Trying to make sense of how to prove this equality but getting a bit confused with the algebra.

Define a = $\frac{1}{m}\sum_{j=1}^{m}x_j$. What I seek to prove is that $\forall y \in V$, where V is a vector space, we have, $$\|y-a\|^2 = \frac{1}{m}\sum_{j=1}^{m}(\|y-x_j\|^2 - \|a-x_j\|^2)$$

The only thing we know about the norm that it is inner product induced. I've tried manipulating (assuming the euclidean norm) but at a complete loss and haven't been able to get anywhere. Would love some help.


For the left-hand side, we have

$$ \begin {align*} \left| y-a \right|^2 &= \left< y-a, y-a \right> \\ &= |y|^2 + |a|^2 - 2\left<y,a\right> \end {align*}$$

Now start with the right-side and see if you can make it look like this:

$$ \begin {align*} \frac{1}{m} &= \sum_j \left( |y-x_j|^2 - |a-x_j|^2 \right) \\ &= \frac{1}{m} \sum_j \left( \left<y-x_j,y-x_j\right> - \left<a-x_j,a-x_j\right> \right) \\ &= \frac{1}{m} \sum_j \left( |y|^2 + |x_j|^2 - 2\left<y,x_j\right> - |a|^2 - |x_j|^2 + 2\left<a,x_j\right> \right) \\ &= \frac{1}{m} \sum_j \left( |y|^2 - 2\left<y,x_j\right> - |a|^2 + 2\left<a,x_j\right> \right) \\ &= |y|^2 - |a|^2 - 2\left<y,a\right>+ \frac{2}{m} \sum_j \left<a,x_j\right> \end {align*} $$

Note that the last sum is equal to $2|a|^2$, so it all adds up.


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