Reputation: 879
I have a matrix M of dimensions 6x6 and it has rank 1. How can I factorize it into two matrices of dimensions 6x1 (say A) and 1x6 (say B) so that M=A*B.
Upvotes: 1
Views: 1440
Reputation: 3326
take the largest eigen vector and multiply it by the largest eigenvalue :
A=[1 2 3 4]'*[1 2 3 4]
A =
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
[v,e] = eigs(A,1);
sqrt(e)*v
ans =
-1.0000
-2.0000
-3.0000
-4.0000
of course, the result is good only up to a sign change.
EDIT: if you assume that the two vectors can be different:
A=[1 2 3 4]'*[5 6 7 8]
[uu,ss,vv]=svd(A);
u=uu(:,1)*ss(1,1)
v=vv(:,1)
assert(norm(u*v'-A)<1E-10)
Now the solution is even less unique. You are determining 2*n values based on only n. This is one solution among many.
For example, look at this other even simpler solution (which will assume your matrix is perfectly rank 1) :
aa=A(:,:)./repmat(A(1,:),[size(A,1),1]);
bb=A(:,:)./repmat(A(:,1),[1,size(A,2)]);
u=aa(:,1);
v=bb(1,:)'*A(1);
assert(norm(u*v'-A)<1E-10)
it produces a totally different result, but that still factorizes the matrix. If you want non-negative factorizations only to slightly reduce the space of possible results, I'd suggest you ask a new question!
Upvotes: 1
Reputation: 272687
If it has rank 1, then all columns/rows are multiples of the first column/row (or indeed of any non-zero column/row). i.e.:
m = M(:,1);
M = [ a*m, b*m, c*m, d*m, e*m, f*m ];
Hopefully you can take it from there.
Upvotes: 0