# Thread: CauchySchwarz Inequality Help :D

1. ## CauchySchwarz Inequality Help :D

||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

2. Hi,
Originally Posted by brd_7
||x||=4, ||y||=5,<x,y>=8. (...)
What is the norm of 4x-4y?
$\displaystyle \|4x-4y\|=4\|x-y\|=4\sqrt{\langle x-y,x-y\rangle}=4\sqrt{\langle x,x\rangle-2\langle x,y\rangle+\langle y,y\rangle}=\ldots$
What is the cosine angle between x and 4x-4y?
Remember that $\displaystyle \langle u,v\rangle = \|u\|\|v\|\cos(u,v)$.
Find λ, such that 2x-y and λx-y are orthogonal?
In other words, solve $\displaystyle \langle 2x-y,\lambda x-y\rangle=0$ for $\displaystyle \lambda$.

3. 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...

4. Originally Posted by brd_7
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 : $\displaystyle \|4x-4y\|=20$.
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 : $\displaystyle \begin{cases}\langle u, v+w\rangle = \langle u, v\rangle + \langle u, w\rangle & (1)\\\langle u+ v, w\rangle=\langle u,w\rangle +\langle v,w\rangle &(2)\\ \langle \alpha u,v\rangle=\langle u, \alpha v\rangle = \alpha\langle u, v\rangle & (3)\end{cases}$ so the LHS of $\displaystyle \langle 2x-y,\lambda x-y\rangle=0$ can be expanded.

\displaystyle \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 $\displaystyle \|x\|^2,\,\|y\|^2$ and $\displaystyle \langle x ,y\rangle$...

5. 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!

6. Originally Posted by brd_7
With λ = -3/8.. eventually
That's it !