0
$\begingroup$

I have a proof that proves the Cauchy Schwarz inequality. I am at the stage where I am trying to show the inequality holds for when $\langle x, y\rangle \neq 0$ and $x$ and $y$ are linearly independent.

For $\alpha \in \mathbb{C}$, $\alpha x +y \neq 0$, then

$$0 < \|\alpha x+y\|^2 = \langle\alpha x+y, \alpha x+y\rangle = |\alpha|^2\|x\|^2 + (\alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle) + \|y\|^2.$$

How do we see that $\langle\alpha x+y, \alpha x+y\rangle = |\alpha|^2\|x\|^2 + (\alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle) + \|y\|^2$?

$\endgroup$
1
$\begingroup$

Recall that a complex inner product in complex linear in the first argument (i.e. $\langle kv, w\rangle = k\langle v, w\rangle$), complex antilinear in the second argument (i.e. $\langle v, kw\rangle = \bar{k}\langle v, w\rangle$), and $\langle v, v\rangle = \|v\|^2$. With all of this in mind, we have

\begin{align*} \langle\alpha x + y, \alpha x + y\rangle &= \langle\alpha x, \alpha x\rangle + \langle\alpha x, y\rangle + \langle y, \alpha x\rangle + \langle y, y\rangle\\ &= \alpha\bar{\alpha}\langle x, x\rangle + \alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle + \|y\|^2\\ &=|\alpha|^2\|x\|^2 + \alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle + \|y\|^2. \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.