1. Cauchy-Schwarz inequality

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

Could someone please clear it up a bit? Thanks.

2. 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

3. 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.

4. $\Rightarrow)$

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

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

Fernando Revilla