Inner product proof

1. Jun 23, 2009

evilpostingmong

1. The problem statement, all variables and given/known data
Prove that
($$\sum$$ajbj)2$$\leq$$$$\sum$$jaj2*$$\sum$$(bj)2/j with j from 1 to n.

for all real numbers a1...an and b1...bn

2. Relevant equations

3. The attempt at a solution
I can prove this using algebra, but how is it done
using inner product concepts? If someone could start me up with a hint,
I could get somewhere. Thank you!
Note: It is apparant that those sums could possibly be inner products of vectors,
but I can't see how going into inner product notation would help me out.

Last edited: Jun 23, 2009
2. Jun 23, 2009

Dick

Define a diagonal matrix N so that the diagonal entries are (sqrt(1),sqrt(2)...sqrt(n)). Can you write the rhs in terms of the transformation N? Think about using the inequality (a.b)^2<=(a.a)^2*(b.b)^2. I don't think it's all that easy to prove using algebra. Can you prove it that way for the n=2 case?

3. Jun 23, 2009

Office_Shredder

Staff Emeritus
Do you know how to prove

$$(\sum a_j b_j)^2 \leq \sum a_j^2 \sum b_j^2$$?

If so, try looking at

$$a_j b_j = \sqrt{j} a_j \frac{b_j}{\sqrt{j}}$$

4. Jun 23, 2009

evilpostingmong

so define v to be=a1v1+a2v2 and u to be b1u1+b2u2 and define N to be a matrix with (sqrt(1)...sqrt(n)) along its diagonal
with n=2.
The dot product of N(v) to itself is <N(v),N(v)>^2=(v,sqrt(2)v)^2=sqrt(a1^2+2*a2^2)^2=a1^2+2a2^2
and same for u except it turns out to be b1^2+b2^2/2
so it is (a1^2+2a2^2)*(b1^2+b2^2/2) how is it so far?
I think at this point, the proof is pretty obvious, but I can't expect it to be that easy,

5. Jun 23, 2009

Dick

The algebraic proof that (a1*b1+a2*b2)^2<=(a1^2+2*a2^2)*(b1^2+b2^2/2) is not particularly obvious. You said you could, that's why I asked. It is pretty obvious using inner products (ha, ha, meaning it took me a while to see it). Can you try phrasing it in that language? I'll give you a hint. What's N^(-1)?

Last edited: Jun 23, 2009
6. Jun 23, 2009

evilpostingmong

you mean the language of inner products, right?

7. Jun 23, 2009

Dick

Are you stalling for time? Of course I mean that. What's the rhs in terms of the linear transformation N? I already asked you this. Yes, <Na,Na> is the first factor. What's the second one? BTW, is this a Knuth problem or is this from the linear algebra book?

8. Jun 23, 2009

evilpostingmong

the rhs is <a1, sqrt(2)a2>^2 *<b1, b2/sqrt(2)>^2 N^-1
takes it back to <a1,a2>^2*<b1,b2>^2

wait a minute, that's the same as <a1,b1>^2*<a2,b2>^2
or <a1,b1>^2*<sqrt(2)a2,b2/sqrt(2)>^2

9. Jun 23, 2009

evilpostingmong

the linear algebra book

10. Jun 23, 2009

Dick

Sorry, but you have a real ability to scramble things up. I only mentioned the n=2 special case because you said you could prove it algebraically. I don't think you can. The sum(j*aj^2) part is <Na,Na> right, do you agree with with that? What the sum(bj^2/j) part in terms of the vector b and N?

11. Jun 23, 2009

Dick

Ok, had a Knuth feeling. But let's continue.

12. Jun 23, 2009

evilpostingmong

<Nb,Nb>

13. Jun 23, 2009

Dick

<Nb,Nb> is b1*b1+2*b2*b2+3*b3*b3... Isn't it? That isn't what you want. Why don't you think about what you write actually means!? You want b1*b1+b2*b2/2+b3*b3/3... Don't you? Consider N^(-1).

14. Jun 23, 2009

evilpostingmong

oh <N^(-1)b,N^(-1)b>
I think I see where this is going...
<Na,Na>*<N^(-1)b,N^(-1)b> = <Na,N^(-1)b><Na,N^(-1)b> = <Na,N(-1)b>^2 = <a,b>^2
or am I jumping the gun as usual?

Last edited: Jun 24, 2009
15. Jun 23, 2009

Dick

Is that an "oh" as in "yes sir, I'll put in just what you say!" or is that an "oh" as in, "ok, I really understand what you are saying. And I realize my previous post was completely thoughtless and in the future I will actually think about what the symbols I'm posting mean before I thoughtlessly post them"? There is a reason why your threads go to high post counts and this is it.

16. Jun 24, 2009

Dick

Jumping the gun. The equality is not true. The inequality is. You seem to be able to prove anything instantly. By moving symbols around without rhyme or reason. Stop it! Slap yourself for me. Sober up! You didn't post one single reason why anyone should think anything you did is true. You just 'rearranged things'. In seriously random ways. That IS NOT A PROOF. A proof involves giving reasons for what you do. On this issue you seem to still not get it.

Last edited: Jun 24, 2009
17. Jun 24, 2009

evilpostingmong

<Na,Na>*<N^(-1)b,N^(-1)b> ok so we have this to work with

18. Jun 24, 2009

Dick

Yes, you have that to work with. Don't forget my hint to use <x,y>^2<=<x,x>^2*<y,y>^2. Office-Shredder made a useful suggestion on the first page. And don't forget my screaming from the first page. If I get one more proof by random symbol transposition, you are dead to me. Try to make it count. Give a reason for every step you take. I'm trusting in you.

19. Jun 24, 2009

evilpostingmong

ok.....
Consider x>y. If x>y, then (<x,x>)^2 is >(<x,y>)^2. Since (<x,x>)^2*(<y,y>)^2 is
> (<x,x>)^2, (<x,x>)^2*(<y,y>)^2 is > (<x,y>)^2 for x>y.

How is this so far? It has "if" "then" and it looks like its trying
form an argument (not random symbol pushing).

heart: ba dub BA DUB BA DUB BA DUB

20. Jun 24, 2009

Dick

Sigh. You are making this up, aren't you? Do you read the text of the book before you start doing proofs or are you just improvising? i) What is x>y supposed to mean when x and y are vectors??? If it's x1>y1, x2>y2, ... then take x=(0,0) and y=(-1,-1). Is <x,x> > <y,y>?? ii) if 'x>y' meant anything, is it something you can assume in the proof. Why?? Write me a 500 word essay on the history and significance of the 'Cauchy-Schwarz inequality, ok? In the essay, speculate on how using it can save you from writing bales of gibberish about vector inequalities.