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T-invariant subspaces

  1. Feb 13, 2007 #1
    Hello,

    can anyone tell me if I understand this right? :rolleyes:

    I have a t-invariant subspace with basis B, and I extend the basis B to be a basis B' for the entire vector space by adding L.I. vectors to it. Then I put B under a linear transformation, T:V --> V, and I will get a set of vectors in the range of T that generates W in that space, i.e. R(T) = span (T(B)). But since the subspace (let's call it v) of V is T-invariant, then the vectors I end up with in R(T) are also in the subspace (minus the ones we added to extend the basis). Is that correct?

    Also, can something only be T-invariant if it's a mapping within the same vector space? Can subspace v of V be t-invariant if the transformation is T: V --> W? I'm not sure if that makes sense, because the generating set in the R(T) couldn't be generating the same subset as in V.
     
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  3. Feb 14, 2007 #2

    HallsofIvy

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    This is remarkably hard to read!

    You have a vector space V and a linear transformation T:V-->V with invariant subspace what? W? "Then I put B under a linear transformation, T:V-->V" Is this the same T as for the "t-invarient subspace"?

    Given a vector space V, a linear transformation T, and a T-invariant subspace U, yes, it is true that any basis for W can be extended to a basis for V. If W is the image of V under T: W= R(T), then clearly V must be a subspace of W. is that what you are asking?

    U can be an invariant subspace for a linear transformation T:V-->W provided that U is in both V and W. For example, let V be all triples (x,y,0), W be all triples (0, y, z) and T(x,y,0)= (0, y, x). Then U= all triples of the form (0, y, 0) is a T- invariant subspace.
     
  4. Feb 14, 2007 #3
    Sorry, I suppose that was a mouthful... :yuck:

    I am still just confused about t-invariance, I guess. What about this is significant? Apparently it's going to be important later on (according to my course instructor), but I'm having trouble understanding what t-invariant means beyond the formal definition.
     
  5. Feb 14, 2007 #4

    matt grime

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    Let V be the vector space of differentiable functions from R to R and T be differentiation. Then P(x) the set of polys is an invariant subspace, E(x) the subspace spanned by the exponential function is an invariant subspace.

    Eigen spaces are invariant subspaces. Generalized eigenspaces are invariant subspaces. Subspaces preserved by T are good because T restricts to a linear map on the subspace.
     
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