional case is the multiplicity of $\gamma$ as root of the caracteristic polynomial. In the infinite dimensional case, we don't have caracteristic polynomial so I don't understand how to use the same definition Commented Nov 25, 2019 at 5:59
• I would guess that he really means the geometric multiplicity :) At least the statement of the theorem would make more sense then. Note that the statement is already wrong in the finite dimensional case. Commented Nov 25, 2019 at 6:28
• See here: tqft.net/web/teaching/2013/Analysis3/Assignments/… Page 7: Apparently one can define $\det (I + z A)$ for a trace class operator A, and this can be used to define what algebraic multiplicity means. Commented Nov 25, 2019 at 7:18
• I believe that is a consequence of the theorem, not a definition. Could work as a definition anyway, but tensor products scares me so I was trying to avoid them. I was reading "Peter Lax - Functional Analyisis" proof but he omits many details (and doesn't talk about algebraic multiplicity, don't know why). Thanks for the Help! Commented Nov 25, 2019 at 7:46

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