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Math Help - Conflicting defintions/explanations

  1. #1
    Senior Member Pinkk's Avatar
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    Conflicting defintions/explanations

    My professor states in one instance that if A is invertible, then B=A^{-1} when AB=BA=I, leading to AA^{-1}=A^{-1}A=I. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that B=A^{-1} implies that B is invertible when BA=AB=I, leading to A^{-1}A=AA^{-1}=I. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is AB=BA=I or BA=AB=I, yet he clearly contradicts himself if order is so important. A similar is in once instance, B=A^{-1} leads to AB=BA=I, and then later on states that it is A=B^{-1} that leads to AB=BA=I. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.

    So I need clarification on the theorem. If AB=BA=I then A is invertible and B=A^{-1}, or does B=A^{-1} imply B is invertible and therefore BA=AB=I. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.
    Last edited by Pinkk; June 9th 2009 at 05:33 PM.
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  2. #2
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    Quote Originally Posted by Pinkk View Post
    My professor states in one instance that if A is invertible, then B=A^{-1} when AB=BA=I, leading to AA^{-1}=A^{-1}A=I. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that B=A^{-1} implies that B is invertible when BA=AB=I, leading to A^{-1}A=AA^{-1}=I. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is AB=BA=I or BA=AB=I, yet he clearly contradicts himself if order is so important. A similar is in once instance, B=A^{-1} leads to AB=BA=I, and then later on states that it is A=B^{-1} that leads to AB=BA=I. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.

    So I need clarification on the theorem. If AB=BA=I then A is invertible and B=A^{-1}, or does B=A^{-1} imply B is invertible and therefore BA=AB=I. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.
    if A is a square matrix, say n \times n, then A is invertible if and only if AB=I_n or BA=I_n, for some n \times n matrix B. the important point, which unfortunately is not mentioned in most elementary

    linear algebra courses, is that for square matrices A,B we have AB=I if and only if BA=I. anyway, in this case we call B the inverse of A and similarly A would be the inverse of B.

    so for square matrices the following four statements are equivalent, meaning any of them implies the other three:

    1. AB=I,

    2. BA=I,

    3. A=B^{-1},

    4. B=A^{-1}.

    for non-square matrices things are different. an n \times m, \ n \neq m, matrix A could have left inverse, i.e. BA=I_m, for some m \times n matrix B, but no right inverse, i.e. AC \neq I_n, for any m \times n

    matrix C. it is also possible that A has a right inverse but not a left inverse (or of course A might have neither left nor right inverse). but if A has both left and right inverse, then it has to be

    a square matrix, i.e. n=m and then the left and right inverse must be equal.
    Last edited by NonCommAlg; June 9th 2009 at 08:24 PM.
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  3. #3
    Senior Member Pinkk's Avatar
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    I understand that A=B^{-1} implies B=A^{-1} given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was AB=BA=I implies A is invertible and thus B=A^{-1}. He uses this defintion to show that A=B^{-1}, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.
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    Quote Originally Posted by Pinkk View Post
    I understand that A=B^{-1} implies B=A^{-1} given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was AB=BA=I implies A is invertible and thus B=A^{-1}. He uses this defintion to show that A=B^{-1}, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.
    in my previous post i actually ignored what your prof said because it's nonsense. i also said that the definition is equivalent to saying that AB=I or BA=I because for square matrices

    AB=I if and only if BA=I. anyway, let's just stick to your prof's definition: then in order to show that A=B^{-1} we need to show that BA=AB=I, which is obviously the same as

    AB=BA=I. so, yes, the order is not important and you're right. another point that i mentioned is the "uniqueness", i.e. if AB=BA=I and AC=CA=I, then B=C. now we can

    talk about "the" inverse.
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  5. #5
    Senior Member Pinkk's Avatar
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    Ah okay, thank you. And yes, I had a hunch that my professor was making it more convoluted than it has to be. The if and only if statement applies only to square matrices since performing BA would result in an m \times m and AC results in an n \times n, and therefore BA \ne AC, and if AB are n \times n square matrices of equal size , then AB=BA since \sum_{r=0}^{n}a_{ir}b_{rj}=\sum_{r=0}^{n}b_{ir}a_{  rj} for square matrices.
    Last edited by Pinkk; June 9th 2009 at 09:16 PM.
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    see how an MIT prof explains the inverse of a matrix. he mentions what i said, i.e. for square matrices AB=I if and only if BA=I but he also says that the proof is not "easy",

    (well, it's not terribly hard either! haha) and so he skips it. i like this guy! his lecture is very accurate, although he talks too much! haha watch the video from 21':30" if you like:

    MIT OpenCourseWare | Mathematics | 18.06 Linear Algebra, Spring 2005 | Video Lectures | detail
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  7. #7
    Member alunw's Avatar
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    No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
    But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
    it can't happen for a finite monoid
    if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alunw View Post
    No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
    But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
    it can't happen for a finite monoid
    if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.
    For instance, the Bicyclic Monoid: <b,c|bc=1>.
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