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Adjoint of a linear operator

  1. Jan 7, 2008 #1
    Let T:V -> V be a linear operator on a finite-dimensional inner product space V.
    Prove that rank(T) = rank(T*).

    So far I've proven that rank (T*T) = rank(T) by showing that ker(T*T) = ker(T). But I can't think of how to go from there.
     
  2. jcsd
  3. Jan 8, 2008 #2
    Oh, should I use the fact that the number of linearly independent rows of a matrix A is equal to the number of linearly independent columns of A*, and that rank(T) = rank of matrix represenation of T wrt to a basis? Or is there a better way?

    rank(T)=rank([T])=rank([T]*)=rank([T*])=rank(T*)
     
    Last edited: Jan 8, 2008
  4. Jan 8, 2008 #3
    well, dimV=dimT+dimKerT=dimT*+dimkerT*
    but kerT=kerT*, one way to prove it is if
    v not zero in ker T, then T(v)=0, so 0=<Tv,v>=<v,T*v>
    but v isn't zero then T*(v)=0, you can do it also vice versa.
     
  5. Jan 8, 2008 #4
    But <v,T*v>=0 implies T*v=0 only if <v,T*v>=0 for ALL v in V, not just for v in kerT.
     
  6. Jan 8, 2008 #5

    morphism

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    imT* is the orthogonal complement of kerT. So dimV=rankT*+dimkerT=rankT+dimkerT.
     
  7. Jan 8, 2008 #6
    Wow! Morphism, do you look these proofs up somewhere, or do you figure it out completely from scratch? If the latter, then you must be a genius!
     
  8. Jan 8, 2008 #7

    morphism

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    It's just experience and not any form of genius - I already knew that fact, and it turned out to be helpful here.
     
  9. Oct 4, 2010 #8
    How is the adjoint of the adjoint of a linear operator, the linear operator itself? i.e. A**=A?
     
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