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 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.
Yes, nice. This also explains the number of positive and negative
eigenvalues: As many positive eigenvalues as there are unknowns
represented by the A block (in the context of the mixed Laplace, as many
as there are 'u' variables), and as many negative eigenvalues as the
number of rows in the bottom left block (as many as there are 'p'
variables).
Best
W.
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Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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