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A non-Euclidean three-point dissimilarity produces a negative MDS eigenvalue

Example

Take the symmetric dissimilarity matrix

D=(011103130).

It is nonnegative with zero diagonal and is not Euclidean because D13+D12=2<D23=3. Its classical-MDS matrix B=−12H3D∘2H3 has a negative eigenvalue, so positive-eigenvalue truncation is not exact distance recovery.

Facts & Assumptions

Given: The stated dissimilarity matrix D and the classical-MDS construction.

[F1]

Classical MDS squares the entries of D and forms B=−12HnD∘2Hn (Classical multidimensional scaling from a finite dissimilarity matrix).

[F2]

A Euclidean distance matrix yields a positive-semidefinite classical-MDS matrix; the general symmetric matrix has a real orthonormal eigendecomposition (Double-centering Euclidean squared distances recovers their centered Gram matrix).

Verification

1.1givenF1algebra

Put u=(1,1,−2)T and v=(1,−1,0)T. Both are zero-sum, so H3u=u and H3v=v. Here D∘2=(011109190), so direct multiplication gives uTBu=−12uTD∘2u=19, vTBv=1, and uTBv=−12uTD∘2v=−8. Thus w=u+8v satisfies wTBw=19+16(−8)+64=−45<0.

2.1step 1.1F2algebra

By [F2], B is symmetric and has an orthonormal eigenbasis. If every eigenvalue were nonnegative, expressing w in that eigenbasis would give wTBw≥0, contradicting step 1.1. Therefore B has a negative eigenvalue.

3.1givenstep 2.1F2∎

The stated triangle inequality fails, since D13+D12=2<D23=3, so no Euclidean point configuration realizes D. This matches the negative eigenvalue and the Euclidean-only exactness hypothesis of [F2].

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