Skew-symmetric matrix representation
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From: Kaba on 16 Jun 2010 19:36 Hi, Given a skew-symmetric real (n x n)-matrix A, and a real orthogonal (n x n)-matrix Q, is it possible to find a real (n x n)-matrix B such that: A = Q^T B - B^T Q ? -- http://kaba.hilvi.org
From: Rob Johnson on 16 Jun 2010 20:16 In article <MPG.26838c5c17078b909898e1(a)news.cc.tut.fi>, Kaba <none(a)here.com> wrote: >Given a skew-symmetric real (n x n)-matrix A, and a real orthogonal >(n x n)-matrix Q, is it possible to find a real (n x n)-matrix B such >that: > >A = Q^T B - B^T Q Yes, there are several. For example, let U be the upper triangular part of A; then, B = (Q^T)^{-1}U. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Kaba on 16 Jun 2010 20:44 Rob Johnson wrote: > In article <MPG.26838c5c17078b909898e1(a)news.cc.tut.fi>, > Kaba <none(a)here.com> wrote: > >Given a skew-symmetric real (n x n)-matrix A, and a real orthogonal > >(n x n)-matrix Q, is it possible to find a real (n x n)-matrix B such > >that: > > > >A = Q^T B - B^T Q > > Yes, there are several. For example, let U be the upper triangular > part of A; then, B = (Q^T)^{-1}U. Right. Meanwhile I found: Q^T B can be decomposed into a sum of a skew-symmetric matrix X and symmetric matrix S: Q^T B = X + S where X = 0.5 [QB - (QB)^T] S = 0.5 [QB + (QB)^T] Therefore B = Q(X + S) Q^T B - B^T Q = Q^T Q (X + S) - (X^T + S^T) Q^T Q = (X + S) - (X^T + S^T) = 2X For equality to hold: A = 2X => X = A / 2 Thus: A = Q^T B - B^T Q <=> B = Q(0.5 A + S) [] This is actually related to reproving the conformal affine regression, the one you wrote about... -- http://kaba.hilvi.org
From: Kaba on 16 Jun 2010 20:46 Kaba wrote: > Q^T B can be decomposed into a sum of a skew-symmetric matrix X and > symmetric matrix S: > > Q^T B = X + S > > where > X = 0.5 [QB - (QB)^T] > S = 0.5 [QB + (QB)^T] That should be X = 0.5 [Q^T B - (Q^T B)^T] S = 0.5 [Q^T B + (Q^T B)^T] But fortunately this error does not affect the proof, since I never expand X or S. -- http://kaba.hilvi.org
From: Kaba on 16 Jun 2010 20:49
Kaba wrote: > Thus: > > A = Q^T B - B^T Q > <=> > B = Q(0.5 A + S) And this should be: A = Q^T B - B^T Q <=> B = Q(0.5 A + R) for any symmetric R, not just for the specified S. -- http://kaba.hilvi.org |