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Linear operator problems

  1. Aug 10, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    1. Let T be a linear operator on an inner product space V. Let U = TT*. Prove that U = TU*.

    2. For a linear operator T on an inner product space V, prove that T*T = T0implies T = T0.


    2. Relevant equations



    3. The attempt at a solution
    1. This appeared at the outset to be a relatively simple one, but for some reason I can't prove it:

    Let [tex]v,x \in V[/tex].
    I know that U* = U, and
    <v,Ux> = <v,TT*x>
    = <T*v,T*x>

    <v,TU*x> = <T*v,U*x>
    =<T*v,TT*x>

    So there's always an extra T I can't get rid of. How do I prove this?

    2. I'll try to prove <v,T*Tx> = <v,T0x>, knowing that (T*T)* = T*T
    But as before this leads nowhere because there is again an extra T I can't get rid off:
    <v,T*Tx> = <Tv,Tx> = <T0v,x>

    <v,Tx> (This one only has one T on the inner product).
     
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  3. Aug 10, 2008 #2

    Dick

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    It may appear simple at the outset, but you can't prove it, because it isn't even true. Let T=[[0,1],[1,0]]. T*=T. TT*=I=U. U*=U=I. U is not equal to TU*. Is there something about these linear operators you aren't telling us about?
     
  4. Aug 11, 2008 #3

    morphism

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    Also, what is T_0, the zero operator?
     
  5. Aug 11, 2008 #4

    Defennder

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    I'm not understanding your notation here. What does [[0,1],[1,0]] mean?

    These are the problems exactly as they given:

    It's easy to do the first part for U1, but it's the U2 linear operator I have trouble with.

    [quote="Linear Algebra 3rd Edn pg 347 by Stephen Friedberg et al]11. For a linear operator T on anner product space V, prove that T*T = T0 implies T = T0. Is the same result true if we assume that TT* = T0[/quote]Hey, T0 actually refers to the zero transformation T0: V -> W, T0(x) = 0 for all [tex]x\in V[/tex]. I didn't know that. Guess this is what happens when you skip chapters.

    Now this can be done easily:
    <v,T*Tx> = <Tv,Tx>= <v,T0x> = <Tv,0>
    Then by the "cancellation law": (<x,y> = <x,z> for all x implies y=z) (Question: does this cancellation also mean that <z,x> = <y,x> implies z=y? I think I can prove it but my proof may be faulty and I need to know if this is true.)
    Tx=0 for all x which imples T=T0
     
  6. Aug 11, 2008 #5

    Dick

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    It means a 2x2 matrix whose first row is [0,1] and second row is [1,0]. U=TT*, U is not equal to TU*. For a general matrix det(U)=det(T)^2 and det(TU*)=det(T)^3. Are you sure that extra T is in the problem isn't a typo?
     
  7. Aug 11, 2008 #6

    Defennder

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    Yeah I double-checked. The question is as stated.
     
  8. Aug 11, 2008 #7

    Dick

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    Then did you figure out what's going on? Is there some assumption from the previous parts of the exercise you are supposed to carry into this one? Because it's certainly not true for general linear transformations.
     
  9. Aug 11, 2008 #8

    Defennder

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    The question is standalone and the previous questions don't appear to have any relation to this one. I don't think it's a problem of notation either since T is used universally as a linear operator. Maybe it's a misprint or something.
     
  10. Aug 11, 2008 #9

    Dick

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    Funny. It's an odd misprint to find in a 3rd edition.
     
  11. Aug 11, 2008 #10

    Defennder

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    There's actually a 4th edn out there. But I haven't managed to find a second-hand copy of it yet, so I'm stuck with the 3rd edn on loan from the library since they only allow a limited 2 hour loan for their 4th edn.
     
  12. Aug 11, 2008 #11

    Dick

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    Check the errata in the 4th edition. Look at the original question, i) U1=T+T*. Prove U1=U1*. Easy. ii) U2=TT*. The obvious question is 'prove U2=U2*'. The T came from nowhere.
     
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