Re: [Pysparse-users] transpose operation

[Dominique O. <dom...@gm...>]

On Tue, Nov 17, 2009 at 6:42 AM, Fergus Gallagher <fe...@go...>wrote:

> > The find/put method is surely the fastest in general. The more
> > fundamental question is do you really need the explicit transpose?
> > Often, algorithms that need the transpose operate directly on the
> > original matrix. If what you need is matrix-vector products of the form
> > A.T*x, you can always use the matvec_transp() method. If you're using
> > the higher-level PysparseMatrix objects, you can do A*x and x*A. The
> > latter actually computes A.T*x.
> >
> > I hope this helps.
> >
>
> Hi,
>
> thanks for all the replies.
>
> What we want to do is generate the "outer product"  A*B^T  (actually A*A^T
> in our case).
>

If you really need A*A^T explicitly, I suggest building A^T and then using
dot().  Often you don't need this matrix explicitly though (e.g., it is easy
to build matrix-vector products with A*A^T without assembling this matrix).
Similarly, if you need to factorize it to solve A*A^T x = b, you can solve
the equivalent system

  [ I  A^T ] [x]   [ 0]
  [ A   0  ] [y] = [-b]

This last system is also much sparser, and it is symmetric. It is indefinite
though.

-- 
Dominique