Lecture#13 Notes

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Unformatted text preview: e we prove this note the following observation. If ||f (t)|| =< f (t), f (t) >= 1, then < f ￿ ( t ) , f ( t ) > + < f ( t ) , f ￿ ( t ) > = 0. Proof. ⇒: Assume Ay = λy and let f : R ￿→ S be differentiable with f (0) = y . Then d < Af (t), f (t) >= < Af ￿ (t), f (t) > + < Af (t), f ￿ (t) > dt =< f ￿ (t), Af (t) > + < Af (t), f ￿ (t) > When t = 0 this equation becomes < f ￿ (0), Af (0) > + < Af (0), f ￿ (0) >= λ[< f ￿ (0), y > + < y, f ￿ (0) >] = 0 By the observation. d ⇐: Assume that dt |t=0 qA (f (t)) = 0 ∀ differentiable f with f (y ) = y. The claim is that y is an eigenvector of A. Let v ∈ Cn be such that v ⊥ y . Now define fv (t) = (cos(t))y + (sin(t))v. Note that ||fv (t)|| = 1. Now by assumption, we have that d ￿ ￿ |t=0 qA (f (t)) = < Afv (t), fv (t) > + < Afv (t), fv (t) > |t=0 dt =< Av, y > + < Ay, v...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.

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