Let $a^{(1)},...,a^{(k)} \in \mathbb{R}^n$.

How can one find the point $x \in \mathbb{R}^n$, which minimizes the sum of distance squares $\sum_{\mathcal{l}=1}^{k} \Vert x-a^{\mathcal{(l)}} \Vert _2^2$

I know that a function $h(t) = \sum_{l=1}^{k}(t-x_i)^2$ has its minimum when $t = \overline{x}$. So for the average $\overline{x}$ of the numbers $x_1,...,x_n$ the sum of squares of the deviations $\overline{x}-x_i$ is minimal.

So to get the center of a set of points

$$ S=\{(x_1,y_1),(x_2,y_2),\dots (x_n,y_n)\}$$ we can get their centroid by $$(\bar x,\bar y) = \left(\frac{1}{n}\sum_{i=0}^n x_i, \frac{1}{n}\sum_{i=0}^n y_i\right).$$

I don't really know if this point actually minimizes the sum of distance squares given above. I'd also like to know if only one such point exists or if there are more.


I believe your intuition is correct.

Consider the following:

$$ \sum_{l=1}^k || x - a^{(l)}||_2^2 = \sum_{l=1}^k \sum_{m=1}^n (x_m - a^{(l)}_m)^2 $$

Where $a_m^{(l)}$ is the $m$th component of the $l$th vector, and $x_m$ is the $m$th component of $x$.

To find the minimum of this with respect to $x$, we can use standard differentiation procedures.

First, differentiate with respect to one direction (i.e. differentiate with respect to some $x_j$): $$ \begin{align} \frac{\partial}{\partial x_j} \sum_{l=1}^k \sum_{m=1}^n (x_i - a^{(l)}_m)^2 &=\sum_{l=1}^k \sum_{m=1}^n \frac{\partial}{\partial x_j} (x_m - a^{(l)}_m)^2 \\ &=\sum_{l=1}^k 2(x_j-a_j^{(l)}) \end{align} $$ Setting this expression to zero you will obtain: $$ \begin{align} \sum_{l=1}^k 2(\hat{x}_j-a_j^{(l)}) &= 0\\ \hat{x}_j &= \frac{1}{k}\sum_{l=1}^k a_j^{(l)} \end{align} $$

This shows that for any direction $j$, there is a stationary point of the function $\sum_{l=1}^k || x - a^{(l)}||_2^2$ (which we will show is a global minimum) given by the average of the $j$th component of the vectors $\{a\}_{l=1}^k$.

Now, to show that this is a unique global minimum, we will use the fact that a function $f(x)$ is strongly convex (which implies that $f(x)$ has a unique minimum point) if its Hessian $H$ (matrix containing the second derivatives) is positive definite (We say that $H$ is positive definite if $\forall c \in \mathbb{R}^n/\{0\}, c^THc > 0$).

Consider the second derivatives of $\sum_{l=1}^k || x - a^{(l)}||_2^2$:

$$ \begin{align} \frac{\partial^2}{\partial_i\partial x_j} \sum_{l=1}^k \sum_{m=1}^n (x_m - a^{(l)}_m)^2 &=\sum_{l=1}^k \frac{\partial}{\partial x_i}2(\hat{x}_j-a_j^{(l)})\\ &=\begin{cases} 2 & i=j\\ 0 & i\neq j \end{cases} \end{align} $$

This means that the Hessian $H$ will be diagonal, and that there will be strictly positive entries on the diagonal. Thus $H$ is positive definite, implying that $\sum_{l=1}^k || x - a^{(l)}||_2^2$ is strongly convex with respect to $x$.

Hence, we can conclude that the stationary point which we found above is indeed a unique minimum.


You can use this relation: $$ \sum_{i=1}^n \|x-a_i\|^2=n\|\bar{a}-x\|^2+\sum_{i=1}^n \|\bar{a}-a_i\|^2 $$ where $\bar{a}=\frac{1}{n}\sum_{i=1}^n a_i$

If you want to minimize w.r.t. $x$, just take $x=\bar{a}$ to cancel the $\|\bar{a}-x\|^2$ term. Therefore the answer is: $$ x=\bar{a}=\frac{1}{n}\sum_{i=1}^n a_i $$

To prove the first relation, you must notice that: $$ \sum_i^{n}\|a_i-\bar{a}\|^2=\dots=\left(\sum_i^{n}\|a_i\|^2\right)-n\|\bar{a}\|^2 $$ then simply develop $$ \sum_{i=1}^n \|x-a_i\|^2=\sum_{i=1}^n (\|x\|^2-2\langle x,a_i \rangle+\|a_i\|^2)=\dots $$ (I can write the details if you want, just ask)


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