CauchySchwarz Inequality Help :D

brd_7 · #1 $Old$ November 15th, 2008, 05:40 AM

||x||=4, ||y||=5,<x,y>=8.
What is the cauchyshwarz inequality? <This was relatively straight forward.. Not too sure about the rest however..
What is the norm of 4x-4y?
What is the cosine angle between x and 4x-4y?
Find λ, such that 2x-y and λx-y are orthogonal?

Many thanks

flyingsquirrel · #2 $Old$ November 15th, 2008, 06:43 AM

Hi,

Quote:

Originally Posted by brd_7 $View Post$

||x||=4, ||y||=5,<x,y>=8. (...)
What is the norm of 4x-4y?

$\|4x-4y\|=4\|x-y\|=4\sqrt{\langle x-y,x-y\rangle}=4\sqrt{\langle x,x\rangle-2\langle x,y\rangle+\langle y,y\rangle}=\ldots$

Quote:

What is the cosine angle between x and 4x-4y?

Remember that $\langle u,v\rangle = \|u\|\|v\|\cos(u,v)$ .

Quote:

Find λ, such that 2x-y and λx-y are orthogonal?

In other words, solve $\langle 2x-y,\lambda x-y\rangle=0$ for $\lambda$ .

brd_7 · #3 $Old$ November 15th, 2008, 07:48 AM

I got 28 for the norm of 4x-4y.. I hope thats right.. I then gained th result of 66.4 for the theta angle in part b.. Thanks for the help.

I think im going to need a litrle bit more help with the last part of the question...

flyingsquirrel · #4 $Old$ November 15th, 2008, 08:53 AM

Quote:

Originally Posted by brd_7 $View Post$

I got 28 for the norm of 4x-4y.. I hope thats right.. I then gained th result of 66.4 for the theta angle in part b.. Thanks for the help.

As the angle you've found is correct, I think you mistyped your answer to the first question : $\|4x-4y\|=20$ .

Quote:

I think im going to need a litrle bit more help with the last part of the question...

The dot product is a bilinear function : $\begin{cases}\langle u, v+w\rangle = \langle u, v\rangle + \langle u, w\rangle & (1)\\\langle u+ v, w\rangle=\langle u,w\rangle +\langle v,w\rangle &(2)\\ \langle \alpha u,v\rangle=\langle u, \alpha v\rangle = \alpha\langle u, v\rangle & (3)\end{cases}$ so the LHS of $\langle 2x-y,\lambda x-y\rangle=0$ can be expanded.

$\begin{aligned} \langle 2x-y,\lambda x-y\rangle &=0\\ \langle 2x-y,\lambda x\rangle+\langle 2x-y,-y\rangle &=0 \text{\,\,\,\,using (1)}\\ \lambda \langle 2x-y, x\rangle-\langle 2x-y,y\rangle &=0 \text{\,\,\,\,using (3)}\\ \end{aligned}$

This can be further expanded using (2) and (3) to make appear $\|x\|^2,\,\|y\|^2$ and $\langle x ,y\rangle$ ...

brd_7 · #5 $Old$ November 15th, 2008, 09:20 AM

Yup, that is what i got sorry

And this is what i got so far, for the next bit..

λ[2<x,x> - <x,y>]- [2<x,y> - <y,y>]

With λ = -3/8.. eventually

Thanks again for all the help!

flyingsquirrel · #6 $Old$ November 15th, 2008, 09:27 AM

Quote:

Originally Posted by brd_7 $View Post$

With λ = -3/8.. eventually

That's it !