Results 1 to 9 of 9

Math Help - Not so easy question (at least I hope) about matrix inverse

  1. #1
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Not so easy question (at least I hope) about matrix inverse

    How do we prove that if a matrix has right inverse, then it's the left inverse too. Is it true in all cases?
    I've read one beginner level book about algebra and it approaches the problem by defining inverse A' of an invertible matrix A to be such matrix that statisfies both AA'=I and A'A=I. Now this type of definition implies that there may be such matrices A and B such that AB=I but BA<>I. I'm not sure how I can discover two such matrices, so providing example by myself is off.
    I've read some abstract algebra too. It seems that proving that all nxn invertible matrices (ones that have right inverse) form a group under matrix multiplication may do the job (we do get cancelation laws), but...
    If A has right inverse A' then A' must have it's own right inverse to statisfy the group axioms. How do I prove that such one does exist?
    Last edited by mrproper; April 3rd 2013 at 12:41 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Not so easy question (at least I hope) about matrix inverse

    I've looked in my abstract algebra text and just remembered that the group axioms may be stated in alternative forms. So my text says:
    1. The set must be closed under the binary operation
    2. For all members of the set associativty of the binary operation must hold
    3. There must be identity el3ment
    4. !! If A is in the group, there must be element A' in the same group such that both (!!) AA'=e and A'A=e hold
    ugh....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Not so easy question (at least I hope) about matrix inverse

    If A is square and AA' = I , multiply both sides on the right by A

    AA'A = IA

    Matrix multiplication is associative and IA = A

    A(A'A) = A

    AX = A

    X must be the identity therefore A'A = I

    This shows that inverses commute

    If A is not square then there is something called 'pseudo inverse'
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Not so easy question (at least I hope) about matrix inverse

    Another stupid question. How do we know there is no other identity. Element X<>I for which AX=A too.
    Last edited by mrproper; April 3rd 2013 at 01:25 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Not so easy question (at least I hope) about matrix inverse

    Suppose A has 2 inverses X and Y then

    AX = I

    AY = I

    using the previous result we know inverses commute

    XA = I

    multiply both sides on the right by Y

    XAY = IY

    matrix multiplication is associative and IY = Y

    X(AY) = Y

    but AY = I

    XI = Y

    X = Y

    This shows the inverse is unique.

    There are no stupid questions, don't be too hard on yourself. It's only easy when we know how.

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Not so easy question (at least I hope) about matrix inverse

    This is actually easy and once I knew the method of proof in relations with the group. It uses subset of group axioms, so it easily applies to our scenario.
    I is identity. I.e. AI=IA=A for every square nxn matrix ....(1)
    I' is identity too. AI'=I'A=A for every square nxn matrix ...(2)

    So In (1) set A=I' to get I'I=I'
    and in (2) set A=I to get I'I=I
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Not so easy question (at least I hope) about matrix inverse

    Oh yes , i just realisd i answered a different question... i thought you were asking why inverses are unique.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Not so easy question (at least I hope) about matrix inverse

    And Mulder, for this to work it might be not enough that we have unique identity.
    We should be sure that if AX=AY then X=Y (right cancelation if i recall?).
    Only then we can cancel A from AX=AI in your proof to get X=I.
    Or we must be sure that AX=A has unique solution.

    Am I correct?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Not so easy question (at least I hope) about matrix inverse

    The above is LEFT cancellation As are on the left. (AX=A).
    And if we have LEFT cancellation we have the AA'=I, AA'A=IA=AI, cancel As to get A'A=I. Unfortunatelly, unwilling to accept too much, we cannot assume right out the cancellation works.

    And from AA'=I I can only prove directly RIGHT cancellation.
    CA=DA , CAA'=DAA', CI=DI, C=D

    ?/????

    probably the way is elementwize proof that if AX=AY then X=Y...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Easy matrix question confusion
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 29th 2012, 08:35 AM
  2. Replies: 2
    Last Post: January 25th 2011, 06:21 AM
  3. Easy matrix question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 25th 2010, 11:13 AM
  4. Matrix Inverse question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 12:59 AM
  5. inverse matrix question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 1st 2008, 10:25 AM

Search Tags


/mathhelpforum @mathhelpforum