### a

Proof of Equation (10.12) $\frac{1}{|C_k|} \sum\limits_{i,i^{\prime} \in C_k} \sum\limits_{j=1}^p (x_{ij} - x_{i^\prime j})^2 = 2 \sum\limits_{i \in C_k} \sum\limits_{j=1}^{p} (x_{ij} - \bar{x}_{kj})^2 \\ = \frac{1}{|C_k|} \sum\limits_{i,i^{\prime} \in C_k} \sum\limits_{j=1}^p ((x_{ij} - \bar{x}_{kj}) - (x_{i^\prime j} - \bar{x}_{kj}))^2 \\ = \frac{1}{|C_k|} \sum\limits_{i,i^{\prime} \in C_k} \sum\limits_{j=1}^p ((x_{ij} - \bar{x}_{kj})^2 - 2 (x_{ij} - \bar{x}_{kj})(x_{i^\prime j} - \bar{x}_{kj}) + (x_{i^\prime j} - \bar{x}_{kj})^2) \\ = \frac{|C_k|}{|C_k|} \sum\limits_{i \in C_k} \sum\limits_{j=1}^p (x_{ij} - \bar{x}_{kj})^2 + \frac{|C_k|}{|C_k|} \sum\limits_{i^{\prime} \in C_k} \sum\limits_{j=1}^p (x_{i^\prime j} - \bar{x}_{kj})^2 - \frac{2}{|C_k|} \sum\limits_{i,i^{\prime} \in C_k} \sum\limits_{j=1}^p (x_{ij} - \bar{x}_{kj})(x_{i^\prime j} - \bar{x}_{kj}) \\ = 2 \sum\limits_{i \in C_k} \sum\limits_{j=1}^p (x_{ij} - \bar{x}_{kj})^2 + 0$

### b

Equation (10.12) shows that minimizing the sum of the squared Euclidean distance for each cluster is the same as minimizing the within-cluster variance for each cluster.