| Re: SVD and Other things Linear [message #14346] |
Thu, 18 February 1999 00:00 |
J.D. Smith
Messages: 214
Registered: August 1996
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Senior Member |
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Lykoan wrote:
>
> Hellu,
> I am trying to find the non trivial solution for the equation Ax=0 where A
> is a known nxm matrix, and x is an unknown 1 x n (or 1xm, I get that
> confused) vector. I was using SVD and SVSOL to get the answer, but then I
> got to this case, and all I get is x = [0, ...., 0]. Does anyone know if I
> can get SVD and SVSOL to find the non-trivial solution, or of any other way
> to do this linearly. I have tried to do it nonlinearly, with BROYDEN, but I
> keep running into local minimums(maximums), or it not even being able to
> converge.
>
> Thank You,
> David Borland
> dborland@egi.com
There may, in fact, only be the trivial solution. The null space of a
matrix with more columns than rows (underdetermined -- fewer equations
than unknowns) is always finite. The spanning vectors of this nullspace
are just the columns of V corresponding to the zeroes on the diagonal of
the diagonal matrix W as returned by SVDC. For non-singular square or
over-determined matrices, the null space is Z, the zero space... i.e.
just x=[0,......,0].
So, the question becomes do you have fewer or more equations than
unknowns? Is A short and wide or tall and skinny?
If you are not expecting a unique solution to begin with (short and
wide), or if you have a singular square matrix, just examine the columns
of the output matrix V corresponding to *small* diagonals in W (they
won't be exactly 0). These will be the set of orthonormal vector(s)
which span A's nullspace.
If you are overdetermined, then really this becomes a least-squares
problem for which the null space is the zero space by definition, since
if there were an non-trivial x as a solution to Ax=0, then this could be
added to any solution of Ax=b for which x was the *best* solution and
change that solution. Another way of seeing that is noting that the
vector which will best approximate a solution of Ax=0 is indeed
x=[0,.....,0], which solves it exactly!
Good Luck,
JD
--
J.D. Smith |*| WORK: (607) 255-5842
Cornell University Dept. of Astronomy |*| (607) 255-6263
304 Space Sciences Bldg. |*| FAX: (607) 255-5875
Ithaca, NY 14853 |*|
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