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Lecture10 - Lecture 10 Bounding Transcriber Andy Parrish...

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Lecture 10: Bounding λ Transcriber: Andy Parrish Goal: Show that (1 - λ/ 2) 2 t i 6 =0 (1 - λ i / 2) 2 t n (1 - α 2 / 4) t , where t is any positive integer. For a positive integer t and a vertex x , let f t,x = χ x Z t - π , where χ x ( v ) = 1 if x = v and 0 otherwise, Z = I + P 2 is the probability transition matrix for a lazy random walker, and π = ~ 1 D vol G is the stationary distribution. Fact 1 : d 1 / 2 x f t,x ( y ) d - 1 / 2 y = d 1 / 2 y f t,y ( x ) d - 1 / 2 x = χ x ( I - L 2 - I 0 ) χ * y , where I 0 = φ 0 φ * 0 = JD vol G is projection onto the 0 eigenvector for L . Proof: d 1 / 2 x f t,x ( y ) d - 1 / 2 y = χ x D 1 / 2 ( Z t - π ) D - 1 / 2 χ * y = χ x D 1 / 2 I + P 2 t - JD vol G ! D - 1 / 2 χ * y = χ x I + D - 1 / 2 AD - 1 / 2 2 ! t - D 1 / 2 JD 1 / 2 vol G ! χ * y = χ x I - L 2 t - I 0 ! χ * y = χ x I - L 2 - I 0 t χ * y because I 0 is idempotent and L I 0 = I 0 L = 0 Since the matrix here is symmetric, multiplying by χ x on the left and χ * y on the right is the same as multiplying by χ y on the left and χ * x on the right. Fact 2 : i 6 =0 (1 - λ i / 2) 2 t = x d - 1 x y d y f 2 t,y ( x ) Proof: { (1 - λ i / 2) 2 t } n - 1 i =0 are the eigenvalues for ( I - L 2 ) 2 t . To exclude the eigenvalue for i = 0, 1
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we should subtract projection onto φ 0 . Thus we get X i 6 =0 (1 - λ i / 2) 2 t = Trace I - L 2 - I 0 2 t !
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