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A*transpose(A) for orthonormal columns ?

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider an n x m matrix A with n >= m. If all columns of A are orthonormal, then A'A = I. What can you say about AA'?

    Where A'A = transpose(A)*A and AA' = A*transpose(A)

    2. Relevant equations

    3. The attempt at a solution
    For the case that n = m:
    A is square. Since the columns of A are normalized, and the set of vectors contained in A is orthogonal, we can call A orthogonal.

    So, A'A = I and AA' = I

    For the case that n > m:
    I'm lost here...

    any hints? What should I be looking up in the books to understand this?
  2. jcsd
  3. Sep 11, 2011 #2
    One thought is...

    For n > m, rank(A) = m.

    So, AA' = L where L is [n x n]. Now I am assuming that the rank(L) = m, but I do not know how to prove this.

    Now if we wanted, AA' = I where I is [n x n], and rank(I) = n.

    Lets try, AA' = I

    -> rank(AA') = rank(I)
    -> rank(L) = rank(I)
    -> m = n

    this is not true, since n > m

    Thanks for any help
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