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Linear algebra- Inverse of a linear mapping

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Let L: V →V be a linear mapping such that L^2+2L+I=0, show that L is invertible (I is the identity mapping)
    I have no idea how to solve this problem or how to start,I mean this problem is different from the ones I solved before, the answer is "The inverse of L is -L-2 "
    If someone please can explain to me how to solve this ,would be great :cool:. Thanks in advance .
     
  2. jcsd
  3. Jul 6, 2013 #2

    micromass

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    Hint: ##I=-L^2-2L##.
     
  4. Jul 6, 2013 #3
    hmm ,that doesn't tell me anything :confused:
     
  5. Jul 6, 2013 #4

    micromass

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    Factor out ##L## in the right hand side.
     
  6. Jul 6, 2013 #5
    hmm ok you mean I=L(-L-2) but can you operate with L^2 like it was a number?? , isn't L^2 =L°L ?? I'm confused:confused:
     
  7. Jul 6, 2013 #6

    micromass

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    Very good remark!!

    The answer you can operate with it like it was a number, but that's perhaps not obvious.
    The property I'm using here is that

    [tex]A\circ (B + C) = A\circ B + A\circ C[/tex]

    and

    [tex]A\circ (\alpha B) = \alpha (A\circ B)[/tex]

    These properties are true, but it requires a separate proof.
     
  8. Jul 6, 2013 #7

    HallsofIvy

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    Those properties are, in fact, the definition of "linear transformation". That's all the proof you need.
     
  9. Jul 6, 2013 #8

    micromass

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    I wouldn't say it's the definition. It still requires a proof.
     
  10. Jul 6, 2013 #9
    ok , I see .Thank you very much :smile:
     
  11. Jul 7, 2013 #10
    Hmm I guess there's something that's still not clear for me , the first property of composition you wrote works for three linear mappings right?, but in this case we have two linear mappings and a number :L°(-L-2) I know I'm wrong somewhere but I don't know where ,I'm confused :confused: .Any help would be appreciated.
     
  12. Jul 7, 2013 #11

    Zondrina

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    Micro stated this here, you should be able to see that :

    ##I = L(-L-2)##

    Then multiplying both sides by ##L^{-1}## on the left yields :

    ##L^{-1}I = L^{-1}L(-L-2)##
    ##L^{-1} = -(L+2)##
     
  13. Jul 7, 2013 #12
    Ok so ##L^{-1}I = L^{-1}L(-L-2)## yields ##I \circ (-L-2)=-L-2## right ?? . Thanks
     
  14. Jul 7, 2013 #13
    No, you cannot do that. You are proving here that ##L^{-1}## exists, so you cannot assume it exists in the proof!
     
  15. Jul 7, 2013 #14
    You are right !:smile: could you explain please why ## L\circ(-L-2)=-L^{2}-2L## ?? if "2" was a linear mapping then the properties would work for this, right? but 2 is a scalar.
     
  16. Jul 7, 2013 #15
    ##-L - 2## is wrong to begin with. You cannot add (or subtract) numbers from linear maps. The correct expression is ##-L - 2I##.
     
  17. Jul 7, 2013 #16
    If there exist such an inverse map, what property does it have i.e. what can you tell about the multiplication of L and [itex]L^{-1}[/itex] ? Or the same question, what is the definition of the inverse map?
     
    Last edited: Jul 7, 2013
  18. Jul 7, 2013 #17
    It makes sense now ,thanks .
     
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