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Orthogonal matrix

  1. Jun 14, 2008 #1
    1. The problem statement, all variables and given/known data
    My book used the term "orthogonal 3 by 3 matrix" and I couldn't find where that was defined. Does that mean that the column vectors are orthogonal or that the row vectors are orthogonal? Or are those two things equivalent?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 14, 2008 #2


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    It just means it's a real unitary matrix.
  4. Jun 14, 2008 #3
    If A is a real matrix, then if A is orthogonal, the rows of A form an orthonormal set as WELL as the columns.
  5. Jun 14, 2008 #4
    How do you prove that the rows (of a real n by n matrix) are orthogonal iff the columns are orthogonal?
  6. Jun 14, 2008 #5
    What's the consequence of a real n by n matrix being orthogonal?

    (A^T)A = I

    What does this say about the columns?
  7. Jun 14, 2008 #6
    Thats orthonormal not orthogonal.
  8. Jun 14, 2008 #7
    Since it's an n x n matrix, it is a square matrix. It is easy to see that any square matrix with orthonormal columns is an orthogonal matrix. By definition of orthogonal matrix, it is a square "blank" matrix U such that [tex]U^{T}U=I[/tex]. What is "blank" and why? Then, show that the rows of U form an orthonormal basis of [tex]R^n[/tex]
  9. Jun 14, 2008 #8
    But not every orthogonal matrix is orthonormal. Its not true that if U is orthogonal then U U^T = I. All we know is that U U^T is diagonal.
  10. Jun 14, 2008 #9
    True, but U is a square matrix. Your original question was "orthogonal 3x3 matrix".

    A square matrix is a orthogonal if and only if the columns form an orthonormal basis in [tex]R^n [/tex], thus it must also be true that the rows of U form an orthonormal basis of [tex]R^n[/tex] (isn't this what you're trying to prove?)

    Theorem: U be a n x n matrix
    1. U is orthogonal
    2. The columns of Q are an orthonormal set of vectors in Rn
    3. The rows of Q are an orthonormal set of vectors in Rn
    Last edited: Jun 14, 2008
  11. Jun 14, 2008 #10


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    No, I think he's trying to prove that any square matrix with orthogonal columns must have orthogonal rows. (I'm not sure if this is true or not, but I haven't put too much thought into it.)
  12. Jun 14, 2008 #11
    That theorem is false.

    Let n = 2 and U =

    1 & 0 \\
    0 & 2 \\
    The second column and the second row are both not normalized.
  13. Jun 14, 2008 #12
    Yes, that is what I am trying to prove.
  14. Jun 14, 2008 #13


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    Your matrix isn't orthogonal since it has a determinant of 2.
  15. Jun 14, 2008 #14
    If A is orthogonal then AA^T = I

    Lets have a real orthogonal 3x3 (square) matrix then with rows u1 = (a1,a2,a3), u2 = (b1,b2,b3), u3 = (c1,c2,c3)

    solve for AA^T = I since it must be invertible.

    You should end up finding that a1^+a2^2+a3^2 =1, b1^2+b2^2+b3^2 = 1 and c1^2+c2^2+c3^2 = 1.

    All the other equations are equal to zero.

    This shows that u1 dot u1 = 1, u2 dot u2 = 1, and u3 dot u3 = 1.

    Thus AA^T = I shows that the rows of A form an orthonormal set of vectors and you just have to solve for A^TA = I since it is true for the columns.
  16. Jun 14, 2008 #15

    I am an idiot. In fact my book does define the term orthogonal and it is just like the Wikipedia page. I thought the definition was just that either the column vectors or the row vectors were pairwise orthogonal and I was trying to figure out which it was. Sorry.
  17. Jun 14, 2008 #16
    Whoops, for the theorem I wrote, what I meant to say that "if one of the following is true, then the rest is true" for (1), (2), and (3)
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