4
$\begingroup$

Show that one gets equality in the Schwarz inequality if and only if $\overrightarrow v$,$\overrightarrow w$ are linearly dependent.

(I am supposing they want me to prove it in an inner product space we call V)

The $\Leftarrow$ part of the proof was pretty straightforward but I don't know how to go about the $\Rightarrow$ part because linear dependence requires us to show $\overrightarrow v$=$\lambda \overrightarrow w$ for some $\lambda \in \mathbb{R}$ but the Cauchy-Schwarz inequality only involves norms\lengths of vectors and not relations between the vectors themselves. I would appreciate tips on starting points.

$\endgroup$
  • $\begingroup$ @MiloBrandt, yes my bad, edited. $\endgroup$ – Omrane Apr 9 '16 at 20:09
  • $\begingroup$ How did you proof of the Cauchy-Schwarz inequality go? $\endgroup$ – Andrew D. Hwang Apr 9 '16 at 20:16
3
$\begingroup$

Assume $\overrightarrow v\neq0$(if $\overrightarrow v=0$ the proof is trivial), and take any norm of $V$. Fix an arbitrary $\lambda$ in the scalar field ($\mathbb{R}$in this case) and note that $$0\leq \|\overrightarrow u-\lambda \overrightarrow v \|^2=\langle \overrightarrow u-\lambda \overrightarrow v ,\overrightarrow u-\lambda \overrightarrow v \rangle$$ Is the first step in deriving the C.S inequality. Thus equality holds iff $$\|\overrightarrow u-\lambda \overrightarrow v \|=0$$ From our properties of norms we know then that $$\overrightarrow u-\lambda \overrightarrow v=0 $$ and we have shown linear dependence.

$\endgroup$
  • $\begingroup$ It'd be more symmetrical (and closer to the definition of linear dependence) to take $0\leq \|\alpha u - \beta v\|^2$ and proceed. Then you don't need to consider the case where one is zero - it'd be built in to the proof naturally. $\endgroup$ – Milo Brandt Apr 9 '16 at 20:18
  • $\begingroup$ This is the proof I'm familiar with, but I'm sure your proof would have been along similar lines and you can adjust your inequality to equality accordingly. $\endgroup$ – K.Power Apr 9 '16 at 20:19
  • $\begingroup$ @MiloBrandt I'd have to look at that. In my proof I end up dividing by $\|v\|$ so that's why I have two cases. $\endgroup$ – K.Power Apr 9 '16 at 20:21
  • $\begingroup$ @K.Power, Thanks for your answer. I don't understand how you use C.S going from first line of operations to second. $|<\overrightarrow u,\overrightarrow v>| \le ||\overrightarrow u|| . ||\overrightarrow v||$ is the equation I have, we need to turn that into an equality as a supposition right? $\endgroup$ – Omrane Apr 9 '16 at 20:33
  • $\begingroup$ @Sadem my first line of operations is how I derive C.S in the first place. It's the starting point of my proof for C.S. What proof did you use to derive C.S? $\endgroup$ – K.Power Apr 9 '16 at 20:37
0
$\begingroup$

From the proof of C-S. With $0\ne x=(x_1,..,x_n)$ and $0\ne y=(y_1,..,y_n).$ $$\|x\|^2\cdot \|y\|^2-(x\cdot y)^2=\sum_{1\leq i<j\leq n}(x_i y_j-x_j y_i)^2.$$ Now $x_i y_j=x_j y_i$ whenever $1\leq i<j\leq n$ iff there exists $k\ne 0$ such that $x_i=k y_i$ for $i=1,..,n.$ Use induction on $n$ to prove this.

The case $x=0\lor y=0$ is trivial.

$\endgroup$

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.