# A*transpose(A) for orthonormal columns ?

1. Sep 11, 2011

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. Sep 11, 2011

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