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

Dim and null question

  1. Feb 13, 2008 #1
    Ok. i have a problem that states: "suppose U and V are finite-dimensional vector spaces and that S [tex]\in[/tex] L(V,W), T [tex]\in[/tex] L(U,V). prove that
    dim(null(ST))[tex]\leq[/tex] dim(null(S)) + dim(null(T)).

    i think that some call the null space the kernel.

    i have tried using the fact that null(S) is a subspace of V, null(T) is a subspace of U, and null(ST) is a subspace of U.
    also, that:
    dim(null(T)) [tex]\geq[/tex] dim(U) - dim(V)
    dim(null(S)) [tex]\geq[/tex] dim(V) - dim(W)
    dim(null(ST)) [tex]\geq[/tex] dim(U) - dim(W)

    the closest that i get is that dim(null(ST))= dim(null(S)) + dim(null(T)).

    Can anyone offer any suggestions?
     
  2. jcsd
  3. Feb 13, 2008 #2
    It's been a while since I took a linear algebra course but maybe it would be helpful to assign a variable call it N to the total number of dimensions.

    BTW what does dim(nul(ST)) mean?
     
  4. Feb 13, 2008 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I hope you didn't get that, because it's patently false.
    For example, let U,V,W = R^2, and let S,T both be the matrix
    [1, 0]
    [0, 0]

    You've thought about dimension and you've thought about nullity... is there any related quantity you haven't thought about?
     
  5. Feb 13, 2008 #4
    So I still want to know what ST is. Is it the inner product, the Cartesian product or is it a matrix product?
     
  6. Feb 13, 2008 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Composition of functions.
     
  7. Feb 13, 2008 #6
    Why should that be the case? I don't see the words composition of functions anywhere in the definition of a vector space.
     
  8. Feb 14, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You also won't see "linear transformation" anywhere in the definition of vector space!

    After you have vector spaces, you can define linear functions on them and the composition is a natural operation on those functions.

    The original post specifically that T was a linear transformation from U to V and that S was a linear transformation from V to W. The composition, ST, is a linear transformation from U to W.

    If S and T are represented as matrices, in some appropriate bases for U, V, and W, then ST would correspond to the matric multiplication.
     
  9. Feb 14, 2008 #8
    yes, sorry if that was unclear. ST is meant to be the composition of the functions.
     
  10. Feb 14, 2008 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Suppose v is in the null space of T. Then Tv= 0 so STv= 0 and v is in the null space of ST. That shows that the null space of T is a subspace of the null space of ST. Now suppose v is in U but NOT in the null space of T. Then Tv is a non-zero vector in V and, in order that STv=0, must be in the null space of T. That is, v is in T-1(null space of V). That set cannot have dimension larger than dim(null space of S). Putting those together, the dimension of the null space of TS cannot be larger than the dim(null space of S)+ dim(null space of T).
     
  11. Feb 14, 2008 #10

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    the null space of S is the space of stuff going to zero, so the null space of ST is the stuff S sends into the null space of T.

    so we need to estimate the size of the pullback K of null(T) under S. this follows from the basic formula relating dimensions of nullspace and image.

    i.e. the image of K is inside null(T), so that formula says the dimension of K is at most the sum of dim null(T) and dim null(S).


    (recall the fundamental formula says that dim im(U) + dim null(U) = dim(dom(U)).

    apply this to S acting on K as domain, and then the image is contained in null(T).))
     
  12. Feb 14, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Mathwonk, I was with you until you swapped T for U:
    "(recall the fundamental formula says that dim im(U) + dim null(U) = dim(dom(U))."

    In the original question, U was a vector space, not a linear transformation!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Dim and null question
  1. Null space (Replies: 8)

  2. The null space (Replies: 3)

  3. Null vector (Replies: 3)

Loading...