<div>Dear sir :</div> <div>I also know these,but can you give some detail procedure for obtaining the corresponding conformations? Thank you in advance! <BR><BR><B><I>Mark Abraham <Mark.Abraham@anu.edu.au></I></B> дµÀ£º</div> <BLOCKQUOTE class=replbq style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid">xi zhao wrote:<BR>> Dear user:<BR>> We know that observing the sampled conformations in the subspace spanned <BR>> by the eigenvectors is a so-called two-dimensional projection(2D <BR>> projection), in 2-D projection, each point represents a snapshot from <BR>> the simulation, and the distribution shows the sampled region along the <BR>> first two eigenvectors during the simulation. But I feel confounded, <BR>> because I do not know to how to obtain corresponding conformation for <BR>> each point in the 2-D projection. Some papers can show these <BR>> corresponding conformation. Please help me!<BR><BR>Well,
if you're after the original 3N-dimensional structure for a given<BR>projected point that is known to correspond to an original structure,<BR>then you need to arrange for there to be a mapping from original to<BR>projected, and then you can apply the reverse mapping. Very likely, once<BR>you've constructed the eigenvectors, the code that is plotting these<BR>projected points in 2D space will just take the original structures in<BR>order and produce the projected points in the same order. Now the<BR>reverse mapping is trivial.<BR><BR>If you're after a 3N-dimensional structure for an arbitrary point (x1,<BR>x2) in projected space which correspond to eigenectors (v1, v2) in<BR>3N-dimensional space, then you need to construct it from the<BR>eigenvectors yourself, as x1*v1 + x2*v2. This structure won't be<BR>physically realistic, however.<BR><BR>The whole point of the eigendecomposition is that any of the structures<BR>used as input can be constructed from a linear combination
of the<BR>eigenvectors. Hopefully, most of the variation is in the first few<BR>eigenvectors so that one can approximate the higher-dimensional space by<BR>a linear combination of a few eigenvectors - and the weights in the<BR>linear combination are members of a low-dimensional space. If this makes<BR>no sense, then reading a few chapters of a linear algebra textbook might<BR>be in order. :-)<BR><BR>Mark<BR>_______________________________________________<BR>gmx-users mailing list gmx-users@gromacs.org<BR>http://www.gromacs.org/mailman/listinfo/gmx-users<BR>Please search the archive at http://www.gromacs.org/search before posting!<BR>Please don't post (un)subscribe requests to the list. Use the <BR>www interface or send it to gmx-users-request@gromacs.org.<BR>Can't post? Read http://www.gromacs.org/mailing_lists/users.php<BR></BLOCKQUOTE><BR><BR><BR><br><a rel="nofollow" class="plink" target="_blank"
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