275A-Solutions-1

275A-Solutions-1 - ECE 275A Homework 1 Solutions 1 To solve...

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ECE 275A Homework # 1 Solutions 1. To solve this problem, we need to start with some known relationship between x , y , and x, y , together with one or more free parameters (if possible) to give us some flexibility in deriving the C-S inequality. Perhaps the simplest such relationship we can envision (having only one free parameter) is 0 ≤ ∥ x αy 2 = x αy, x αy = x 2 2 Re α x, y + | α | 2 y 2 . (1) Note that this relationship involves the squares of the norms of x and y , whereas the C-S inequaliy involves the norms themselves. Thus, in order to understand better where we are heading, let us rewrite the C-S inequality as |⟨ x, y ⟩| 2 y 2 ≤ ∥ x 2 . (2) We can now try to find a value for α which will place equation (1) into the form (2). Without too much work, one can show that the value, α = x, y y 2 , will do the job. Note the combination of educated guessing and luck involved in this proof, which is a standard one encountered in many textbooks. Comment: Don’t feel at a loss when studying the optimization-based proof used in the Moon text. Although the proof he gives is valid qua proof, it is not one that is natural to use at this stage of knowledge. It would be pedagogically useful to be given an
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275A-Solutions-1 - ECE 275A Homework 1 Solutions 1 To solve...

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