Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two linear transformations agree, subspace

  1. Nov 14, 2013 #1
    I've been up way too long, so pardon me if this doesn't make sense, but..

    Let V and W be vector spaces.
    Let T and U be linear transformations from V to W.

    Consider the set of all x in V such that T(x) = U(x)

    1.) I think that this is a subspace of V.
    2.) Can I say anything about its dimension?

    The dimension is at most V, when the two transformations are the same. What is the minimum dimension in terms of dimensions for V, W, null spaces, or ranges?

    Is it AT LEAST the smaller of the two nullities?
     
  2. jcsd
  3. Nov 14, 2013 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are lots of examples for T and U where the only element they map equally is 0. If U = aT (a not = 1) for example.
     
  4. Nov 14, 2013 #3
    Surely..

    But if T and U only map from 0 equally, then the nullspace of either T or U is just {0}.

    That's why I think the minimum dimension of this space can be related to the nullities of T and U.
     
  5. Nov 14, 2013 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's only if the nullities of T & U exceed the dimension of V that they must have more than 0 in common! And, then you could have anything from NullT + NullU - DimV to min{NullT, Null U}
     
  6. Nov 14, 2013 #5
    OK, I understand. But how could we have nullities that exceed the dimension of V? Isn't the nullspace a subspace of V?

    EDIT or do you mean the sum of the two nullities?
     
  7. Nov 14, 2013 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes the sum of the nullities. And, what goes on outside the null spaces is equally independent. They may have no null vectors in common (except 0) but be equivalent outside of NullT U NullU
     
  8. Nov 14, 2013 #7
    I think I am confusing myself. Could you maybe see something wrong with the following?

    Let T and R be arbitrary linear transformations.

    rank(T+R) <= rank(T) + rank(R)

    (dim-null)(T+R) <= (dim-null)(T) + (dim-null)(R)
    null(T+R) >= null(T) + null(R)
    null(T+R) >= max{null(T), null(R)}

    Let (T+R)(x0) = 0
    Then T(x0) + R(x0) = 0
    So T(x0) = -R(x0)
    So nullspace(T+R) = {k | T(k) = -R(k)}

    Thus

    dim{k | T(k) = -R(k)} >= max{null(T), null(U)}

    Now let U(x) = -R(x)

    dim{k | T(k) = U(k)} >= max{null(T), null(-U)}

    And, null(-U) = null(U)

    dim{k | T(k) = U(k)} >= max{null(T), null(U)}


    Thanks again.
     
  9. Nov 14, 2013 #8
    You can think of your subspace as the kernel of the linear map [itex]T-U[/itex].
     
  10. Nov 14, 2013 #9
    Whoops. I see you were already basically doing that. My bad.
     
  11. Nov 14, 2013 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    T & U could each be null on half the basis vectors (different halves), so be different on every vector except 0.

    There is no relationship between what the LT's do outside their null spaces and how big their null spaces are.
     
  12. Nov 14, 2013 #11
    I greatly appreciate the insight, PeroK. I will think about these things.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Two linear transformations agree, subspace
Loading...