Minimum Sum of Squared Distances

Let's prove that the minimum is at $\mathbf{x} = \overline{\mathbf{v}}$ .

Let $\{\hat{e_1}, \hat{e_2}, \ldots, \hat{e_m}\}$ be the canonical basis for $\mathbb{R}^m$ . Then $\mathbf{x}=\displaystyle \sum_{j=1}^{m} x_j \hat{e_j}$ and $\mathbf{v}_i=\displaystyle \sum_{j=1}^{m} v_{i,j} \hat{e_j}$ .

We have: $\begin{aligned} f(\mathbf{x}) &= \sum_{i=1}^{n} (\mathbf{x} - \mathbf{v}_i) \cdot (\mathbf{x} - \mathbf{v}_i) \\ &= n \mathbf{x} \cdot \mathbf{x} - 2\mathbf{x} \cdot \mathbf{v}_i + \mathbf{v}_i \cdot \mathbf{v}_i \\ &= n \sum_{j=1}^{m} x_j^2 - 2\sum_{i=1}^{n} \sum_{j=1}^{m} x_j v_{i,j} + \sum_{i=1}^{n} \sum_{j=1}^{m} v_{i,j}^2 \end{aligned}$ Take the partial derivative with respect to $x_k$ and set it equal to zero, where $1 \leq k \leq m$ : $\begin{aligned} \dfrac{\partial f}{\partial x_k} &= 2n x_k - 2\sum_{i=1}^{n} v_{i,k} = 0 \\ \implies x_k &= \dfrac{1}{n} \sum_{i=1}^{n} v_{i,k} \end{aligned}$ Finally, the point that will minimize $f(\mathbf{x})$ will be: $\begin{aligned} \overline{\mathbf{v}} &= \sum_{j=1}^{m} x_j \hat{e_j} \\ &= \sum_{j=1}^{m} \left( \dfrac{1}{n} \sum_{i=1}^{n} v_{i,j} \right) \hat{e_j} \\ &= \dfrac{1}{n} \sum_{i=1}^{n} \sum_{j=1}^{m} v_{i,j} \hat{e_j} \\ &= \dfrac{1}{n} \sum_{i=1}^{n} \mathbf{v}_i \end{aligned}$