[petsc-users] Getting access to matrix rows without and setting values simultaneously

[Alexander Grayver]

Jed, I'm afraid we are talking about different "grids", by the grid I 
meant that this matrix is 2d array, not a real grid of some physical 
parameters or whatever and it's also not a linear operator itself. Sorry 
that I possibly made you confused.
It's very big and that is why I chose PETSc. I compute this matrix as a 
product of three other petsc matrices, one of which is also dense.
Now I need this transformation to be applied and then I'm planning to 
use this matrix to create my operator, more precisely I need to use A'*A 
as the linear operator and Matrix-Free notation since A'*A is dense and 
even larger than original A.
So I guess I need A as a matrix, since (A'*A)*v operations will be 
required further.

I completely understand that petsc Mat has certain field of applications 
and my current problem is out of it, but still I wonder if there is a 
way to apply a scalar operation to each element of the matrix?
Could you maybe give me an idea how to implement it? Any workaround?

Regards,
Alexander

On 16.06.2011 14:30, Jed Brown wrote:
> On Thu, Jun 16, 2011 at 14:08, Alexander Grayver 
> <agrayver at gfz-potsdam.de <mailto:agrayver at gfz-potsdam.de>> wrote:
>
>     This matrix is rather grid o values.
>
>
> Okay, then you can manage it with a DA (DMDA in petsc-dev). This will 
> give a better parallel distribution and give you access to entries in 
> convenient ways (e.g. using arrays). See the section of the user's 
> manual Structured Grids Using Distributed Arrays.
>
> In PETSc, the Mat type is for linear operators. A Mat might represent 
> a transformation from one of your grids to another. A given state on 
> the grid is given by a Vec, and DMDA will help you manage that.
>
>     The operation is very simply, it's scalar. Let's say we have this
>     matrix A of size MxN and I want to apply thw following operation
>     to each element of the matrix:
>
>     A(i,j) = log( (A(i,j) - a(j)) /  (b(j) - A(i,j)) ),
>     i=0..M-1 j=0..N-1
>
>     the vectors a,b are just arrays size of N and have them on each
>     processor.
>
>
> This will be very easy using DMDA.

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