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:V→Vbean operator satisfying |Tv|=|v|for all
v∈V.Provethat Tv·Tw=v·wfor all v,w∈V.
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−|v−w|2)/4expressing
dot products in terms of distances. The proof of this identityis astraight-
Subtracting these equations weget the desired result.
Now wecan use this identity:
Tricks that connect dot-products to distances are often called “polariza-
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:
For afunction of twovariables whose ﬁrst derivatives at apoint (x0,y0)
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.