3
$\begingroup$

Let $(V,\langle,\rangle)$ be a euclidean vector space and $u, v \in V$ two vectors. With $||x||=\sqrt{\langle x,x\rangle}$ the parallelogram law $||u+v||^{2} + ||u-v||^{2} = 2||u||^{2} + 2||w||^{2}$ is valid.

(a)Prove the parallelogram law.

(b)Let $V=\mathbb{R}^{2}$ with the canonical scalar product. Provide a geometric proof of the parallelogram law (Pythagoras).

(c)Show that the norm $||x|| = \displaystyle\sum\limits_{i=1}^n|x_i|$ on $\mathbb{R}^{2}$ does not come from a scalar product.

I guess (a) was just a matter of expanding the terms and using bilinearity/symmetry:

\begin{align*} ||u+v||^{2}+||u-v||^{2} &= \langle u+v,u+v\rangle+\langle u-v,u-v\rangle\\ & = \langle u,u+v\rangle+\langle v,u+v\rangle+\langle u,u-v\rangle-\langle v,u-v\rangle\\ &= \langle u+v,u\rangle+\langle u+v,v\rangle+\langle u-v,u\rangle-\langle u-v,v\rangle \\ &= \langle u,u\rangle+\langle v,u\rangle+\langle u,v\rangle+\langle v,v\rangle+\langle u,u\rangle-\langle v,u\rangle-\langle u,v\rangle+\langle v,v\rangle \\ &= 2\langle u,u\rangle+2\langle v,v \rangle\\ &= 2||u||^{2}+2||v||^{2}. \end{align*}

For(c): is my goal to show that the norm is either not symmetric, bilinear, or positive definite? I am not sure how to show this since it does not say what to do with $y$, (my understanding is that a scalar product is a numerically valued function of ordered pairs of vectors $(x,y)$ which satisfy those three conditions). I am guessing however, that it is not enough to just write that this norm is not a scalar product since it doesn't tell me what to do with $y$, right?

With (b): all I know is that the canonical scalar product is $\langle x,y\rangle = \displaystyle\sum\limits_{i=1}^nx_iy_i$, but what exactly do they mean by a "geometric" proof?

Thank you in advance for any help!

  • 0
    Ah, there was one more internet-y issue: since html is delimited by `<` and `>`, typing those symbols can interfere with the MathJaX rendering engine. If you need these symbols it is safer to use `\lt` for $\lt$ and `\gt` for $\gt$.2011-04-17

2 Answers 2

3
  1. Your solution of (a) is entirely correct.

  2. The points $0,u,v,u+v$ are the corners of a parallelogram with sides $u$ and $v$ and diagonals $u+v$ and $u-v$.

  3. Does the parallelogram law hold? (here you probably intend $2 = n$)

  • 0
    @ghsthalt: You'$r$e welcome. I hope $y$ou could solve the problem now. See you around!2011-04-17
3

For (c): a norm can not be symmetric, bilinear or positive definite; these are properties that a bilinear form can have (in which case we call it an inner product). You need to show that there does not exist an inner product

$(-,-):\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$

such that $\sqrt{(x,x)}=\sum_{i=1}^n|x_i|=\|x\|$ for all $x$.

By part (a) of the question (which you solved correctly) you know that if such an inner product were to exist, then this norm necessarily satisfies the parallelogram law. So it is suggested that you inspect whether it holds.

  • 0
    @Theo: Re-reading my reply, I see that it may look like I first $c$ommented on his attempt at (c) and then on his attempt at (a). I meant only to comment on his attempt at (c). Part of this comment is a reference to part (a) of the question (which he has succesfully solved, as you say) which can be used to solve part (c). I have rephrased a bit to make this more clear.2011-04-17