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#1 2007-09-27 15:46:22

mikau
Member
Registered: 2005-08-22
Posts: 1,504

dot product proof, is this valid?

one of my homework problems asked me to prove the  Shwarz inequality, after messing around for a while i eventually determined if i could do the following, i could prove it fairly easily:

If v and u are vectors in R^n and u*u = 1 (dot product) and v*v =1, prove u*v <= 1.

i was messing with this all day, whenever i got a spare moment, and at the train station i think i came up with a solution, but for some reason i don't trust it.

Here's my proof.

We have u*u = 1 and v*v =1, suppose that u*v > 1, this implies u*v > u*u and u*v > v*v, expanding out the elements of the vectors we can write the first inequality as

u1*v1 + u2*v2 ... + u_n * v_n > (u1)^2 + (u2)^2 + ... + (u_n)^2 if we transfer all elements to the right side and factor we obtian

0 >  u1(u1 - v1) + u2(u2 - v2) + ... + u_n(u_n - v_n)

by the same steps we can obtain

0 > v1(v1 - u1) + v2(v2 - u2) + ... + v_n(v_n - u_n) now factor out a -1

0 > -v1(u1 - v1) - v2(u2 - v2) - ... - v_n(u_n - v_n)

now we have two equalities which follow from the premise, if we add the larger side to the larger side and the smaller side to the smaller side, the following inequality must be true, we get

0 + 0 > -v1(u1 - v1) - v2(u2 - v2) - ... - v_n(u_n - v_n) + u1(u1 - v1) + u2(u2 - v2) + ... + u_n(u_n - v_n)

simplifying

0 >  (u1 - v1)^2 + (u2 - v2)^2 + ... + (u_n - v_n)^2, we have that the sum of a set of square numbers is less than zero so clearly the premise is incorrect. Therefore u*v must be less than or equal to 1 and the possibility of either occurance can be shown by example.

DONE!

Brilliant or bogus?


A logarithm is just a misspelled algorithm.

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#2 2007-09-27 18:18:16

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: dot product proof, is this valid?

but isnt dot product also defined as u.v = |u||v|cosP in which case:

u.u = |u|^2cosP since its the same vector cosP is 1, and so |u|^2 = 1, |u| = 1 same for v.v
and then u.v = |u||v|cosP = cosP, and cosP is always <= 1?


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The End Of All Things To Come.

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#3 2007-09-28 03:11:54

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

I guess, but thats not something i was informed of, my book hasn't even introduced that definition yet so i should be able to do it without it. Of course, if you can prove that equation using only the basic definition of dot product, then that would work as well.

and i realized why i didn't trust my above proof. Its not done yet! i have to show that |u*v| <= 1, so now i must show (u*v) >= -1

We have u*u = 1 and v*v = 1.
Suppose u*v < -1,
then u*v < -u*u and u*v < - v*v
thus

u1*v1 + u2*v2 + ... + u_n*v_n < -v1^2 - v2^2 - ... - (v_n)^2

so 0 < -v1(v1 + u1) - v2(v2 + u2) ... - v_n(v_n + u_n) likewise from the second inequality we can obtain:
0 <  -u1(v1 + u1) - u2(v2 + u2) ... - u_n(v_n + u_n) adding larger to larger and smaller to smaller we get

0 < -(u1 + v1)^2 - (u2 + v2)^2 - ... - (u_n + v_n)^2, and so

0 > (u1 + v1)^2 + (u2 + v2)^2 + ... + (u_n + v_n)^2, and again the contradiction.

Last edited by mikau (2007-09-28 03:22:33)


A logarithm is just a misspelled algorithm.

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#4 2007-09-28 05:27:00

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: dot product proof, is this valid?

You could also do it this way.

Assume to the contrary that u·v > 1. Then

Contradiction.

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#5 2007-09-28 06:23:29

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

thats a nice way of doing it! smile darn it, why didn't I think of that?

Actually  thats kind of like a consdensed version of what i did. I knew there had to be a way you could show u*u = 1, v*v = 1 and u*v > 1 was a contradiction without expanding. swear

Last edited by mikau (2007-09-28 06:31:55)


A logarithm is just a misspelled algorithm.

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#6 2007-09-28 07:24:25

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

well since you got that so fast, Jane, lets see if you can get THIS one! cool

use the schwarz inequality (  |u*v| <= ||u|| ||v|| ) to prove that   || u + v|| <= ||u|| + ||v||

and no googling. wink


A logarithm is just a misspelled algorithm.

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#7 2007-09-28 07:58:12

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: dot product proof, is this valid?

Same trick, really. wink


since the norm of a vector is non-negative.

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#8 2007-09-28 07:59:33

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

yep. Thats the one the book used.  smile


A logarithm is just a misspelled algorithm.

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#9 2007-09-28 08:03:38

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

...how did you get that so quick?


A logarithm is just a misspelled algorithm.

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#10 2007-09-28 14:53:34

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

hey! do you know of more problems like these? I really need more practice with vectors and proofs.


A logarithm is just a misspelled algorithm.

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#11 2007-09-29 00:44:20

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: dot product proof, is this valid?

Prove that | ||u|| - ||v|| | ≤ ||u - v||


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#12 2007-09-29 06:32:01

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: dot product proof, is this valid?

suppose
| ||u|| - ||v|| | > ||u - v||

then
sqrt(u*u) - sqrt(v*V) > sqrt((u - v)*(u -v))

squaring both sides

u*u - 2sqrt((u*u)(v*v)) + v*v > v*v - 2u*v + u*u

so u*u - 2||u|| ||v|| + u*u > v*v - 2u*v + u*u

subtracting u*u + v*v from both sides we obtain

- 2||u|| ||v|| > - 2u*v  so

u*v > ||u||||v||  this is false if u*v is negative, and contradicts the schwarz inequality if its positive. so it can't be greater.

aaand the possibility of less then or equal to can be shown by example. We leave this up to the reader as an exersize wink

Thanks ricky! I WANT MORE!

We can stray away from dot products and or vector norms of course.

Last edited by mikau (2007-09-29 06:36:41)


A logarithm is just a misspelled algorithm.

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