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Direct Sum

  1. Apr 12, 2006 #1
    "If V is finite dimensional, and W is a subspace of V, prove that if [tex]T(W)\subset W[/tex], there's always a W' such that W' (direct sum) W = V and [tex]T(W')\subset W'[/tex]."

    I can't find such a W'.
     
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  3. Apr 12, 2006 #2

    Hurkyl

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    The first thing that springs to my mind is to look at eigenvectors. It won't cover the general case, but maybe it will suggest an idea?
     
  4. Apr 12, 2006 #3
    Acutally it's "prove or disprove". Unfortunately I can't do either.
     
  5. Apr 13, 2006 #4

    AKG

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    Am I missing something? If W = span{v1, ..., vk}, then we can write V = span{v1, ..., vk, vk+1, ..., vn}. The only W' that will satisfy [itex]V = W \oplus W'[/itex] is W' = span{vk+1, ..., vn}, is it not? If so, then it should be easy to think of a counter example, i.e. one such that T(W') is not contained in W'.

    Let V = R4 = span{e1, ..., e4}. Let T(a,b,c,d) = (b,c,d,0). Let W = span{e1, e2}. T(W) = span{e1}, so T(W) is contained in W. Now if W' is to be a subspace such that its direct sum with W is V, then W' would have to be span{e3, e4}, would it not? But then T(W') = span{e2, e3} which is not contained in W'.
     
  6. Apr 13, 2006 #5

    HallsofIvy

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    AKG, you'd better state explicitely that you are assuming that all of the vectors v1, ..., vk, vk+1,... vn are independent and further that all of the v1,... vk are orthogonal to all of the vk+1, ... , vn.
     
  7. Apr 13, 2006 #6

    matt grime

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    AKG: let V=R^2, and v_1 be a unit direction in the x direction. Now, you're honestly telling me that you think there is exactly one complementary subspace? That there is only one other line through the origin in R^2? I don't think you do think that.
     
  8. Apr 13, 2006 #7

    AKG

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    Sorry guys, my mistake.
     
  9. Apr 14, 2006 #8
    I think I got it. Suppose T^2 = 0 but T != 0, let W be the nullspace of T. Then any other subspaces that intersect with W trivially that is also T-invariant must be the 0 space. Since null T != V, no such W' in question exists.
     
  10. Apr 14, 2006 #9

    matt grime

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    Look, if you can split the space into two T invariant subspaces, then what is the minimal number of eigenvectors T must have?
     
    Last edited: Apr 14, 2006
  11. Apr 14, 2006 #10
    You can ALWAYS split the space into 2 T-invariant subspaces. The point of the question is that if one of them is given to you, can you always find another that complements it? I constructed the counterexample such that 'no'.
     
  12. Apr 14, 2006 #11

    matt grime

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    Firstly, unless you have some odd notion of what 'split' must mean then that is false (to me it would most naturally mean 'write as a dircet sum of two proper subspaces'), and secondly I'd've thought it bloody obvious that the when I refer to a 'splitting' it to is the decomposition as given in your original post.

    Theorem: If T is a linear map from V to V and V is the non-trivial direct sum of two T invariant proper subspaces then T must have at least two eigenvectors.

    Corollary: there is a counter example to your original hypothesis.
     
  13. Apr 14, 2006 #12
    It isn't more natural for "split" to mean the decomposition of the direct sum of two PROPER subspaces, but that's irrelevant. In the original quesiton, no assumptions that W or W' must be proper subspaces are made. It is entirely possible that W = V and W' = 0. Your theorem assumes that T has a priori proper, invariant subspaces. Does any of this have to do with my counterexample?
     
  14. Apr 14, 2006 #13

    matt grime

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    I fail to see your problem. There is a mistake in my argument, not one you've pointed out, and not a fatal one: I tacitly assumed an algebraically closed field. I fail to see why it is important whether all T have or have not got invariant subspaces, the point is if T has then it has extra properties, and since we can easily find T with invariant proper subspaces, and without these extra properties it must be a false conjecture.

    Let us correct it: suppose that T has eigenvalues in the base field (this is no loss), and that there is a decomposition of V into non-trivial invariant subspaces, then T has two eigenvectors.

    Since it is easy to find T with e-values in the base field and with exactly one e-vector (which is trivially an invariant subspace), the original conjecture is false for very general reasons, not just because of your argument, which is in some sense a special case when the e-values are zero.
     
    Last edited: Apr 14, 2006
  15. Apr 14, 2006 #14

    AKG

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    matt grime, are you still assuming the base field to be complex? Consider the rotation in R³ about the z-axis. The xy-plane and the z-axis are two proper, T-invariant subspaces whose direct sum is V. T has only one eigenvector, and its span is the z-axis (with e-value 1).
     
  16. Apr 14, 2006 #15

    matt grime

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    Ah, I see the issue here. God this is getting bloody complicated for a very trivial fact. If you think geometrically it is clear why the hypothesis can't be true.

    Let's restate it as "suppose all of the roots of the characteristic polynomial of T lie in the base field", or how about 'after tensoring with the algebraic closure of the basefield (extending scalars)'

    Wonder if eljose would like to read these unrigorous arguments of mine....

    The point it it is very easy to manufacture T with an invariant subspace that has no invariant complement. The simplest way to do this is to pick a T with exactly one eigenvector, and whose characteristic poly splits over the field. Any complementary subspace must possess another eigenvector which is impossible.

    Another way of thinking about it is that the hypothesis is stating something about the generalized eigenspaces of T needing to split into invariant sums.
     
    Last edited: Apr 14, 2006
  17. Apr 14, 2006 #16
    Alright then. Thanks for the help.
     
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