Cauchy-Schwarz inequality

**Mollier** · February 8th 2011, 11:49 PM

Hi,

one of my books states the inequality as

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

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

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

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

$\begin{aligned}<br /> |x^*\alpha x| \leq& ||x|| ||\alpha x|| \\<br /> |\alpha| |x^*x| \leq& |\alpha| ||x||^2\\<br /> |x^*x| =& ||x||^2<br /> \end{aligned}$

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

Could someone please clear it up a bit? Thanks.

**FernandoRevilla** · February 9th 2011, 12:36 AM

The result is the following:

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

You have proven $\Leftarrow)$ .

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

Fernando Revilla

**Mollier** · February 9th 2011, 02:04 AM

Originally Posted by FernandoRevilla

The result is the following:

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

You have proven $\Leftarrow)$ .

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

Fernando Revilla

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

Thanks.

**FernandoRevilla** · February 9th 2011, 04:25 AM

$\Rightarrow)$

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

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

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

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

Fernando Revilla

Thread: Cauchy-Schwarz inequality

LinkBack

Thread Tools

Display

Cauchy-Schwarz inequality

Similar Math Help Forum Discussions

Cauchy-Schwarz inequality

Cauchy-Schwarz Inequality

cauchy schwarz inequality

Cauchy-Schwarz inequality for Integrals

Cauchy-Schwarz Inequality

Search Tags