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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!

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if $u$ and $v$ are the sides of a parallelogram, then $|u+v|^2, |u-v|^2$ are the squares of the lengths of the diagonals. the sum of the squares of the four sides is $2(|u|^2+|v|^2)$. –  yoyo Apr 17 '11 at 16:37
2  
I fixed your TeX. Two remarks: First, the symbols < and > are relations and thus result in bad spacing, hence it is better to use $\langle$ \langle and $\rangle$ \rangle which also look better for other reasons. Second, you need to type \\\ in order to get a line break. –  t.b. Apr 17 '11 at 16:40
    
But of course +1 for showing your work. –  t.b. Apr 17 '11 at 16:42
    
@Theo: thank you so much for that! I was trying to figure out why it was such a mess... definitely valuable remarks! –  ghshtalt Apr 17 '11 at 16:44
    
One more thing. You didn't define $\alpha$. You probably intend $\alpha(x,y) = \langle x, y\rangle$, don't you? –  t.b. Apr 17 '11 at 16:50
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2 Answers 2

up vote 3 down vote accepted
  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 $2 = n$)

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Thank you for this answer. I think I should be able to manage (3) now. With the geometric interpretation I am still a bit stuck. When you mention those 4 corners of the parallelogram, is each one an ordered pair, like $u= (u_1,u_2)$? Also, is "Pythagoras" written in the problem as a pointer to use the Pythagorean theorem somehow? I just don't know exactly where to apply it, I try to 'extend' the parallelogram horizontally to form right angles, but then the lengths become unclear... –  ghshtalt Apr 17 '11 at 19:14
    
@ghshtalt: Yes, $0 = (0,0)$, $u = (u_1, u_2)$, $v = (v_1,v_2)$ and $u+v = (u_1+v_1, u_2 + v_2)$ are the corners of a parallelogram. The diagonals are $u+v$ and $u-v$ (remember vector addition and subtraction). Yes, the classical Pythagorean theorem is intended. Draw the height of the parallelogram at an obtuse angle of the parallelogram. This will split the other side in two parts. You can then express the length of the diagonals using the height and the two parts of the other side. Squaring and taking the sum will eliminate all unknowns and what remains is the parallelogram law. –  t.b. Apr 17 '11 at 19:28
    
thanks for all your help –  ghshtalt Apr 17 '11 at 19:30
    
@ghsthalt: You're welcome. I hope you could solve the problem now. See you around! –  t.b. Apr 17 '11 at 19:41
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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.

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+1, but for (a) you could have said that what @ghstalt did is exactly the intention. –  t.b. Apr 17 '11 at 17:06
    
@Theo: Re-reading my reply, I see that it may look like I first commented 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. –  wildildildlife Apr 17 '11 at 19:28
    
thank you for this answer –  ghshtalt Apr 17 '11 at 19:31
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