The Condition Number of the PageRank Problem

We determine analytically the condition number of the PageRank problem. Specifically, we prove the following statement: “Let P be an n × n row-stochastic matrix whose diagonal elements Pii = 0. Let c be a real number such that 0 ≤ c < 1. Let E be the n × n rank-one row-stochastic matrix E = ev , where e is the n-vector whose elements are all ei = 1, and v is an n-vector that represents a probability distribution. Define the matrix A = [cP + (1 − c)E] . The problem Ax = x has condition number κ = (1 + c)/(1− c).” This statement has implications for the accuracy to which PageRank can be computed, currently and as the web scales. Furthermore, it provides a simple proof that, for values of c that are used by Google, small changes in the link structure of the web do not cause large changes in the PageRanks of pages of the web. 1 Theorem Theorem 1. Let P be an n× n row-stochastic matrix whose diagonal elements Pii = 0.. Let c be a real number such that 0 ≤ c ≤ 1. Let E be the n × n rank-one rowstochastic matrix E = ev , where e is the n-vector whose elements are all ei = 1, and v is an n-vector that represents a probability distribution1. Define the matrix A = [cP + (1− c)E] . The problem Ax = x has condition number κ = (1 + c)/(1− c). 2 Notation and Preliminaries P is an n× n row-stochastic matrix whose diagonal elements Pii = 0. E is the n× n rank-one row-stochastic matrix E = ev , where e is the n-vector whose elements are all ei = 1 and v is an n-vector whose elements are all non-negative and sum to 1. A is the n× n column-stochastic matrix: A = [cP + (1− c)E] (1) We let x be the dominant eigenvector of A. By convention, we choose eigenvectors x such that ||x||1 = 1. Since A is a non-negative matrix, the dominant eigenvector x is also non-negative. Therefore, ex = ||x||1 = 1 (2) Since A is column-stochastic, it’s dominant eigenvalue λ1 = 1, 1 ≥ |λ2| ≥ . . . ≥ |λn| ≥ 0. That is, Ax = x (3) 1 i.e., a vector whose elements are nonnegative and whose L1 norm is 1. 3 Proof of Theorem 1 We prove this case via a series of lemmas.