# Trick that allows expression of angles

37 views4 pages
Published on 23 Jun 2011
School
UTSG
Department
Mathematics
Course
MAT224H1
Professor
Lecture 12 notes byY. Burda
Westarted with adigression, describing atrickthat allows to express
angles in terms of distances (a sort of cosine law):
Problem Let T:VVbean operator satisfying |Tv|=|v|for all
vV.Provethat Tv·Tw=v·wfor all v,wV.
The meaning of this problem is that an operator that preserves distances
must also preservedot products and hence angles (v·w=|v||w|cos 6v,w).
Solution Wewill use the identityv·w=(|v+w|2|vw|2)/4expressing
dot products in terms of distances. The proof of this identityis astraight-
forward calculation:
|v+w|2=(v+w)·(v+w)=|v|2+|w|2+2v·w
|vw|2=(vw)·(vw)=|v|2+|w|22v·w
Subtracting these equations weget the desired result.
Now wecan use this identity:
Tv·Tw=|Tv+Tw|2|TvTw|2
4=|T(v+w)|2|T(vw)|2
4=
=|v+w|2|vw|2
4=v·w
Tricks that connect dot-products to distances are often called “polariza-
tion tricks”.
One application of the notions westudied about quadratic forms is to
investigation of behaviour of afunction near its critical point.
Recall that for agoodenough function of one variable wehavethe fol-
lowing test for amaximum/minimum: at apointof maximum or minimum
the ﬁrst derivativeof the function vanishes and if the second derivativeat
the same pointis positive, then this is aminimum point. Similarly if the
second derivativeis negative, then the pointis amaximum point.
The proof of this criterion is based on the following approximation:
f(x0+x)f(x0)+f(x0)∆x+f′′(x0)
2x2
For afunction of twovariables whose ﬁrst derivatives at apoint (x0,y0)
1
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## Document Summary

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 |t v| = |v| for all v v . Prove that t v t w = 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 = |v||w| cos 6 v, w). Solution we will use the identity v w = (|v +w|2 |v w|2)/4 expressing dot products in terms of distances. The proof of this identity is a straight- forward calculation: |v + w|2 = (v + w) (v + w) = |v|2 + |w|2 + 2v w. |v w|2 = (v w) (v w) = |v|2 + |w|2 2v w. Subtracting these equations we get the desired result. = |t (v + w)|2 |t (v w)|2.

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