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Is this a complete test to show that a matrix is orthogonal?

  1. Jul 30, 2012 #1
    I used to test orthogonality by using the definition MT = M-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MMT = I?

    If we multiply both side by M-1, we get MT = M-1.

    Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?
     
  2. jcsd
  3. Jul 30, 2012 #2


    Of course...it's exactly the same, right?!

    DonAntonio
     
  4. Jul 30, 2012 #3
    This might be a dumb question, but would it be possible to have a matrix M where MMT = I, yet MTM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M-1 = MT, but I wanted to make sure.
     
  5. Jul 30, 2012 #4

    micromass

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    That is actually a very good question. There is no reason why [itex]MM^T=I[/itex] should imply [itex]M^TM=I[/itex].
    But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
    So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.
     
  6. Jul 30, 2012 #5

    Well, that holds in any group and in any monoid where every element has a right inverse, since then ( with [itex]\,a'=\,[/itex] right inverse of [itex]\,a[/itex] ):
    [tex]a'a=a'a(a'a'')=a'(aa')a''=a'ea''=a'a''=e[/tex]
    so the right inverse is also the left one.

    DonAntonio
     
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