- #1
DiscipulusHum
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Homework Statement
Prove that |x'y| <= ||x|| ||y|| for vectors x, y
Homework Equations
||x|| is the norm of x
x' is the transpose of x
The Attempt at a Solution
||(x/||x||)-(y/||y||)|| = [ (x'/||x|| - y'/||y||)(x/||x|| - y/||y||) ]^1/2 = [-1/(||x||*||y||) (x'y + y'x) +2]^1/2
we know that this norm is positive or zero, so:
[-1/(||x||*||y||) (x'y + y'x) +2]^1/2 > 0
-> x'y + y'x < 2||x||||y||
x'y = y'x because they are scalars and the transpose of a scalar doesn't change its value
-> 2x'y < 2||x||||y||
-> x'y < ||x||||y||
My result doesn't have absolute values around x'y; what am I doing wrong?