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Mathematics 307|December 6, 1995 Generalized eigenvalues and conservative systems Assume M positive denite. To solve Mx00 + Kx = 0 we convert it into a rst order system by setting y = Mx and getting 0 x = 0 M ,1 x y ,K 0 y or v0 = Av; v = eAt v0 We apply Gauss elimination to factor M = L tL; M ,1 = tL,1 L,1 ; tL M ,1 L = I and then multiply t L 0 0 M ,1 tL,1 0 = 0 I 0 L,1 ,K 0 0 L ,L,1 K tL,1 0 The matrix K = L,1 K tL,1 is still symmetric. We can nd an orthogonal X such that X ,1 K X = D where D is diagonal. So X ,1 0 0 I X 0 = 0 I 0 X ,1 ,K 0 0 X ,D 0 Finally we use p ! 0 0 1 if ! = k. So we set 0 1 ,k 0 Thus if 0 0 = 0 ! 1 ,! 0 2! = 64 and get ,1 ! 0 0 I 0 I ,D 0 0 0 3 0 !2 0 75 0 0 !3 1 ,1 0 0 0 I = , 0 = C ,1 t 0 X 0 L 0 Y= 0 I 0 X ,1 0 L,1 ,1 tL X 0 = 0 X ,1 L,1 t ,1 ,1 0 L X , 1 Y = 0 LX Solving the generalized eigenvalue problem then Finally 2 Y A Y ,1 = , 0 0 ; A = Y ,1 AY; eAt = Y ,1eCt Y !it , sin !i t eCt = cos sin !i t cos !i t Lower triangular systems We want to nd x such that Lx = c. We nd x0, x1 etc. `0;0x0 = c0 `i;0x0 + `i;1x1 + + `i;i xi = ci x0 = c0 =`0;0 xi = ci , `i;0 x0 , `i;1 x1 , We will call this with a parameter n, the dimension, so we can use partial vectors anmd matrices. Cholesky factorization We have which we solve by induction. 0 ` t` = ta 0 t a A 2 = ` = a ` = ,1 a ` t` + t = A t = A , ` t` Note that ` t` is a square matrix with entries `i `j . For i = 0 to n ,p2 = `i;i := mi;i for j = i + 1 to n , 1 `j;i := mj;i= for j = i + 1 to n , 1 for k = i + 1 to j pmj;k := mj;k , `j;i `k;i `n,1;n,1 := mn,1;n,1 Lower triangular inverses To solve or 0 x X 0 =I ` = 1; x + X` = 0 is the easiest way to do it inductively. Note that we should also have a routine for nding Lx where L is lower triangular.