Re: [eigen] Fast method for computing Gramian Matrix

[Jim Bosch <talljimbo@xxxxxxxxx>]

Is it necessary to do something like this to form the full dense matrix 
when using SelfAdjointEigenSolver?  I noticed that it doesn't have a 
UpLo template parameter to tell it which triangle to use.

Thanks!

Jim



On 03/29/2012 06:55 PM, Gael Guennebaud wrote:
> with a copy:
>
> C = B.selfadjointView<Upper>();
>
> or in-place using triangularView as a writing mask:
>
> B.triangularView<StrictlyLower>() = B.adjoint();
>
> cheers,
> gael
>
> On Fri, Mar 30, 2012 at 12:18 AM, Mark Borgerding<mark@xxxxxxxxxxxxxx>  wrote:
> > On 03/29/2012 01:18 AM, Douglas Bates wrote:
> > > On Thu, Mar 29, 2012 at 3:50 AM, Mark Borgerding<mark@xxxxxxxxxxxxxx>
> > >   wrote:
> > > > (sorry for hitting send too soon)
> > > >
> > > > http://en.wikipedia.org/wiki/Gramian_matrix
> > > >
> > > > What is the fastest way to compute MM' when M is wide?
> > > > The diagonal elements are the norm-squared of the rows of M. The result
> > > > is
> > > > Hermitian.
> > > >
> > > > Is there an optimized method in Eigen or otherwise?
> > >
> > > I have used the rankUpdate method for a selfadjoint view to calculate
> > > this.  I assume that it is an efficient way to calculate the result
> > > but I haven't done any timings.
> > >
> > >
> > > MatrixXcd MMt(M.rows(), M.rows();
> > > MMt.setZero().selfadjointView<Eigen::Upper>().rankUpdate(M);
> > Thanks.  That cut the time to about 80% of the full matrix product.
> >
> > Is there an elegant way to recover the full, square self-adjoint matrix from
> > a SelfAdjointView ?
> > I've been adding the SA view to its own adjoint and then subtracting the
> > original diagonal.