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Invariant subspaces

  1. Nov 11, 2004 #1
    Does every linear operator have a nontrivial invariant subspace? My professor mentioned this question in class, but never actually answered it. I am curious if this is true or not and why.
     
  2. jcsd
  3. Nov 11, 2004 #2

    AKG

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    Consider the operator T : [itex]\mathbb{R}^2 \to \mathbb{R}^2[/itex] that rotates any vector about the origin by [itex]\pi /2[/itex]. If [itex]W \subset \mathbb{R}^2[/itex] is non-trivial, it contains some [itex]x \in \mathbb{R}^2,\ x \neq 0[/itex]. Let [itex]x = (x_1, x_2)[/itex]. Now, W must have also:

    [tex]T(x) = (-x_2, x_1),\ T^2(x) = (-x_1, -x_2),\ T^3(x) = (x_2, -x_1)[/tex]

    Since a subspace must be closed under scalar multiplication, it must contain the two perpendicular lines passing through the origin where the first line passes through [itex](x_1, x_2)[/itex] and the second passes through [itex](-x_2, x_1)[/itex]. The simplest case of this would be the x- and y-axes. However, since W must also be closed under addition, you can clearly see that it must contain all of [itex]\mathbb{R}^2[/itex], which is indeed trivial (I would hope that's what you mean by trivial).

    EDIT: If it wasn't clear, the conclusion is that for every operator there need not exist a non-trivial subspace invariant with respect to that operator.
     
  4. Nov 12, 2004 #3
    Okay I will buy that if you work with R^n, but what if you start going into the complex number land as your vector space? Then T(a, ia)=-(ia,a)=-i(a,ia). I reread my notes and my professor changed the original question to an the equivalent question of Given T an element of L(V); where dim(V)>=1 Does there exist a nonzero vector v and element of V such that T(v)=(lambda)v for lambda an element of the field which the vector space is over. If you use C then the example you gave works.
     
  5. Nov 12, 2004 #4

    matt grime

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    If the field is algebraically closed then the characteristic polynomial will have a root (which is why the first example doesn't have invariant proper subspaces) and will have an eigenvalue, and there is an eigenvector associated to it.
     
  6. Nov 12, 2004 #5

    AKG

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    Sounds like first year algebra. One of the most important things you'll be dealing with in this algebra course, and probably the next (and you'll see it in a number of other places as well) are eigenvalues and eigenvectors. Given some operator T, we call [itex]\lambda[/itex] and eigenvalue of T if there exists some vector [itex]v \in V[/itex] such that [itex]T(v) = \lambda v[/itex], and [itex]v[/itex] is called the eigenvector of T corresponding to [itex]\lambda[/itex]. If T has an eigenvalue, then it has an eigenvector, and if we let that vector be v, then clearly Span{v} is a T-invariant subspace (you can easily check this for yourself).

    How do you find eigenvalues? The eigenvalues are exactly the roots of the characteristic polynomial of t, which we can denote g(t), where:

    [tex]g(t) = \det (T - tI)[/tex]

    Where I is the identity operator. In order to compute this, express T as a matrix with respect to any basis, and then you can easily see how this is done. This will give you a polynomial in the parameter t. Then, you find the roots:

    [tex]g(t) = 0[/tex]

    The solutions, t, are the eigenvalues. You should know how to find the matrix representation of basic transformations in [itex]\mathbb{R}^2[/itex], like the rotation by [itex]\pi /2[/itex]. At any rate, if you don't know it, it is:
    Code (Text):
    [ 0 -1 ]
    [ 1  0 ]
    Compute the characteristic polynomial, and you'll see it has no real roots, thus no eigenvalues, and so the subspaces in the form Span{v}, where v is an eigenvector, called the eigenspaces corresponding to [itex]\lambda[/itex] don't exist. I'm not sure whether it's necessary for an operator to have eigenvalues in order for it to have an invariant subspace, but it is sufficient (although matt grime's post suggests that it is necessary).

    Now, by the fundamental theorem of algebra, an n-degree polynomial over the complex field will have n (not-necessarily-distinct) roots, and so it certainly has eigenspaces, which, of course, are invariant.
     
  7. Nov 13, 2004 #6

    matt grime

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    It certainly isn't necessary for a matrix to have eigenvalues in order to have an invariant subspace, and I'd hope I didn't suggest that - if the field is algebraically closed then it will necessarily have an invariant line, since it will necessarily have an eigenvalue.

    Consider any rotation T on R^2 without eigenvalues (over R), and consider the operator

    [tex]T\oplus T : \mathbb{R}^2\oplus\mathbb{R}^2 \to\mathbb{R}^2\oplus\mathbb{R}^2[/tex]

    Then it has two invariant and complementary subspaces, but no eigenvalues (over R).
     
  8. Nov 28, 2004 #7

    mathwonk

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    I suspect his teacher referred to the famous open problem: the "invariant subspace problem", posed over infinite dimensional complex Hilbert spaces where determinants are unavailable. The analogous problem over a Banach, non Hilbert space was solved (in the negative) in 1984 by Read, in the Bulletin of the London Math Society. I.e, there does exist a bounded linear operator on a Banach space B with no invariant subspace except {0} and B.

    (A Banach space is a linear space with a concept of length, and in which Cauchy's convergence criterioon is satisifed. A Hilbert space is a Banach space in which the length arises from a dot product, so that angles and hence perpendicularity, also make sense.
     
    Last edited: Nov 28, 2004
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