Lecture14

# Lecture14 - Lecture 14 Theorem 1 Let X be a Banach space...

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Unformatted text preview: Lecture 14 Theorem 1. Let X be a Banach space. The map exp : L ( X ) → L ( X ) : A 7→ e A is differentiable at 0 and exp = id L ( X ) . Proof. We must estimate k e H- e- id L ( X ) H k k H k . Because e H = ∞ X n =0 1 n ! H n , we have that e H- I- H = ∞ X n =2 1 n ! H n = H 2 ∞ X n =0 1 ( n + 2)! H n ∴ k e H- I- H k ≤ k H k 2 ∞ X n =0 1 ( n + 2)! k H k n ≤ k H k 2 e k H k ∴ k e H- I- H k k H k ≤ k H k e k H k . As H → 0, k H k → 0 and e k H k → 1. Hence k e H- I- H k k H k → 0. Fact : The function exp is differentiable everywhere. Theorem 2. Let A ∈ L ( X ) and define f ( t ) = e tA for t ∈ R . Then f ( t ) = Ae tA for all t ∈ R . Theorem 3. Let dim X < ∞ . If A ∈ L ( X ) then Det ( e A ) = e TrA . 1 Proof. Two methods. Algebra : There exists a basis for X relative to which the matrix M of A is triangular. Now Det ( e A ) = Det ( e M ) = Y e diag M TrA = TrM = X diag M ∴ Det ( e A ) = Y e diag M = e ∑ diag M = e TrA ....
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## This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.

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Lecture14 - Lecture 14 Theorem 1 Let X be a Banach space...

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