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Unformatted text preview: â‰¤ k a k 22 k a k 2 + k a k 4 k b k 2 ( a Â· b ) 2 = k a k 21 + k a k 2 k b k 2 ( a Â· b ) 2 ! . 1 Since a Â· b 6 = 0, a 6 = 0, hence k a k > 0. Therefore, the above can be rearranged to yield ( a Â· b ) 2 â‰¤ k a k 2 k b k 2 . Taking the square root yields the result. Â± From the proof, it is clear that the CauchSchwartz inequality is, in fact, strict unless for some Î» , a = Î»b , meaning that a and b are collinear. CauchySchwartz, in turn, implies the following. Theorem 2. For any a,b âˆˆ R N , k a + b k â‰¤ k a k + k b k . Proof k a + b k 2 = ( a + b ) Â· ( a + b ) = k a k 2 + 2 a Â· b k b k 2 â‰¤ k a k 2 + 2  a Â· b  + k b k 2 â‰¤ k a k 2 + 2 k a kk b k + k b k 2 = ( k a k + k b k ) 2 . The second inequality follows form Theorem 1. Taking square roots yields the result. Â± 2...
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 Fall '08
 JohnNachbar
 Economics, Pythagorean Theorem, Hypotenuse, Euclidean norm

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