Simultaneously orthogonally transform two SPD matrices to tridiagonal form? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:21:38Z http://mathoverflow.net/feeds/question/36537 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36537/simultaneously-orthogonally-transform-two-spd-matrices-to-tridiagonal-form Simultaneously orthogonally transform two SPD matrices to tridiagonal form? Greg 2010-08-24T09:50:46Z 2010-08-24T14:22:59Z <p>Supposing you have two SPD matrices $A,B\in\mathbb{R}^{n\times n}$ are there any known results on the existence or non-existence of a unitary matrix $Q$ such that $Q^\top A Q=T_A$ and $Q^\top B Q=T_B$ are both tridiagonal. If such a transformation exists in general, it is not required for my purposes that it be computable in finitely many steps.</p> <p>I am aware of non-orthogonal congruence transformations which tridiagonalize two matrices. </p> <p>Thanks!</p> <hr> <p>Edit:</p> <p>Thanks for the response. I am familiar with the papers of Tisseur and Garvey et. al, but they are using non-orthogonal transformations. In one paper they use alternating 1D Householder reflectors and matrices of the form $L=I+xy^\top$ to force portions of the leading columns to be in the same space. </p> <p>I tried finding a counter-example from the 3x3 case, but it looks like I have plenty of degrees of freedom to play with and higher dimensions become treacherously difficult to manage individual elements. </p> <p>Maybe this question is equivalent to finding a $Q$ such that for an arbitrary matrix $V$ that $Q^\top V$ is bidiagonal, which certainly looks hopeless to me. </p> http://mathoverflow.net/questions/36537/simultaneously-orthogonally-transform-two-spd-matrices-to-tridiagonal-form/36557#36557 Answer by J. M. for Simultaneously orthogonally transform two SPD matrices to tridiagonal form? J. M. 2010-08-24T14:22:59Z 2010-08-24T14:22:59Z <p>To elaborate on what I said in the comments... the reason I feel there might not be a way to do this is as follows: the problem of simultaneously tridiagonalizing two SPD matrices $A$ and $B$ is equivalent to tridiagonalizing either of $AB^{-1}$ or $A^{-1}B$; unfortunately both of these can be unsymmetric (the product of two symmetric matrices need not be symmetric), and orthogonal (unitary) similarity transformations of an unsymmetric matrix only manage to reduce to Hessenberg form.</p>