Wolfgang pointed out to me that the pivot based answer is wrong -
fortunately I have another explanation :)
If we have the block matrix
A B^T
B 0
where A is SPD and B has linearly independent rows, this matrix has
equal eigenvalues to
I 0 A B^T A B^T
* =
-B A^-1 I B 0 0 -B A^-1 B^T
since the eigenvalues of the leftmost matrix are all 1 (its triangular
with 1s on the main diagonal). The eigenvalues of the rightmost matrix
are the eigenvalues of A and the eigenvalues of -B A^-1 B^T. Since A
is SPD, we can rewrite A^-1 = L L^T (its Cholesky factorization) so -B
A^-1 B^T = -B L L^T B^T = -(B L) (B L)^T: a second Cholesky
factorization. Hence the bottom right is negative definite and
therefore the matrix as a whole is indefinite.
Thanks,
David
On Sun, Feb 2, 2020 at 9:41 PM Wolfgang Bangerth <bange...@colostate.edu> wrote:
>
> On 2/1/20 12:44 PM, Krishnakumar Gopalakrishnan wrote:
> >
> >
> > "_Two difficulties_ arise due to the _*zero pivots*_ in the bottom-right
> > block
> > of the matrix.
> >
> > 1. Firstly, following a classical result from linear algebra, such matrices
> > are indefinite and the conjugate gradient solver cannot be applied. We
> > would have to resort to other iterative solvers instead, such as MinRes,
> > SymmLQ, or GMRES, that can deal with indefinite systems.
> > 2. Secondly, due to the//zero block, there are zeros on the diagonal and
> > none
> > of the usual, "simple" preconditioners (Jacobi, SSOR) will work as they
> > require division by diagonal elements"
>
> How about this:
> https://github.com/dealii/dealii/pull/9470
>
> Best
> W.
>
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bange...@colostate.edu
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