On Jun 4, 10:56 pm, Feng <[EMAIL PROTECTED]> wrote:
> Hi Kunzmilan, thanks for your idea of using distance matrices. But one
> of my friends came up with a seemly counter-example:
>
> Take 3 collinear points in 2D: (0,0), (1,0), (2,0).
> The distance matrix is:
> 0 1 4
> 1 0 1
> 4 1 0,
> whose eigenvalues are -4, -0.4495, 4.4495. It means that they form a
> 2D shape, but they make a line (1D shape).
>
> Is there anything wrong in it?
>
I tried to answer the question five days ago, but my answer did not
appeared.
The eigenvalue a at straight chains is produced by the reflexion plane
(elements of the of the eigenvector are symmetrical to the center of
the chain) and its rotation tensor b = (a + W/2) = [\sum d^4 - 3/4
W^2]^1/2, where W is The Wiener index (half of the sum of distance
matrix elements.
The sum of squared eigenvalues must be equal to the trace of the
squared matrix.
Solving the quadratic equation gives four eigenvalues (including zero)
as W/2 +/- (b or W/2).
kunzmilan
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