# 回复： Re: [gmx-users] 2D projection and corresponding conformation

xi zhao zhaoxiitc2002 at yahoo.com.cn
Thu Jun 12 02:31:52 CEST 2008

```Dear sir :
I also know these,but can you give some detail procedure for obtaining the corresponding conformations? Thank you in advance!

Mark Abraham <Mark.Abraham at anu.edu.au> 写道：
xi zhao wrote:
> Dear user:
> We know that observing the sampled conformations in the subspace spanned
> by the eigenvectors is a so-called two-dimensional projection(2D
> projection), in 2-D projection, each point represents a snapshot from
> the simulation, and the distribution shows the sampled region along the
> first two eigenvectors during the simulation. But I feel confounded,
> because I do not know to how to obtain corresponding conformation for
> each point in the 2-D projection. Some papers can show these
> corresponding conformation. Please help me!

Well, if you're after the original 3N-dimensional structure for a given
projected point that is known to correspond to an original structure,
then you need to arrange for there to be a mapping from original to
projected, and then you can apply the reverse mapping. Very likely, once
you've constructed the eigenvectors, the code that is plotting these
projected points in 2D space will just take the original structures in
order and produce the projected points in the same order. Now the
reverse mapping is trivial.

If you're after a 3N-dimensional structure for an arbitrary point (x1,
x2) in projected space which correspond to eigenectors (v1, v2) in
3N-dimensional space, then you need to construct it from the
eigenvectors yourself, as x1*v1 + x2*v2. This structure won't be
physically realistic, however.

The whole point of the eigendecomposition is that any of the structures
used as input can be constructed from a linear combination of the
eigenvectors. Hopefully, most of the variation is in the first few
eigenvectors so that one can approximate the higher-dimensional space by
a linear combination of a few eigenvectors - and the weights in the
linear combination are members of a low-dimensional space. If this makes
no sense, then reading a few chapters of a linear algebra textbook might
be in order. :-)

Mark
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