Bandwidth reduction of multiple matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T18:00:46Z http://mathoverflow.net/feeds/question/93337 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93337/bandwidth-reduction-of-multiple-matrices Bandwidth reduction of multiple matrices Greg von Winckel 2012-04-06T16:35:06Z 2012-06-16T13:22:00Z <p>Suppose I have a symmetric matrix A, and several diagonal matrices \$D_1,D_2,...\$ </p> <p>Are there any matrix transformations, such as \$P^\top A P\$ so that \$P^\top AP\$, \$P^\top D_1 P\$, \$P^\top D_2 P\$, etc are either all tridiagonal, or all have minimimal bandwidth in some sense? If for example, I only had \$D_1\$ then solving the generalized eigenvalue problem for the matrix pencil \$A,D_1\$ would give me a basis that simultaneously diagonalizes both \$A\$ and \$D_1\$. Obviously, one basis will not simultaneously diagonalize more matrices in general, but a set of banded matrices would also be pretty nice. </p> <p>I am aware of Tisseur and Garvey's papers on simultaneous tridiagonalization of two matrices. </p> http://mathoverflow.net/questions/93337/bandwidth-reduction-of-multiple-matrices/93395#93395 Answer by Bart for Bandwidth reduction of multiple matrices Bart 2012-04-07T08:25:00Z 2012-04-07T08:25:00Z <p>If you use the typical reorderings (like reverse Cuthill-McKee ordering), \$P^T A P\$ will have a smaller bandwidth (in general, not tridiagonal though). Since \$P\$ is a permutation matrix, all \$P^T D_i P\$ will remain diagonal too.</p> <p>Bart</p>