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The invariant subspace problem states something along the following lines: Let $H$ be a Banach space of dimension greater than one. Does every bounded linear operator $T:H \to H$ have a non-trivial closed $T$-invariant subspace?

Now my question is: Why is this a big problem? What does the solution give us? It has been solved in most of the special cases which occur in practice.

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"In practice" many people study operators on Hilbert space, and the special case of infinite dimensional separable Hilbert space is the big open problem. It is known that there are examples of operators on Banach spaces without invariant subspaces, and that there are examples of Banach spaces on which every operator has an invariant subspace (see the link in Andrey Rekalo's answer). This is all nice, but doesn't seem to shed light on the Hilbert space case. –  Jonas Meyer Dec 19 '10 at 6:37
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One of the primary reasons for studying the invariant subspace problem is the hope that it will lead to useful structural theories. For example, in their interesting expository paper $\ $ The invariant subspace problem, $\ $ Radjavi and Rosenthal wrote:

It is hoped that knowledge of the invariant subspaces of operators will shed light on their structure. In the case of operators on finite-dimensional spaces, for example, the Jordan canonical form theorem shows that operators are direct sums of particularly well-behaved operators on certain invariant subspaces. Also, normal operators on infinite-dimensional spaces can be represented as integrals with respect to some invariant subspaces, (their spectral subspaces - this is the spectral theorem). To prove that every operator has a non-trivial invariant subspace might be the beginning of a general structure theory. $\ \:$ On the other hand, $\ $ a concrete counterexample would also be very interesting. To say that $A$ has no non-trivial invariant subspace is equivalent to saying that linear combinations of $\ {A^n\:f\:: n \in \mathbb N}\ $ are dense in the space for each nonzero vector $f$. Representing such an $A$ on various spaces would give a number of approximation theorems.

This paper, combined with Yadav's $\ $ The Invariant Subspace Problem $\ $provide nice expositions on the problem and its rich history. Neither of these introductions are mentioned on said MO page. IMO they provide a much nicer overview of the problem than the little that is said there.

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A similar question was asked on MathOverflow a week ago. Take a look at the enlightening answer by Bill Johnson and the discussion in the comments here.

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Interesting, thank you. –  Jonas Teuwen Dec 18 '10 at 13:59
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The invariant subspace problem asks whether every bounded linear operator has a non-trivial invariant subspace. The existence of some operator with invariant subspaces is trivial. (Consider finite-rank operators.)

And it is known that the theorem is not true for all Banach spaces so the current formulation asks whether it is true for Hilbert spaces. I do not have any good references however.

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Oops, yes, of course for every. So I want to know why we would like to find the solution to this at is has been solved for a lot of operators and there has been given a counterexample for some Banach spaces. –  Jonas Teuwen Dec 18 '10 at 13:27
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