Path: news.mathworks.com!not-for-mail From: "Bruno Luong" <b.luong@fogale.findmycountry> Newsgroups: comp.soft-sys.matlab Subject: Re: intersection of two complex matrices Date: Tue, 23 Nov 2010 13:04:03 +0000 (UTC) Organization: FOGALE nanotech Lines: 33 Message-ID: <icge43$kc6$1@fred.mathworks.com> References: <icg3d0$41j$1@fred.mathworks.com> Reply-To: "Bruno Luong" <b.luong@fogale.findmycountry> NNTP-Posting-Host: webapp-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1290517443 20870 172.30.248.38 (23 Nov 2010 13:04:03 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 23 Nov 2010 13:04:03 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 390839 Xref: news.mathworks.com comp.soft-sys.matlab:689296 "Frank " <allinone_2003@yahoo.com.hk> wrote in message <icg3d0$41j$1@fred.mathworks.com>... > Hi all, > > I have a question about the common subspaces of two complex matrices and hope that you can help. > > I have two complex matrices, A and B both with size K by (K-1). > How can I obtain the intersection subspace of them? > In other words, I want to obtain a complex matrix R with size K by C satisfying the following: > For all x, there is a y, such that Rx=Ay and similarly, for all z, there is a w such that Rz = Bw. > Furthermore, if both A and B have full rank, can we say that C is at least (K-2) or somethings like that? Take QA/QB respectively orthonormal basis of A and B (using orth()), then use null on Q = [QA QB] The nullspace of N=null([QA QB]) has k columns, it's the dimension of span<A> intersect span B has two parts NA=N(size(QA,2),:) NB=N(size(QA,2)+1:end,:) and QA*NA+QB*NB = 0 i.e. span<QA*NA> is the span <QB*B> QA*NA = A*(A\QA)*NA QB*NB = B*(A\QB)*NB If {y} = (A\QA)*NA and {w} = (A\QB)*NB is what you are looking for. Bruno