On 2/4/20 8:50 AM, David Wells wrote:
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
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 0A B^T
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
Dear Krishna,
Allow me to clarify my previous statement. From Strang (intro to
linear algebra, 5th edition, page 352):
When a symmetric matrix S has one of these five properties, it has them all:
1. all n pivots of S are positive.
2. all n upper left determinants are positive.
3. all n
Dear Dr David Wells,
Thank you for the explanation. However, this only satisfies me partially
because the very next statement in that tutorial says:
*"We would have to resort to other iterative solvers instead, such as
MinRes, SymmLQ, or GMRES, that can deal with indefinite systems. However,
Hi Krishna,
This is a classic linear algebra result - a symmetric matrix is
positive definite if and only if it has positive pivots. Since this
matrix has a zero block it does not even have a full set of pivots so
it cannot be positive definite.
Thanks,
David Wells
On Sat, Feb 1, 2020 at 9:49
I realize that this question is not exactly about the code/concepts behind
deal.II library itself, but rather about a mathematical statement from
step-20.
*"After assembling the linear system we are faced with the task of solving
it. The problem here is: the matrix has a zero block at the
