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 ³ 0

so ||u||²+2áu,vñt+||v||² t² ³ 0

so (||v|| t + áu,vñ/||v||)²+||u||² -áu,vñ²/ ||v|| ³ 0

So ||u||² ³ áu,vñ²/||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=áu, vñ², 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