Consider a matrix A∈Rm×n In the case when rank(A)=n then this implies the existence of a left inverse: A−1l=(A⊤A)−1A⊤ such that A−1lA=In×n.
Similarly if rank(A)=m then this implies the existence of a right inverse: A−1r=A⊤(AA⊤)−1 such that AA−1r=Im×m.
I understand how the concept of a right inverse is naturally follows in say solution of a least squares problem: Ax=b, $rank(A)=n
I expect it to involve the fact that in this case rank(A)=m<n and so there are now infinitely many solutions to Ax=b and that the right inverse now someone seeks the solution which minimises the length of the solution vector?
No comments:
Post a Comment