rec10 - EE221A: Discussion 9 Lillian Ratliff November 5,...

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Unformatted text preview: EE221A: Discussion 9 Lillian Ratliff November 5, 2010 1 Eigenvalue-vector discussion n mk : k=1 (s − λk ) FACT 1.0.1 Minimum polynomial: Ψ(A) = 0. σ (A) = {λk }n Then, Ψ(s) = 1 largest Jordam block for λk . mk is the dimension o fhte FACT 1.0.2 T = (λk I − A) ⇒ ∃!ν ∈ N : N (T ν ) = N (T ν +1 ), N (T ν +1 ) = N (T ν +2 ), ν is called the ascent and ν = mk . Definition f : C → C complex-analytic (i.e. exists a convergent taylor series to f ). on a connected open ∆ ⊃ σ (A). Where connected means it cannot be represented as the union of two or more disjoint nonempty open subsets. if p is a polynomial satisfying spectral interpolation conditions p( ) (λk ) = f ( ) (λk ), k = 1, . . . n, = 0, . . . , mk − 1 then we define f (A) := p(A) = an−1 An−1 + · · · + a0 I . (pg. 129 C & D) Superscripts are derivatives! 1.1 How to calculate the pk n k=1 mk −1 =0 Theorem f (A) = f ( ) (λk )pk (A); pk (A) independent of f . Recall that we have f (T −1 AT ) = T −1 f (A)T so assume the above equality is true then n m k −1 f (T −1 AT ) = T −1 f (A)T = k=1 =0 f ( ) (λk )T −1 pk (A)T Example 1.1.1 Calculate pk (Jk ): Recall from class the computation of f (Jk ) where Jk is a jordan block. This implies that ï ò 1 0 I(n− )×(n− ) pk (Jk ) = 0 ! 0× Idea is that you equate the LHS and RHS then calculate the pk FACT 1.1.1 vk an e-vec for A ⇒ vk e-vec for f (A). Proof: f (A) = p(A) = an−1 An−1 + · · · + a0 I ⇒ Avk = λk vk ⇒ f (A)vk = (an−1 λn−1 + · · · a0 )vk = f (λk )vk k FACT 1.1.2 v an e-vec for f (A) ⇒ v an e-vec for A? Proof: Let f be identical to some constant c ∈ C. Then, f (A) = T −1 (cI )T (he had f (A) = cI ) (since f (Jk ) = cI ) so the answer is no. 1 ...
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This note was uploaded on 04/11/2011 for the course EE 221A taught by Professor Clairetomlin during the Fall '10 term at Berkeley.

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