| T.R | Title | User | Personal Name
 | Date | Lines | 
|---|
| 1760.1 | Overdetermined. | CADSYS::COOPER | Topher Cooper | Thu May 27 1993 16:45 | 11 | 
|  | >    offcourse the solution is not unique.
    My linear algebra is rusty and never was too great, but it sems to me
    that you have N� equations (one for each element of A) and 2*N unknowns
    (one for each element of u and one for each element of v) so for N
    greater than 2 you've got in general an overdetermined problem, not an
    underdetermined one.  I.e., for most A there will be no solution.
    Maple can probably be coaxed into finding solutions when there are any.
				    Topher
 | 
| 1760.2 | you're both right | AUSSIE::GARSON | nouveau pauvre | Thu May 27 1993 18:39 | 14 | 
|  |     re .0
    
    Just off the top of my head, a solution exists iff each row of the
    matrix is a (real) multiple of the first row that is not all zeros,
    in which case the solution is non-unique and takes the form
    
    u = t * (a column giving the multiples)
    
    v = 1/t * (the first row that is not all zeros)
    
    where t is an arbitrary non-zero real number.
    
    [The special case where the matrix is entirely zeros, not covered by the
     above, has obvious solutions and many of them.]
 | 
| 1760.3 | SVD will test for this | GAUSS::ROTH | Geometry is the real life! | Thu May 27 1993 18:39 | 16 | 
|  |     I assume you mean the matrix is the outer product of the vectors.
	A[i,j] = u[i]*v[j]
    The criteria for this is for A to be rank 1, since all rows or
    columns are scalar multiples of each other.
    The most numerically reliable way to get this would be based
    on a singular value decomposition since this would also indicate
    just how close A is to really satisfying this criteria.
    SVD routines are available in most linear algebra packages, and
    I have a copies of these.  (See Numerical Recipes for another one;
    the latest edition fixes some bugs in the first edition.)
    - Jim
 | 
| 1760.4 |  | GAUSS::ROTH | Geometry is the real life! | Thu May 27 1993 18:45 | 7 | 
|  |     By the way, if the matrix *really* is rank 1, then you could simply
    normalize each row or column to be a unit vector.
    Then the norms will form one of your vectors (up to sign) and
    one of the unit vectors will form the other.
    - Jim
 |