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If the dot product is defined as below, how can it be proven that it is distributive over addition?

$$ \mathbf{u} \cdot \mathbf{v} = \frac{1}{2}\bigg(\Vert \mathbf{u} \Vert^2 + \Vert \mathbf{v} \Vert^2 - \Vert \mathbf{u} - \mathbf{v} \Vert^2\bigg) $$

I tried using the polarization identity and using the fact that

$$ \left\{ \begin{array}{c} - \mathbf{u} + \mathbf{v_1} + \mathbf{v_2} = \bigg(\mathbf{v_1} - \frac{1}{2}\mathbf{u}\bigg) + \bigg(\mathbf{v_2} - \frac{1}{2}\mathbf{u}\bigg) \\ \quad \mathbf{u} + \mathbf{v_1} + \mathbf{v_2} = \bigg(\mathbf{v_1} + \frac{1}{2}\mathbf{u}\bigg) + \bigg(\mathbf{v_2} + \frac{1}{2}\mathbf{u}\bigg) \end{array} \right. $$

But still couldn't figure out how to prove it.

Lucas Cruz
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    Hi, welcome. On this site it's expected that you show your work so and explain where you are confused, instead of pasting the problem and requesting a solution. A hint: do some research on the “polarization identity” for normed linear spaces – Matthew Leingang Jul 26 '21 at 13:55
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    Hi @MatthewLeingang, thank you for the helpful comments. I included what I've tried so far to the original question. – Lucas Cruz Jul 26 '21 at 14:09
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    A normed real vector space is a vector space $V$ over $\Bbb R$ with a function $f:V\to [0,\infty)$ such that (i) $f(v)=0$ iff $v=0, ; $(ii) $f(rv)=|r|f(v)$ when $r\in \Bbb R;$ (iii) $f(u+v)\le f(u)+f(v).$ We usually write $|u|$ for $f(u).$ The formula for $u\cdot v$ in your Q will often NOT be linear in $u$ or $v$ and hence not distributive over $+$. Without more info about which vector space and which norm, we cannot proceed. – DanielWainfleet Jul 26 '21 at 14:30
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    You have to assume the norm satisfies the parallelogram law. Even then, the claim is not obvious. See Norms Induced by Inner Products and the Parallelogram Law. – peek-a-boo Jul 26 '21 at 14:35
  • @DanielWainfleet This just happens to be an exercise I came across in an undergraduate textbook on vectors and tensors. I can't quite tell whether the definitions given there would be sufficient for result to follow. There a vector was defined as the set of all arrows equivalent to a given arrow. An arrow was defined as an ordered pair of points, and two arrows were defined equivalent if one could be brought into coincidence with the other by a parallel translation. Finally, the length of a vector was defined as the length of any of its arrows. – Lucas Cruz Jul 26 '21 at 14:45
  • @peek-a-boo Thanks for referencing that question. I noticed that Step 2 of the answer given by t.b. answers my question. – Lucas Cruz Jul 26 '21 at 14:47

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