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Orthogonal Matrices

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data

    1. If I got a square orthogonal matrix, then if I make up a new matrix from that by rearranging its rows, then will it also be orthogonal?

    2. True/false: a square matrix is orthogonal if and only if its determinant is equal to + or - 1

    2. Relevant equations

    no equations

    3. The attempt at a solution

    1. I think it should also be orthogonal since it forms a basis, and the basis would be the same, but just a linear combination of the previous, right?

    2. false, its determinant doesn't necessarily ensure it is orthogonal. So, how would/should I correct that statement?
  2. jcsd
  3. Jan 16, 2009 #2
  4. Jan 16, 2009 #3
    An nxn matrix is orthogonal iff its rows form an orthormal basis for [tex]\mathbb{R}^n[/tex] (note the symmetry of [tex]AA^T=A^TA=I[/tex] for an orthogonal matrix A). The linear independence of a collection of vectors doesn't depend on the order in which you write them, so the rows of the new matrix still form an orthonormal basis.

    Just be careful your language: a linear combination of a basis reads as a linear combination of its vectors, which gives just one vector.
  5. Jan 16, 2009 #4
    I forgot about the linear independence part of it for #1.

    As for #2, I took a counter-example from wikipedia XD
    [ 2 0 ]
    [ 0 0.5 ]
    where its determinant = 1
    but the length of each column is not 1 (not orthonormal)
    I guess counter-examples should be enough?

    Thanks for the help you two. :)
  6. Jan 16, 2009 #5
    The statement #2 is (colloquially) of the form "(property X implies property Y) AND (property Y implies property X)" (*). If all you want to do is show that (*) is false (e.g., if you were asked to prove or disprove the statement), then it suffices to show that property Y does not imply property X.

    To show that property Y does not imply property X, it suffices to give an example for which property Y holds but X does not. Why? Because it definitively answers the question as to whether Y implies X. There is no guessing about it!
  7. Jan 16, 2009 #6
    Yea, you're right Unco. I need to work on my logic a bit more. I'm very grateful for your help :D
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