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 June 29th, 2017, 01:54 PM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 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
 June 30th, 2017, 06:26 AM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 "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)." From: http://math.ucr.edu/~res/math133/fal...es1insert1.pdf 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 07:00 AM.

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