Cauchy-Schwarz Inequality and Equality

By Jason Wallace on Monday, February 04, 2002 - 07:17 pm:

"Show that if u and v are linearly dependent, then equality holds in the Cauchy-Schwarz inequality"

Any help in explaining this question would be greatly appreciated.

By David Loeffler on Tuesday, February 05, 2002 - 12:24 am:

Well, the general proof of C-S runs like this: assume

v

isn't the zero vector. Then for any scalar

t

and vectors

u

v

in our vectors space,

$‖ u + t v ‖^{2} \geq 0$

so $‖ u ‖^{2} + 2 〈 u, v 〉 t + ‖ v ‖^{2} t^{2} \geq 0$

so $(‖ v ‖ t + 〈 u, v 〉 / ‖ v ‖)^{2} + ‖ u ‖^{2} - 〈 u, v 〉^{2} / ‖ v ‖ \geq 0$

So $‖ u ‖^{2} \geq 〈 u, v 〉^{2} / ‖ v ‖$ , which is just the standard C-S.

However, equality can only occur if there is some scalar $t$ for which $‖ (u + v t) ‖ = 0$ .

So if $‖ u ‖^{2} ‖ v ‖^{2} = 〈 u, v 〉^{2}$ , there is some $t$ such that $u + v t = 0$ , or $v$ is the zero vector in which case both sides are 0, so it is always true that they are equal.)

This condition - that $u = v t$ for some scalar $t$ or $v = 0$ - is precisely the condition for two vectors to be linearly dependent.

David