Let V be an inner product space. Prove that for every $u,v\in V: (u,v)=\frac{1}{4}\left(||u+v||^{2}-||u-v||^{2}\right)$

Question

Let $V$ be an inner product space. Prove that for every $u,v\in V$: $(u,v)=\frac{1}{4}\left(||u+v||^{2}-||u-v||^{2}\right)$

Answer 1

See that $$|u+v|^2=\langle u+v,u+v\rangle$$

Now expand this inner product...

Do you now see what is $|u-v|^2$??

Can you complete now??

Answer 2

Hint

$$||u\pm v||^2=||u||^2\pm2(u,v)+||v||^2$$

Answer 3

Rough Analogy: $$(a+b)^2=a^2+b^2+2ab\\(a-b)^2=a^2+b^2-2ab\\ab=\frac14[(a+b)^2-(a-b)^2]$$

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$$(\vec a+\vec b)^2=a^2+b^2+2(a,b)\\(\vec a-\vec b)^2=a^2+b^2-2(a,b)\\(a,b)=\frac14[(\vec a+\vec b)^2-(\vec a-\vec b)^2]$$