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For a finite dataset, alternate nearest-centre assignment and replacement of each centre by the centroid of its assigned observations. Keep the current assignment whenever an observation is tied between nearest centres. A changed assignment then strictly decreases the within-cluster sum of squared Euclidean distances; updating centres cannot increase it by the centroid sum-of-squares identity.

After each centre update, the centres are determined by the current partition of a set of observation indices. There are only finitely many such partitions, so a strictly improving sequence cannot revisit one and must stop. A convenient exact stopping test is to check after each update whether every observation is still assigned to a nearest centre. This proof is for algorithm termination; it does not guarantee a global minimum.

Empty clusters need an explicit convention. To retain exactly $K\le n$ nonempty clusters, transfer an observation from a cluster with at least two members to each empty cluster, making it a singleton centred at that observation. This cannot increase the objective: the transferred observation now has squared distance zero, and no other contribution changes before centre updating. A deterministic choice is an eligible observation farthest from its currently assigned centre, with ties resolved by index. After the first round, an empty cluster can arise only during a strictly improving reassignment, so these repairs preserve the finite-termination argument.

Write $n_k=|C_k|$ and suppose the cluster is nonempty. Expanding the Euclidean norm through its inner product givesSum over $x\in C_k$. The final term does not depend on $x$, so its sum is $n_k\Vert z-{\bar{x}}_k{\Vert }^2$, which proves the requested identity. This expansion is the starting point for the centroid sum-of-squares identity.

Combining the first two parts gives the centroid sum-of-squares identityThe added term is nonnegative and, since $n_k>0$, vanishes exactly at $z={\bar{x}}_k$. Thus the unique minimizing centroid in this step of K-means clustering isAn empty cluster needs a separate algorithmic convention because its sample mean is undefined and its objective is identically zero.

Use the corrected decompositionThe hint's minus sign before $({\mu }_j-w_j)$ is a typographical error. Squaring, summing, and using ${\sum }_i(y_{ij}-{\mu }_j)=0$ for each coordinate gives the centroid sum-of-squares identitySince $m>0$ and $w\ne \mu$, the final term is strictly positive. HenceThus the centroid uniquely minimizes the sum of squared Euclidean distances to the observations.

Write $x_i\in {\mathbb{R}}^p$ for row $i$ of the data matrix. The k-means clustering objective is the within-cluster sum of squared Euclidean distances:The nonempty $C_k$ form a partition of a set of observation indices. For fixed clusters, part (a) shows that the best $m_k$ is their centroid.

Initialize distinct centres, for example by choosing $K\le n$ distinct observations. Then repeat the following steps:

The empty-cluster repair is possible for $K\le n$ and cannot increase $J$: the transferred observation now has squared distance zero, while all other squared distances stay unchanged before the centre update.

Assignment cannot increase $J$, and any changed assignment strictly decreases it under the tie rule. Updating centres cannot increase $J$ by the centroid sum-of-squares identity. After an update, the centres are determined by the partition of a set. There are only finitely many possible assignments. Unless this is the initial round, any nonterminal round starts with an assignment that strictly reduces $J$; repairing empty clusters and updating cannot undo that decrease. Thus a partition cannot recur, proving finite termination of K-means clustering. The final result need not be a global minimum; different initial centres can lead to different final clusterings.

For observations ${\mathbf{x}}_i\in {\mathbb{R}}^p$, K-means clustering seeks nonempty clusters in cluster analysis $C_1,\dots ,C_K$ and centroids ${\mathbf{m}}_k$ minimizingAt fixed centroids, assigning each point to a nearest centroid minimizes its squared Euclidean distance. At fixed assignments, the centroid sum-of-squares identity shows that the sample mean uniquely minimizes each group's contribution. These alternating steps cannot increase $W$.

For finite termination of K-means clustering, more than a lower bound on $W$ is needed. Keep an assignment unchanged when nearest centroids tie, and stop when assignments no longer change. Every change then strictly decreases $W$. After updating the centroids, $W$ is determined by the partition of a set of $n$ indices; there are only finitely many choices for that partition of a set, so no strictly improving partition of a set can recur. Empty clusters in cluster analysis require a convention, such as moving a point from a nonsingleton cluster in cluster analysis to become its own centroid, which cannot increase $W$. With such a consistent convention the algorithm terminates, although it may reach only a local minimum.