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Homework Statement
Does the Cauchy Schwarz inequality hold if we define the dot product of two vectors [itex] A,B \in V_n [/itex] by [itex] \sum_{k=1}^n |a_ib_i| [/itex]? If so, prove it.
Homework Equations
The Cauchy-Schwarz inequality: [itex] (A\cdot B)^2 \leq (A\cdot A)(B\cdot B) [/itex]. Equality holds iff one of the vectors is a scalar multiple of the other.
The Attempt at a Solution
The result is trivial if either vector is the zero vector. Assume [itex] A,B\neq O [/itex].
Let [itex] A',B'\in V_n [/itex] s.t. [itex] a'_i = |a_i| [/itex] and [itex] b'_i = |b_i| [/itex] for each [itex] i=1,2,...,n [/itex]. Let [itex] \theta [/itex] be the angle between [itex] A' [/itex] and [itex] B' [/itex]. Notice that [itex] A\cdot B = A'\cdot B' [/itex], [itex] ||A|| = ||A'|| [/itex], [itex] ||B|| = ||B'|| [/itex], and [itex] A\cdot B > 0 [/itex]. Then [tex] A\cdot B = ||A||\,||B||\,\cos\theta \implies A\cdot B = ||A||\,||B||\,|\cos\theta| \leq ||A||\,||B|| [/tex] since [itex] 0\leq |\cos\theta|\leq 1 [/itex]. Then [itex] (A\cdot B)^2 \leq ||A||^2||B||^2 = (A\cdot A)(B\cdot B) [/itex]. This proves the inequality.
We now show that equality holds iff [itex] B = kA [/itex] for some [itex] k\in\mathbb{R} [/itex].
([itex] \Longrightarrow [/itex]) We prove this direction by the contrapositive. If [itex] B\neq k A [/itex] for any [itex] k\in\mathbb{R} [/itex], then [itex] B' \neq |k| A' [/itex]. Hence [itex] \theta \neq \alpha \pi [/itex] for any [itex] \alpha \in \mathbb{Z} [/itex]. Thus [itex] 0 < |\cos\theta| < 1 [/itex] and therefore [itex] A\cdot B < ||A||\,||B|| [/itex]. In other words, equality does not hold.
([itex] \Longleftarrow [/itex]) If [itex] A = kB [/itex] for some [itex] k\in\mathbb{R} [/itex], then [itex] B' = |k| A' [/itex]. Hence [itex] \theta = \alpha \pi [/itex] for some [itex] \alpha \in \mathbb{Z} [/itex]. Therefore [itex] \cos\theta = \pm 1 [/itex]. Thus equality holds.
Does this look okay?
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