508 insures ann utu for a unitary u and an upper

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Unformatted text preview: . , λk } , then the spectral projector onto N (A − λi I) along R (A − λi I) is given by k k (A − λj I) Gi = O C j =1 j =i (λi − λj ) for i = 1, 2, . . . , k. (7.3.11) j =1 j =i Consequently, if f (z ) is defined on σ (A) , then f (A) = is a polynomial in A of degree at most k − 1. k i=1 f (λi )Gi Problem: For a scalar t, determine the matrix exponential eAt , where A= −α α β −β with α + β = 0. Solution 1: The characteristic equation for A is λ2 + (α + β )λ = 0, so the eigenvalues of A are λ1 = 0 and λ2 = −(α + β ). Note that A is diagonalizable Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 530 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 because no eigenvalue is repeated—recall (7.2.6). Using the function f (z ) = ezt , the spectral representation (7.3.6) says that eAt = f (A) = f (λ1 )G1 + f (λ2 )G2 = eλ1 t G1 + eλ2 t G2 . It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu The spectral projectors G1 and G2 are determined from (7.3.11) to be G1 = A − λ2 I 1 = −λ2 α+β β α β α and G2 = A 1 = λ2 α+β α −α 1 α+β β α β α + e−(α+β )t α −α , D E so eAt = G1 + e−(α+β )t G2 = −β β T H −β β . Solution 2: Compute eigenpairs (λ1 , x1 ) and (λ2 , x2 ), construct P = x1 | x2 , and compute IG R 0 f (λ1 ) 0 f (λ2 ) eAt = P eλ1 t 0 P−1 = P 0 eλ2 t P−1 . The computational details are called for in Exercise 7.3.2. Example 7.3.4 Y P Problem: For T = 1 /2 1 /4 1 /2 3 /4 , evaluate limk→∞ Tk . Solution 1: Compute two eigenpairs, λ1 = 1, x1 = (1, 1)T , and λ2 = 1/4, x2 = (−2, 1)T . If P = [x1 | x2 ], then T = P 1 104 P−1 , so 0 / O C Tk = P 1k 0 0 1/4k P−1 → P 1 0 0 0 P−1 = 1 3 1 1 2 2 . (7.3.12) Solution 2: We know from (7.3.6) that Tk = 1k G1 + (1/4)k G2 → G1 . Since λ1 = 1 is a simple eigenvalue, formula (7.2.12) on p. 518 can be used to compute T T T G1 = x1 y1 /y1 x1 , where x1 and y1 are any right- and left-hand eigenvectors associated with λ1 = 1. A right-hand e...
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