Another question on ranks (linear algebra).

You are trying to show that the transpose of an nxn multiplied by itself has the same or less rank? I would use the definition of matrix multiplication, and then transpose, and do Gauss elimination (probably on a 2x2). There may be a more subtle way, but the forceful approach should work.

 

Let U = {x : AA^t x = 0}, and V = {x : A^t x = 0}. Assume you have already shown that dim V <= dim U (which implies r(A) >= r(AA^t) ), and now assume dim U > dim V (which would imply r(A) > r(AA^t) ) and find a counterexample (it's simple). When you have shown that dim U > dim V can't hold, then dim U = dim V must hold, and hence r(A^t) = r(A) = r(AA^t). Hope this works.

 

Actually, that's what's bothering me, too. If we assume it holds for any matrix, could a 'proof by counterexample' work? Guess some of the PF mathematicians should take over on this one.

 

You can show this directly from the singular value decomposition of A.

Let the SVD of A be U.D.V^T where D is the diagonal matrix of singular values. Write the SVD of A.A^T in terms of U D and V, and show that A and A.A^T have the same number of zero singular values, hence the same rank.

There is a similar proof involving eigenvalues and eigenvectors.

There is probably a similar but more "elementary" proof involving choosing a set of basis vectors where a subset of the basis vectors spans the null space of A - but I haven't thought idea that right through.