K-means 是关于方差分析范式的。ANOVA - 单变量和多变量 - 是基于这样一个事实，即关于大质心的平方偏差总和由关于组质心的这种散布和关于大质心的散布组成：SStotal=SSwithin+SSbetween。因此，如果SSwithin被最小化，那么SSbetween被最大化。

已知某些点关于其质心（算术平均值）的偏差 SS 与点之间的总平方欧几里德距离直接相关：与质心的平方偏差之和等于两两平方欧几里德距离之和除以数量的点。（这是centroid的三角特性的直接扩展。并且这种关系也被用于距离矩阵的双中心化。）

因此，说“质心（作为点）的 SSbetween 被最大化”是说“质心之间的（加权）平方距离集被最大化”的别名。

Ni注意：在 SSbetween 每个质心都由该集群中的点数加权i。也就是说，每个质心都被计数Ni。例如，数据中有两个质心，1和2，SSbetween =N1*D1^2+N2*D2^2其中D1和D2是质心与总均值的偏差。前一段中“加权”一词的来源。

例子

Data (N=6: N1=3, N2=2, N3=1)
        V1            V2       Group
       2.06          7.73          1
        .67          5.27          1
       6.62          9.36          1
       3.16          5.23          2
       7.66          1.27          2
       5.59          9.83          3

SSdeviations
        V1            V2            Overall
SSt   37.82993333 + 51.24408333 = 89.07401666
SSw   29.50106667 + 16.31966667 = 45.82073333
SSb   8.328866667 + 34.92441667 = 43.25328333

SSt is directly related to the squared Euclidean distances between the data points:

Matrix of squared Euclidean distances
     .00000000    7.98370000   23.45050000    7.46000000   73.09160000   16.87090000 
    7.98370000     .00000000   52.13060000    6.20170000   64.86010000   45.00000000 
   23.45050000   52.13060000     .00000000   29.02850000   66.52970000    1.28180000 
    7.46000000    6.20170000   29.02850000     .00000000   35.93160000   27.06490000 
   73.09160000   64.86010000   66.52970000   35.93160000     .00000000   77.55850000 
   16.87090000   45.00000000    1.28180000   27.06490000   77.55850000     .00000000 

Its sum/2, the sum of the distances = 534.4441000
534.4441000 / N = 89.07401666 = SSt

The same reasoning holds for SSb.

Matrix of squared Euclidean distances between the 3 group centroids (see https://stats.stackexchange.com/q/148847/3277)
     .00000000   22.92738889   11.76592222 
   22.92738889     .00000000   43.32880000 
   11.76592222   43.32880000     .00000000 

3 centroids are 3 points, but SSb is based on N points (propagated centroids):
N1 points representing centroid1, N2 points representing centroid2 and N3 representing centroid3.
Therefore the sum of the distances must be weighted accordingly:
N1*N2*22.92738889 + N1*N3*11.76592222 + N2*N3*43.32880000 = 259.51969998
259.51969998 / N = 43.25328333 = SSb

Moral in words: maximizing SSb is equivalent to maximizing
the weighted sum of pairwise squared distances between the centroids.
(And maximizing SSb corresponds to minimizing SSw, since SSt is constant.)

您正在寻找的是König-Huygens 定理。