Question

I'm clueless on the reasoning of this:
We can define norm of a vector v as:
||v||=sqrt() (Where is an inner product of v and v.) only if this is true: ||u+v||^2+||u-v||^2=2(||u||^2+||v||^2) I can sort of see the geometry behind it, but I'd like to understand it more and would be very grateful if anybody pointed me in the right direction.

Answer 1

I'm trying to see the relationship between \(\vec u\) and \(\vec v\).

Answer 2

u and v are any two vectors in the inner product space

Answer 3

In that case, the identity\[\vec v\cdot\vec v=|\vec v|^2\]comes in handy:\[|\vec u+\vec v|^2+|\vec u-\vec v|^2=2(|\vec u|^2+|\vec v|^2)\\(\vec u+\vec v)\cdot(\vec u+\vec v)+(\vec u-\vec v)\cdot(\vec u-\vec v)=2(\vec u\cdot\vec u+\vec v\cdot\vec v)\\\vec u\cdot\vec u+2\vec u\cdot\vec v+\vec v\cdot\vec v+\vec u\cdot\vec u-2\vec u\cdot\vec v+\vec v\cdot\vec v=2\vec u\cdot\vec u+2\vec v\cdot\vec v\\2\vec u\cdot\vec u+2\vec v\cdot\vec v=2\vec u\cdot\vec u+2\vec v\cdot\vec v\]

Answer 4

it is just a guess, but maybe this ondition has somehing to do about the vector space being closed under + and - operations

Answer 5

Then you can extrapolate this idea into your case.

Answer 6

across: Thanks for the input. I'm guessing my question is why wouldn't that be true? Under what conditions?

Answer 7

That's an interesting question. I can't think of an example for which the equality doesn't hold. I'm thinking perhaps this is a restriction to Euclidean geometry.

Answer 8

I just found a comment on this in one textbook, actually (my translation to english): notice that this condition is a generalisation of Pythagorean theorem. It only says that the sum of lengths of both diagonals squared of any rhomboid (that's what a dictionary told me, never head of that word) equals the sum of lengths of all its sides squared.
That is, the squaring occurs before, pardon my english - lengths a^2+b^2+c^2+d^2. Same goes for diagonals.