# Cauchy-Schwarz inequality

• Feb 8th 2011, 11:49 PM
Mollier
Cauchy-Schwarz inequality
Hi,

one of my books states the inequality as

$\displaystyle |x^*y| \leq ||x|| ||y||$ for all $\displaystyle x,y\in\mathbb{C}^{n\times 1}$,

and says that equality holds if and only if $\displaystyle y=\alpha x$ for

$\displaystyle \alpha =x^*y/x^*x$.

To me it looks like equality holds if $\displaystyle y=\alpha x$ for any $\displaystyle \alpha\in\mathbb{C}$:

\displaystyle \begin{aligned} |x^*\alpha x| \leq& ||x|| ||\alpha x|| \\ |\alpha| |x^*x| \leq& |\alpha| ||x||^2\\ |x^*x| =& ||x||^2 \end{aligned}

I do not understand why it has to be $\displaystyle \alpha =x^*y/x^*x$.

Could someone please clear it up a bit? Thanks.
• Feb 9th 2011, 12:36 AM
FernandoRevilla
The result is the following:

$\displaystyle |x^*y|= \left\|{x}\right\| \left\|{y}\right\|\Leftrightarrow \{x,y\}\textrm{\;lin.\;dep.}$

You have proven $\displaystyle \Leftarrow)$ .

To prove $\displaystyle \Rightarrow)$ you'll need to choose a particular $\displaystyle \alpha$ .

Fernando Revilla
• Feb 9th 2011, 02:04 AM
Mollier
Quote:

Originally Posted by FernandoRevilla
The result is the following:

$\displaystyle |x^*y|= \left\|{x}\right\| \left\|{y}\right\|\Leftrightarrow \{x,y\}\textrm{\;lin.\;dep.}$

You have proven $\displaystyle \Leftarrow)$ .

To prove $\displaystyle \Rightarrow)$ you'll need to choose a particular $\displaystyle \alpha$ .

Fernando Revilla

So if $\displaystyle |x^*y|=||x|| ||y||$ then $\displaystyle ||\alpha x-y||=0$ and so $\displaystyle \alpha x-y=0$.

Thanks.
• Feb 9th 2011, 04:25 AM
FernandoRevilla
$\displaystyle \Rightarrow)$

If $\displaystyle x=0$ the equality is trivial and $\displaystyle \{x,y\}$ is linearly dependent . If $\displaystyle x\neq 0$ decompose:

$\displaystyle y=\alpha x+u,\quad (x\perp{u})$

Using $\displaystyle |<x,y>|= \left\|{x}\right\| \left\|{y}\right\|$ , prove that $\displaystyle <x,y>=\alpha <x,x>$ and $\displaystyle 0=<u,u>$.

That is $\displaystyle u=0$ or equivalently $\displaystyle y=\alpha x$ being $\displaystyle \alpha=<x,y>/<x,x>$ .

Fernando Revilla