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Homework Help: Why A'*A=I but A*A' !=I

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I encounter a strange problem.

    Let A= [1.0000 0 0
    0 0 0
    0 0 0.4472
    0 0.3162 0
    0 0.9487 0
    0 0 0.8944]

    I am surprise to find that A'*A=I, but A*A'<>I . Can anyone give me an explanation?
     
  2. jcsd
  3. Jan 22, 2010 #2

    rock.freak667

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    Homework Helper

    A is a 6x3 matrix, so A is not square. A is not invertible.

    If AB=I, then A=B-1 provided that A and B are invertible to begin with.
     
  4. Jan 23, 2010 #3
    Thanks for ur kind reply. but must A and B be square such that the following equation holds ?

    (A'B)'=B'*A

    I think that if A*A'=I (wheter is square or not), we take transpose operator on both sides of the equation, and obtain that

    (A*A')' = I'

    then

    A*A' = I
     
  5. Jan 23, 2010 #4

    rock.freak667

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    Homework Helper

    Sorry I was reading A' as A-1 not AT

    But read this, orthogonal matrices
     
  6. Jan 23, 2010 #5
    thank u so much.

    BTW, is there an upper bound for |A'*A-A*A'|

    when A is a rectangluar column orthogonal matrix?
     
  7. Jan 23, 2010 #6

    vela

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    Staff Emeritus
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    If you have

    [tex]A^TA = I[/tex]

    then

    [tex](A^TA)^T=A^T(A^T)^T=A^TA=I[/tex].

    You're starting with [itex]AA^T[/itex], which isn't equal to the identity matrix, so its transpose won't be either.
     
  8. Jan 23, 2010 #7
    Thank u, vela and rock. i realize my mistake.
     
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