1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: A*transpose(A) for orthonormal columns ?
  1. Orthonormal basis (Replies: 1)

  2. Orthonormal basis (Replies: 1)

  3. Orthonormal vectors (Replies: 13)