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One task in linear algebra

  1. Sep 26, 2012 #1
    A is a square matrix n*n with the following properties: A*A=A and A not equal I (identity matrix).
    How to prove the following equation:
    (I+A)^-1=I+A/2 ?
     
  2. jcsd
  3. Sep 26, 2012 #2

    HallsofIvy

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    You can't. It's not true:
    (I+ A)(I+ A/2)= I+ A+ A/2+ A^2/2= I+ A+ A/2+ A/2= I+ 2A, not I.

    If you mean (I+ A)/2,
    (I+ A)(I+ A)/2= (I+ A+ A+ A^2)/2= (I+ 3A)/2.

    I thought perhaps you had left out a "-" but that doesn't see to work either so I have no idea what you are trying to prove.
     
  4. Sep 26, 2012 #3
    I mean with this (I+A)^-1, inverse of the matrix (I + A).
    You have understood correctly.
    As I tried to solve the problem I came to the same answer.
    Perhaps there is a technical error, but I'm not sure.
    Thanks for answer.
     
  5. Sep 26, 2012 #4
    Hi isa_vita, maybe this is coming from language issues, or maybe there is an error in the exercise you want to do, could you post the text as it comes ? (translated) ?
    you equation does have a solution: A=0 works, but you would say 'solve' instead of 'prove', so maybe you didn't word it correctly (but it sounds fishy, this would be a weird exercise anyway), so, anyway, do you have the full text of the exercise ?
    Cheers...
     
  6. Sep 26, 2012 #5
    Hi oli4e, my mistake, I mean prove not solve. I gave full text of the exercise(in my translation :)).
    I know for a zero solution, but it isn't the task. I think it is right HallsofIvy. It is a technical error.
    Thanks for a help. Cheers...
     
  7. Sep 26, 2012 #6

    HallsofIvy

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    I still don't understand what you are asking. Given that [itex]A^2= A[/itex], it does NOT follow that I+ A/2 is the inverse of I+ A.
     
  8. Sep 28, 2012 #7

    chiro

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    Assuming that the inverse even exists why not just multiply both sides by (I + A) and collect terms? (Remember I'm assuming that the inverse of this exists which it may not, since I haven't checked it).
     
  9. Sep 28, 2012 #8

    HallsofIvy

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    That's exactly what I did- and did NOT get the identity matrix.
     
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