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Homework Help: Vector Space

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Let U and V be vector spaces of dimensions of n and m over K and let Hom(subscriptK)(U,V) be the vector space over K of all linear maps from U to V. Find the dimension and describe a basis of Hom(subscriptK)(U,V). (You may find it helpful to use the correspondence with mxn matrices over K)


    2. Relevant equations



    3. The attempt at a solution
    is Hom(subscriptK)(U,V) a matrix that maps U to V?
    I don't get what Hom(subscriptK)(U,V) is...
     
  2. jcsd
  3. Feb 17, 2010 #2

    Dick

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    Hom_K(U,V) isn't 'a' matrix mapping U to V. It's the set of ALL matrices mapping U to V. K indicates the field the entries in the matrices are from (i.e. real, complex, etc, etc).
     
  4. Feb 17, 2010 #3
    How does one describe the basis in this case?
     
  5. Feb 17, 2010 #4

    Dick

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    Pick a basis {u1...un} for U and a basis {v1...vm} for V and think about a how to make a simple set of independent linear transformations whose span is all linear transformations.
     
  6. Feb 17, 2010 #5
    is the dimension of Hom_K(U,V) = mxn?
     
  7. Feb 17, 2010 #6

    Dick

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    Yes. Can you show that by producing a basis?
     
  8. Feb 18, 2010 #7
    ^ no... can you please show me how?
     
    Last edited: Feb 18, 2010
  9. Feb 18, 2010 #8

    Dick

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    Describe SOME linear transformation from U->V in terms of the basis. ANY one.
     
  10. Feb 19, 2010 #9
    I am trying to prove this without using matrix form. I am using the transformation:
    [tex]
    $ T_{ij} $ ( i= 1, \ldots, n and j= 1,\ldots,m) as the linear transformation that does the following: \\
    $ u_i \mapsto v_j \quad \text{ and } \quad u_{k \neq i} \mapsto 0 $
    [/tex]
    However I do not know where to start in proving these transformations are linearly independent. I am used to dealing with vectors and doing this kind of thing with a vector space of linear transformations is throwing me off.
     
    Last edited: Feb 19, 2010
  11. Feb 19, 2010 #10

    Dick

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    Ok, so linear independence means that sum over all i and j of a_ij*T_ij=0 implies ALL a_ij=0, right? Suppose a_KL is not zero. Put u_K into the transformation sum(a_ij*T_ij). That turns it sum(a_ij*T_ij(u_K)). What does that look like if you simplify to a single sum over j?
     
  12. Feb 19, 2010 #11
    Excellent, I understand much better how to work with linear combinations of these transformations. I get it now. Thanks
     
  13. Feb 19, 2010 #12
    Oh one last question, if I wanted to use einstein summation notation here could I just leave off both sum symbols?
    ie. is
    [tex]
    \alpha_{ij}T_{ij} = 0
    [/tex]
    the same as
    [tex]
    \sum_{i=1}^{n} \sum_{j=1}^{m} \alpha_{ij}T_{ij} = 0
    [/tex]
    using einstein summation notation?
     
  14. Feb 19, 2010 #13

    Dick

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    Right. That's actually what I was writing. I put the 'sum' in to make the sums were understood. Be sure you say you are using the Einstein summation convention, though. That's not automatically understood.
     
  15. Feb 19, 2010 #14
    cool, thanks again!
     
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