# 2 Main Result

Stationary points can be local minimum points or saddle points. The only local maximum point of stress is at \(X=0\) (De Leeuw (1993)). Among the local minimum points there are one or more global minimum points. A sufficient condition for a local minimum to be global is that at the stationary point we have \(V-B(X)\gtrsim 0\), or \(V^+B(X)\lesssim I\) (De Leeuw (2016)). This is a very restrictive condition which we generally do not expect to to be true. There is, however, a much weaker relation between stationary points and global minima on a subspace.

**Theorem 1: [Local-Global]** If \(X\in\mathbb{R}^{n\times p}\) is a regular stationary point of the MDS problem then \[
\min_{T\in\mathbb{R}^{p\times p}}\sigma(XT)=\sigma(X).
\] **Proof:** First, observe that \[
d_{ij}(XT)=\sqrt{\text{tr}\ X'A_{ij}XTT'}
\] which is the square root of a non-negative linear function of \(S=TT'\), and is consequently concave in \(S\). It follows that \[
\sigma(XT)=1-\text{tr}\ X'B(XT)XS+\frac12\text{tr}\ X'VXS
\] is convex in \(S\). From Rockafellar (1970), theorem 31.4, the minimum over \(S\gtrsim 0\) is attained at a unique point where

- \(S\gtrsim 0\).
- \(X'(V-B(XT))X\gtrsim 0\).
- \(\text{tr}\ X'(V-B(XT)X)XS=0\).

But if \(X\) is a stationary point of the MDS problem we have \((V-B(X))X=0\). Thus the minimum over \(S\) is attained at \(S=I\), and the minimum over \(T\) is attained at any rotation matrix \(T\) with \(T'T=TT'=I\), which is what the theorem says. \(\blacksquare\)

The part of theorem 1 where it is shown that \(\sigma(XT)\) has a unique (and thus global) minimum over \(T\) for fixed \(X\) is mentioned in Borg and Groenen (2005), p 283. I merely added the result that the unique minimizer \(T\) is necessarily a rotation matrix if \(X\) is a stationary point. Note that the stationary point \(X\) may be a saddle point, it does not have to be a local minimum point.

An important special case of the theorem is full-dimensional scaling (De Leeuw (1993), De Leeuw, Groenen, and Mair (2016)), in which \(p=n\).

**Corollary 1: [Full]** If \(X\in\mathbb{R}^{n\times n}\) is a stationary point of the MDS problem then it is the unique global minimum.

**Proof:** In this case \[
\min_{T\in\mathbb{R}^{n\times n}}\sigma(XT)=\min_{Z\in\mathbb{R}^{n\times p}}\sigma(Z).
\] By theorem 1 consequently at a stationary point \(X\) \[
\sigma(X)=\min_{Z\in\mathbb{R}^{n\times p}}\sigma(Z).
\] \(\blacksquare\)