Proof of Cauchy Schwarz for complex numbers

AI Thread Summary
The discussion revolves around proving the Cauchy-Schwarz inequality for complex numbers by manipulating the expression <x - ty, x - ty>, where t is defined as <x,y>/<y,y>. Participants are attempting to simplify the equation but face challenges in eliminating the conjugate notation represented by "*". Suggestions include further simplification of the inner product <x,ty> and making the substitution for t to facilitate the proof. Clarifications are made regarding the notation, specifically that "*" denotes the complex conjugate. The conversation highlights the complexities involved in handling inner products in the context of complex vector spaces.
dispiriton
Messages
8
Reaction score
0

Homework Statement


I am told to try and solve <x - ty, x - ty> where t = <x,y>/<y,y>
However, I am stuck at that equation and unable to manipulate it to get rid of the *

Homework Equations





The Attempt at a Solution


<x - ty, x - ty> = <x,x> - <x,ty> - <ty,x> + <ty,ty>
= mod(x)^2 + mod(y)^2 (t^2) - t*<x,y> - (<x,ty>)*
 
Physics news on Phys.org
dispiriton said:

Homework Statement


I am told to try and solve <x - ty, x - ty> where t = <x,y>/<y,y>
However, I am stuck at that equation and unable to manipulate it to get rid of the *

Greetings dispiriton! What do you mean by "the *"?
 
dispiriton said:
<x - ty, x - ty> = <x,x> - <x,ty> - <ty,x> + <ty,ty>
= mod(x)^2 + mod(y)^2 (t^2) - t*<x,y> - (<x,ty>)*

You could simmplify this further by working out <x,ty>*. Try to get thet out of the inproduct.

Then, make the substitution t=<x,y>/<y,y>...
 
Undoubtedly0 said:
Greetings dispiriton! What do you mean by "the *"?

Its sort of like the "bar" where it is the conjugate.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Replies
14
Views
2K
Replies
26
Views
691
Replies
8
Views
1K
Replies
18
Views
3K
Replies
6
Views
3K
Replies
20
Views
2K
Replies
5
Views
1K
Replies
5
Views
1K
Back
Top