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June 29th, 2017, 02:54 PM   #1
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Cauchy Schwarz Inequality Equality

What does Cauchy Schwarz imply for equality?

Cauchy Schwarz: $\displaystyle |x \cdot y|\leq |x||y|$, Always for all x and y
Equality: $\displaystyle |x \cdot y|= |x||y|$

$\displaystyle \left | \frac{x}{|x|}\cdot\frac{y}{|y|} \right | =1$
$\displaystyle |u \cdot v|=1, \quad |u|=|v|=1$
Assume $\displaystyle u=\pm v +k$
$\displaystyle |u|=|\pm v +k|\leq |v|+|k|\rightarrow k=0$
$\displaystyle u=\pm v$

$\displaystyle \frac{x}{|x|}=\pm\frac{y}{|y|}$
$\displaystyle x=\pm\frac{|x|}{|y|}y$

x=cy implies Equality (trivial), but does Equality imply x=cy, c arbitrary?

x and y in Rn
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June 30th, 2017, 07:26 AM   #2
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"Conversely, suppose that we have

$\displaystyle |x\cdot y|=|x||y|$

Then direct calculation shows that

$\displaystyle 0=\left | x-\frac{x\cdot y}{y\cdot y}y \right |^{2}$

and this equation implies that x is a multiple of y (since both vectors are nonzero, x must in fact be a nonzero multiple of y)."


Can anyone get this? It's better than OP. Looks like straight-forward vector algebra.

EDIT Rather than trying to find how he got the answer, the logical way to look at this is: given an equation in x and y, solve for x in terms of y.
EDIT Got it (one step):

$\displaystyle \left | x\cdot \frac{y}{|y|} \right |=|x|$

Since y/|y| is a unit vector, this is only possible if x and y are collinear.
To appreciate this, google the subject

Last edited by zylo; June 30th, 2017 at 08:00 AM.
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