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Trick that allows expression of angles

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Martin, Burda

Lecture 12 notes by Y. Burda We started with a digression, describing a trick that allows to express angles in terms of distances (a sort of cosine law): Problem Let T : V V be an operator satisfying Tv = v for all v V . Prove that Tv Tw = v w for all v,w V . The meaning of this problem is that an operator that preserves distances must also preserve dot products and hence angles (v w = vwcos v,w). 2 2 Solution We will use the identity vw = (v+w vw )4 expressing dot products in terms of distances. The proof of this identity is a straight- forward calculation: v + w = (v + w) (v + w) = v + w + 2v w v w = (v w) (v w) = v + w 2v w Subtracting these equations we get the desired result. Now we can use this identity: Tv + Tw Tv Tw 2 T(v + w) T(v w) 2 Tv Tw = = = 4 4 v + w v w 2 = = v w 4 Tricks that connect dot-products to distances are often called polariza- tion tricks. One application of the notions we studied about quadratic forms is to investigation of behaviour of a function near its critical point. Recall that for a good enough function of one variable we have the fol- lowing test for a maximumminimum: at a point of maximum or minimum the rst derivative of the function vanishes and if the second derivative at the same point is positive, then this is a minimum point. Similarly if the second derivative is negative, then the point is a maximum point. The proof of this criterion is based on the following approximation: f (x ) f(x + x) f(x ) + f (x )x + 0 x 2 0 0 0 2 For a function of two variables whose rst derivatives at a point (x ,y ) 0 0 1
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