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Given a subalgebra E of $M_n$ (nxn complex valued matrices), what can we say about the subspaces F of $M_n$ such that $EF \subset F$? Googling for an answer gives me the reference:

Israel Gohberg, Peter Lancaster, and Leiba Rodman (2006). Invariant Subspaces of Matrices with Applications.

However, my library doesn't have this book. Is there a nice survey article available anywhere on this?

Thanks

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  • $\begingroup$ I think you are just asking for left E-submodules of V^n where V is the natural left module of E, V=C^n. If modules over finite dimensional algebras are not familiar, then I can recommend some textbooks. If modules are familiar, it might help to explain how this description is not helpful enough. $\endgroup$
    – Jack Schmidt
    Commented Jun 23, 2010 at 1:15
  • $\begingroup$ The book Phil mentions does indeed seem very close to Jack's comment. But I'm not actually sure that this is what Phil wants. I suppose you could split up a matrix in F into it's columns, and then look at invariant subspaces for each column, but that's not really the same. My guess is that this is quite a specialised question (as opposed to looking at E acting on C^n). Phil: do you have a more specific question?? $\endgroup$
    – Matthew Daws
    Commented Jun 23, 2010 at 6:52
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    $\begingroup$ I agree with Jack, Matthew, and Bruce: you should explain the context. For example, there is a sharp description if the subalgebra $E$ is semisimple. $\endgroup$
    – Victor Protsak
    Commented Jun 23, 2010 at 7:45
  • $\begingroup$ Matthew: The subspaces F in his problem are exactly the E-submodules of Mn(C). However, the E-module Mn(C) is isomorphic to the E-module V^n. Submodules of V^n can be quite a big more complicated than just "coordinate submodules" that work entry-by-entry in V or column-by-column in Mn(C). Submodules of the natural module and its direct powers are often studied, so this might help. In particular, the collection of all such F is a modular lattice since it is just a submodule lattice. However, submodules of E^n are very complicated over some fields, so maybe he needs something else. $\endgroup$
    – Jack Schmidt
    Commented Jun 23, 2010 at 13:10
  • $\begingroup$ Sorry about the vague question. Matt's correct that I was interested in the case that E was a sub $C^*$ algebra of $M_n$ and that the book I mentioned wasn't what I was after at all! $\endgroup$
    – Phil Ellison
    Commented Jun 23, 2010 at 16:22

2 Answers 2

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This was originally tagged fa.functional analysis, I think. So he's an Operator Algebraic answer. I'm going to make the strong assumption that E is self-adjoint (i.e. closed under taking the hermitian transpose). If not, then really this is an algebraic question, and it's probably irrelevant that you are working with the complex numbers...

Anyway, then E is a finite-dimensional von Neumann algebra. The action of E on M_n is the same as identifying M_n with $\mathbb C^n \otimes \mathbb C^n = \ell^2_n \otimes \ell^2_n$ and letting E act as $E \otimes 1$. Then invariant subspaces for $E$ correspond to orthongonal projections in the commutant of E, which by Tomita is $E' \otimes M_n$ where $E' = \{ A\in M_n : AB=BA (B\in E)\}$ the commutant of $E$ in $M_n$. We identify $E'\otimes M_n$ with $M_n(E')$, and then it's just (ahem!) a case of working out the projections (self-adjoint idempotents) here. In concrete cases, this is probably not too hard...

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    $\begingroup$ Thanks. This is the sort of answer I was after. My question arose from a conversation with Naz... I'm sure he'll try and find you at some stage to learn more! $\endgroup$
    – Phil Ellison
    Commented Jun 23, 2010 at 16:23
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You can say (without fear of contradiction) that they form a modular lattice. I learnt this from "Algebra" (Second Edition) by MacLane and Birkoff (which is showing my age) in XIV 5. There may be other things you can say. You would get a better response if you provided some context to your question.

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