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Does AB=I imply BA=I?

  1. Feb 21, 2012 #1
    Hi all.

    Does there exist a field F such that AB=I doesn't imply BA=I, where A and B are square matrices (both n×n) over the field F?

    Thanks in advance.
     
  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

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    hi asmani! :smile:

    hint: BAB ? :wink:
     
  4. Feb 21, 2012 #3

    morphism

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    AB=I always implies BA=I regardless of the field. This is because the rank-nullity theorem holds over any field, so an injective linear map (between vector spaces of the same dimension n) is necessarily surjective (and vice-verse).

    In fact AB=I will imply BA=I even over a commutative ring. Probably the quickest way to see this is to use the characterization "An nxn matrix with entries in a commutative ring R is invertible (i.e. has a left and right inverse) iff detA is a unit in R". Note that AB=I implies that detA and detB are units.

    P.S. Here I'm addressing the problem of showing that "left invertible" implies "right invertible". Once you have this, you can use tiny-tim's hint to show that the left inverse is equal to the right inverse.
     
    Last edited: Feb 21, 2012
  5. Feb 22, 2012 #4
    AB=I does not always imply BA=I . There are somethings called left inverses and right inverses. You can look it up when you see the least square projections of a vector which does not lie on a column space S onto S.

    Then if there is a left inverse C , then, though , CA = I but AC ≠ I .
    A similar story goes for the right inverse.

    Good Luck
     
  6. Feb 22, 2012 #5

    micromass

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    We are talking about square matrices. In that case, AB=I does certainly imply BA=I.
     
  7. Feb 22, 2012 #6
    oops :) yep . Sorry , i was talking in general terms :) thanks for quoting !
     
  8. Feb 22, 2012 #7

    D H

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    Not always. There are certain mathematical characteristics that are required to be able to say that. I would have agreed completely had you instead said "We are talking about square matrices over a field."


    This is a very nice hint. What mathematical characteristics let tiny-tim get away with writing BAB without using any parentheses? What characteristics do you get "for free" just by saying the word "field"?
     
  9. Feb 24, 2012 #8
    Thanks a lot for the replies.
    Actually the original problem was to show that "left invertible" implies "right invertible" and vice versa.
    I can't see how to work out this yet. I know that we can write B=BI and since AB=I, we get to B=B(AB) which is the same as B=(BA)B (commutativity of multiplication) and then we have B(I-BA)=0. Does it help?
     
  10. Feb 24, 2012 #9

    Bacle2

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    Careful! If you had commutativity _matrix-wise_ (you do have it entry/element-wise),
    AB=I would automatically imply BA=I.
     
  11. Feb 24, 2012 #10

    mathwonk

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    this is not a trivial result. it is equivalent to showing that any set of n independent vectors in k^n must span it.
     
  12. Feb 25, 2012 #11
    Sorry, I meant associativity, which leads to B(AB)=(BA)B.
    Thanks.
     
  13. Feb 25, 2012 #12
    Here is the proof I found:

    AB=I implies det(A)det(B)=1, which means that det(A)≠0. It's a known fact that A is invertible iff det(A)≠0. Thus A is invertible and has both right and left inverses and they are equal.

    Is there something wrong?
     
  14. Feb 25, 2012 #13

    morphism

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    No, there is nothing wrong with this. In fact I already gave you this exact same proof in my post above.
     
  15. Feb 25, 2012 #14
    You're right. I didn't notice that, because I wasn't familiar with rings.
    Thanks.
     
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