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L.a. proof

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data
    prove that dimV=dimU[tex]\bot[/tex]+dimU


    2. Relevant equations



    3. The attempt at a solution
    I've done this on paper, and set V=nullT+rangeT where T maps a vector from
    V to U. Is it safe to assume that nullT and U[tex]\bot[/tex] are the
    same? Reasoning is that <T(wi), T(uj)>=0 with wi in nullT and uj in U. Since T(wi)=0,
    wi gets mapped to 0 in U, and given that the dot product=0, wi is orthogonal to uj. Also consider mapping
    from say the xy plane to x. The y component is not in x, and it is at a right angle to all vectors in the
    range x itself, so it is orthogonal to x. Same for mapping from x y z to x y. All z components are
    orthogonal to all x and y components. This isn't the actual proof btw. If I'm wrong here, then I'll try a different
    approach.
     
    Last edited: Jul 5, 2009
  2. jcsd
  3. Jul 5, 2009 #2

    HallsofIvy

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    What T d you want to use?

    I think better is this: select a basis [itex]\left{ u_1, u_2, \cdot\cdot\cdot, \u_n\right}[/itex] for U. Let [itex]\left{v_1, v_2, \cdot\cdot\cdot, \v_n\right}[/itex] be a basis for [itex]U^\perp[/itex]. Now show that the union of the two sets is a basis for V.
     
    Last edited: Jul 5, 2009
  4. Jul 5, 2009 #3
    That's a much better approach! T maps v in V from V to U. dimV=n, dimU=m
    m<n (in this case). v=a1v1+...+anvn. Tv=a1v1+...+amvm. Just wondering whether
    or not nullT=U[tex]\bot[/tex]. I can't prove anything without knowing what
    I'm talking about. Good work, Halls!
     
  5. Jul 5, 2009 #4

    Dick

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    That's fine as far as it goes. Sure null(T) is U^perp as you've defined it. But Halls also said you should prove the span of U and U^perp is V. You didn't do that. Do you see why that's necessary?
     
  6. Jul 6, 2009 #5

    HallsofIvy

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    I can't answer your question because you still haven't said what T is! All you have said is that "T maps v in V from V to U". There are many linear transformations that do that. What is T?
     
  7. Jul 6, 2009 #6
    basis for V is {v1....vn} dimV=n dimU=m m<=n.T is the transformation that takes v (v=a1v1+...+anvn) and sends it to U so Tv=a1v1+...+amvm.
     
  8. Jul 7, 2009 #7
    Ok let dimV=n, dim U=m m<=n, Let V=nullT+rangeT, in that dimV=dimnullT+dimrangeT. Let T be the transformation that takes v in V and maps it to U so that Tv=a1v1+..+amvm=a1u1+...+amum. Now split the basis for V into {w1...wk} for nullT and {u1...um} for rangeT=U. Since wi is in nullT, T(wi)=0, which shows that wi is orthogonal to uj (note: 1<=j<=m, 1<=i<=k). Since wi is orthogonal to uj, and wi is in nullT, nullT=Uperp and since dimV=dimnullT+dimrangeT, dimnullT=dimUperp. And since rangeT=U, dimrangeT=dimU. Therefore dimV=dimUperp+dimU.

    I figured that since we already know that dimV=dimnullT+dimrangeT (since we know that it is true with any transformation from
    any space to any other space: dim spaceX=dimnullT+dimrangeT for any
    T mapping from X--->Y and Y is always the rangeT) we can find a way to show that nullT=Uperp. Now that we have shown that nullT=Uperp, its obvious that
    their dimensions equate. And since rangeT=U, their dimensions equate. Therefore, dimV=dimnullT+dimrangeT=dimUperp+dimU
     
    Last edited: Jul 7, 2009
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