most recent 30 from http://mathoverflow.net 2013-06-19T03:50:12Z http://mathoverflow.net/feeds/question/56641 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56641/decomposition-of-an-orthogonal-matrix GuillaumeThomas 2011-02-25T15:46:34Z 2011-02-25T22:07:59Z

<p>Hi,</p> <p>I have a matrix : $W=I+U^TV$ </p> <ul> <li>$dim(W)=(D,D)$ </li> <li>$dim(U)=dim(V)=(N,D)$ with $N &lt; &lt; D$ </li> </ul> <p>I need it to be orthogonal ie $W^TW=I$</p> <p>which gives me : $V^TU+U^TV+V^TUU^TV=0$ </p> <p>From that point, i don't know where to go. Have anyone got some ideas about that issue ?</p> <p>Cheers</p> <p>Guillaume</p>

http://mathoverflow.net/questions/56641/decomposition-of-an-orthogonal-matrix/56652#56652 Jeff Schenker 2011-02-25T17:18:44Z 2011-02-25T17:18:44Z

<p>Do you want a general solution or simply a good family of examples? If you want examples, let $V=-U=\sqrt{2} S^T$ where $S$ is an isometry from $\mathbb{R}^N$ into $\mathbb{R}^D$.</p>

http://mathoverflow.net/questions/56641/decomposition-of-an-orthogonal-matrix/56684#56684 user11000 2011-02-25T22:07:59Z 2011-02-25T22:07:59Z

<p>This should be a comment. I think this question is not apt here. You should consider posting questions like these at <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange</a>.</p> <p>Given $U$ and if it is full rank, we can choose $$V = -2 \left(UU^T \right)^{-1}U$$ and the matrix $$W = I - 2 U^T \left(UU^T \right)^{-1}U$$ is an orthonormal matrix. A special case of this is provided by Jeff.</p>