Possible webpage title: Expressing Dot Product in Terms of Norms

  • Thread starter Thread starter theneedtoknow
  • Start date Start date
  • Tags Tags
    Dot
Click For Summary
SUMMARY

The discussion focuses on expressing the dot product in terms of norms for vectors in Rn. Part (a) confirms the identity 2||x||² + 2||y||² = ||x + y||² + ||x - y||² using properties of the dot product. Part (b) explores expressing the dot product x • y using norms without coordinates, highlighting that the inner product is distributive and can be derived from the norms of the vectors involved. The discussion also notes that a norm must satisfy the parallelogram law to ensure the existence of a corresponding dot product.

PREREQUISITES
  • Understanding of vector norms and the dot product in Rn
  • Familiarity with properties of inner products and their distributive nature
  • Knowledge of the parallelogram law in vector spaces
  • Basic algebraic manipulation skills for expressions involving sums
NEXT STEPS
  • Study the proof of the parallelogram law and its implications for norms
  • Learn about the relationship between norms and inner products in vector spaces
  • Explore examples of norms that do not satisfy the parallelogram law
  • Investigate the derivation of the dot product from norms in various contexts
USEFUL FOR

Mathematicians, physics students, and anyone studying linear algebra or vector calculus who seeks to deepen their understanding of the relationships between dot products and norms.

theneedtoknow
Messages
169
Reaction score
0

Homework Statement



(a) Using the dot product, show that for x, y ∈ Rn, the formula
2||x||^2 + 2||y||^2 = ||x + y||^ 2 + ||x − y||^2 holds.

(b) The norm on Rn can be defined in terms of the dot product by
the formula ||x|| = √(x • x). Show that the reverse is true. That is,
find a formula for x • y involving the norms of vectors (||x||, ||y||,
||x+y||, and ||x−y|| for example), and without using coordinates.


Homework Equations



http://upload.wikimedia.org/math/e/9/0/e901236dd90219902035e3479b31a2ee.png


The Attempt at a Solution



for part a, i started with the right side
(i use the word sum to represent the summation from 1 to n of each variable
||x+y||^2 + ||x-y||^2 = (x+y)•(x+y) + (x-y)•(x-y) = sum[(x+y)(x+y)] + sum[(x-y)(x-y)] = sum x^2 +sum (2xy) + sum y^2 + sum x^2 - sum (2xy) + sum y^2 = 2sumx^2 + 2sumy^2 = 2||x||^2 + 2||y||^2 = left side

i'm pretty sure I am doing this part right but it never hurts to double check

now part b i just have no idea where to start with
im supposed to express the dot product x•y in terms of norms ... but x•y= sum from i=1 to n of (xi*yi) = x1y1 + x2y2+...+xnyn
I can't think of anything to factor or do to that last expression in order to get it looking like something I can express in terms of norms
i know that x•y = ||x|| ||y|| cos(theta) where theta is the angle between x and y, but i think i need a way to express it using only norms
 
Physics news on Phys.org
A is correct but you don't need to use sums. The inner product is distributive so you can say (x+y,x+y)=(x,x)+(y,y)+2(x,y). Saves some time.

I find b a bit vague but you could calculate (x+y,x+y) and (x-y,x-y) and solve for (x,y).
 
(b) is indeed a bit vague. There are two possible interpretations that come to mind:

(1) Assuming that the norm is induced by a dot product, i.e.,

||v|| = <v,v>^{1/2}

then find an expression for <x,y> in terms of the norm. This isn't too hard but involves a bit of trial and error to find something that works.

(2) Assuming only that you are given a norm, prove that there exists a dot product that induces it, and find a formula for it.

Of note is that (2) is NOT TRUE in general. In fact, it happens to be possible for a given norm IF AND ONLY IF that norm satisfies the parallelogram law, i.e., part (a) of the OP's question.

Somewhat incongruously, (2) is given as an exercise in Axler's "Linear Algebra Done Right." The only proof I've been able to find is at least an order of magnitude harder than the level of the typical exercise in that book. I posted on that very subject this past weekend:

https://www.physicsforums.com/showthread.php?t=314503

Perhaps someone knows of a more elementary solution.
 
Last edited:

Similar threads

  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 43 ·
2
Replies
43
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K