ALGORITHMS FOR PAGERANK SENSITIVITY DISSERTATION

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This last step completes the derivation of the Jordan form. Note, however, that the eigenvalues of G, which are also the eigenvalues of M, are given by the diagonal of αJ + (1 − α)e eT . Recall that J = 1, which corresponds to a 11 11 semi-simple eigenvalue of S. Write J = [ 1 αJ+(1−α)e eT =[1 10 Note that J1 is different from J2 in the previous section. This result is a corollary of the old theo- rem from Brauer about eigenvalues of combinations of matrices. Elden’s proof is specific to the PageRank case and more modern. 11 In the remainder of the document, x(1) means this limiting value. J1 αJ1 This analysis confirms the following theorem. ]. +(α)T =+(α)TG, =+(α)TVR(α)(αJ+(1−α)e eT)R(α)−1V−1, which implies +(α)T =eTR(α)−1V−1, 1 (2.39) (2.40) 11 Theorem 5 (Eldén [2004]; Brauer [1952] theorem 29). Pare1,λ2,...,λn thentheeigenvaluesofM(α)are1,αλ2,...,αλn. finding the limit WiththeJordanformfromeqs.(2.27)to(2.30)we can work out the PageRank vector x(1) in the limit sense.11 For α < 1, the PageRank vector +(α) satisfies (2.37) (2.38) w(α) solves the system w(α)T (I − αJ) = (1 − α)(eT − vT V), but J has 1 structure that makes the solution obvious. An additional concern is that +(α)T includes the first row of V−1, a vector that may be arbitrary! = e1T V−1 − w(α)T V−1 . The only dependence on α is in w(α). This fact appears unfortunate because Let’s resolve these concerns. We can decompose12 12 You might think this next step is incorrect because it asserts a form on J that possibly requires reordering V. The algebra, however, still works if we assert this form on step 1. 13 Remember that X and V are going to slightly different because they are Jordan forms of transposed matrices. andV=(V0 V1 V2)13 then w(α)T=e−vTV, (2.42) w (α)T =−(1−α)vTV (I−αD )−1, and (2.43) 111 w (α)T =−(1−α)vTV (I−αJ )−1. (2.44) 222 010 ⎡⎢I ⎤⎥ J = ⎢ D ⎥ , (2.41) ⎢1J⎥ ⎢⎥ ⎣ 2⎦ where D1 is a diagonal matrix of all eigenvalues on the unit circle not equal to 1 and J2 is a Jordan matrix with all eigenvalues on the interior of the unit circle. If we conformally partition w(α)T = (w (α)T w (α)T w (α)T ) 012 Both of the linear systems for w1 (α) and w2 (α) are non-singular for 0 ≤ α ≤ 1 and w0(α) is a constant function! ] so that10 11 If the eigenvalues of 2.7 ⋅ the limit case 33

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