2
$\begingroup$

I just studied the inner product space and wanted to find out any geometric explanation of the equality:

$\langle x,y\rangle=\frac{||x||^2+||y||^2-||x-y||^2}{2}. $

I know that the scalar product gives the length of the orthogonal projection of one vector $x$ over the other vector $y$. Furthermore I understand the geometric picture of the distributive law, which has to be applied twice:

$||x-y||^2=\langle x-y,x-y\rangle =\langle x,x-y\rangle-\langle y ,x-y\rangle= ||x||^2+||y||^2-2\langle x,y\rangle.$

Nevertheless, this does not convince myself to get an global geometric picture/explanation of my first equality.

Is there anyone who has it clear? Thank you very much.

$\endgroup$
  • $\begingroup$ Not $\langle x-y\rangle$ but $\langle x,y\rangle$ for the ending term. $\endgroup$ – Jean Marie Nov 8 '16 at 9:04
  • $\begingroup$ Having an intuitive idea of what is the scalar product even in $\mathbb{R^2}$ is hard. It is a tool, that's my vision. The best proof that it is not "evident" is that it has appeared very lately as a mathematical concept (circa 1900). $\endgroup$ – Jean Marie Nov 8 '16 at 9:08
  • $\begingroup$ @JeanMarie Thank you I have changed it. And yes, I totally agree with you. $\endgroup$ – Manuel Nov 8 '16 at 10:07
3
$\begingroup$

That formula is nothing but the cosine law, because vector $x-y$ is the third side of a triangle of sides $x$ and $y$: $$ |x-y|^2=|x|^2+|y|^2-2|x||y|\cos\theta= |x|^2+|y|^2-2\langle x,y\rangle. $$ And cosine law has a geometric explanation, see here for instance.

$\endgroup$
0
$\begingroup$

Please bear in mind that while the scalar product has a geometric motivation, it is generally hard to understand why $\langle x, y \rangle = k $ except for when $x \perp y $ and when $x $ is a scalar multiple of $y $.

Having that said, remember that if you want to project $x $ over some vector $u $ then the projection $p $ is:

$$\frac {\langle x, u \rangle}{||u||^2}u = p $$

And therefore the only way your identity could make sense is, in my opinion, by substituting in the formula above and check that it can reduce to the same thing. Perhaps you could start by checking heuristically that the formula holds for orthogonal $x, y $ and for parallel $x, y $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.