# THE \$25,000,000,000 EIGENVECTOR THE LINEAR ALGEBRA BEHIND GOOGLE

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4 4 K. BRYAN AND T. LEISE a strongly connected web (that is, you can get from any page to any other page in a finite number of steps); see Exercise below. Unfortunately, it is not always true that the link matrix A will yield a unique ranking for all webs. Consider the web in Figure., for which the link matrix is A =. We find here that V (A) is two-dimensional; one possible pair of basis vectors is x = [/, /,,, ] T and y = [,, /, /, ] T. But note that any linear combination of these two vectors yields another vector in V (A), e.g., 4 x + 4 y = [/8, /8, /8, /8, ]T. It is not clear which, if any, of these eigenvectors we should use for the rankings! It is no coincidence that for the web of Figure. we find that dim(v (A)) >. It is a consequence of the fact that if a web W, considered as an undirected graph (ignoring which direction each arrows points), consists of r disconnected subwebs W,..., W r, then dim(v (A)) r, and hence there is no unique importance score vector x V (A) with i x i =. This makes intuitive sense: if a web W consists of r disconnected subwebs W,..., W r then one would expect difficulty in finding a common reference frame for comparing the scores of pages in one subweb with those in another subweb. Indeed, it is not hard to see why a web W consisting of r disconnected subwebs forces dim(v (A)) r. Suppose a web W has n pages and r component subwebs W,..., W r. Let n i denote the number of pages in W i. Index the pages in W with indices through n, the pages in W with indices n + through n + n, the pages in W with n + n + through n + n + n, etc. In general, let N i = i j= n j for i, with N =, so W i contains pages N i + through N i. For example, in the web of Figure we can take N = and N = 5, so W contains pages and, W contains pages, 4, and 5. The web in Figure. is a particular example of the general case, in which the matrix A assumes a block diagonal structure A = A... A.... A r, where A i denotes the link matrix for W i. In fact, W i can be considered as a web in its own right. Each n i n i matrix A i is column-stochastic, and hence possesses some eigenvector v i lr ni with eigenvector. For each i between and r construct a vector w i lr n which has components for all elements corresponding to blocks other than block i. For example, w = v. v, w =,.... Then it is easy to see that the vectors w i, i r, are linearly independent eigenvectors for A

5 THE \$5,,, EIGENVECTOR 5 with eigenvalue because. Aw i = A v i = w i.. Thus V (A) has dimension at least r.... Dangling Nodes. Another difficulty may arise when using the matrix A to generate rankings. A web with dangling nodes produces a matrix A which contains one or more columns of all zeros. In this case A is column-substochastic, that is, the column sums of A are all less than or equal to one. Such a matrix must have all eigenvalues less than or equal to in magnitude, but need not actually be an eigenvalue for A. Nevertheless, the pages in a web with dangling nodes can still be ranked use a similar technique. The corresponding substochastic matrix must have a positive eigenvalue λ and a corresponding eigenvector x with non-negative entries (called the Perron eigenvector) that can be used to rank the web pages. See Exercise 4 below. We will not further consider the problem of dangling nodes here, however. Exercise. Suppose the people who own page in the web of Figure are infuriated by the fact that its importance score, computed using formula (.), is lower than the score of page. In an attempt to boost page s score, they create a page 5 that links to page ; page also links to page 5. Does this boost page s score above that of page? Exercise. Construct a web consisting of three or more subwebs and verify that dim(v (A)) equals (or exceeds) the number of the components in the web. Exercise. Add a link from page 5 to page in the web of Figure. The resulting web, considered as an undirected graph, is connected. What is the dimension of V (A)? Exercise 4. In the web of Figure., remove the link from page to page. In the resulting web page is now a dangling node. Set up the corresponding substochastic matrix and find its largest positive (Perron) eigenvalue. Find a non-negative Perron eigenvector for this eigenvalue, and scale the vector so that components sum to one. Does the resulting ranking seem reasonable? Exercise 5. Prove that in any web the importance score of a page with no backlinks is zero. Exercise 6. Implicit in our analysis up to this point is the assertion that the manner in which the pages of a web W are indexed has no effect on the importance score assigned to any given page. Prove this, as follows: Let W contains n pages, each page assigned an index through n, and let A be the resulting link matrix. Suppose we then transpose the indices of pages i and j (so page i is now page j and vice-versa). Let Ã be the link matrix for the relabelled web. Argue that Ã = PAP, where P is the elementary matrix obtained by transposing rows i and j of the n n identity matrix. Note that the operation A PA has the effect of swapping rows i and j of A, while A AP swaps columns i and j. Also, P = I, the identity matrix. Suppose that x is an eigenvector for A, so Ax = λx for some λ. Show that y = Px is an eigenvector for Ã with eigenvalue λ. Explain why this shows that transposing the indices of any two pages leaves the importance scores unchanged, and use this result to argue that any permutation of the page indices leaves the importance scores unchanged.. A remedy for dim(v (A)) >. An enormous amount of computing resources are needed to determine an eigenvector for the link matrix corresponding to a web containing billions of pages. It is thus important to know that our algorithm will yield a unique set of sensible web rankings. The analysis above shows that our first attempt to rank web pages leads to difficulties if the web isn t connected. And the worldwide web, treated as an undirected graph, contains many disjoint components; see [9] for some interesting statistics concerning the structure of the web.

6 6 K. BRYAN AND T. LEISE Below we present and analyze a modification of the above method that is guaranteed to overcome this shortcoming. The analysis that follows is basically a special case of the Perron-Frobenius theorem, and we only prove what we need for this application. For a full statement and proof of the Perron-Frobenius theorem, see chapter 8 in []... A modification to the link matrix A. For an n page web with no dangling nodes we can generate unambiguous importance scores as follows, including cases of web with multiple subwebs. Let S denote an n n matrix with all entries /n. The matrix S is column-stochastic, and it is easy to check that V (S) is one-dimensional. We will replace the matrix A with the matrix M = ( m)a + ms, (.) where m. M is a weighted average of A and S. The value of m originally used by Google is reportedly.5 [9, ]. For any m [, ] the matrix M is column-stochastic and we show below that V (M) is always one-dimensional if m (, ]. Thus M can be used to compute unambiguous importance scores. In the case when m = we have the original problem, for then M = A. At the other extreme is m =, yielding M = S. This is the ultimately egalitarian case: the only normalized eigenvector x with eigenvalue has x i = /n for all i and all web pages are rated equally important. Using M in place of A gives a web page with no backlinks (a dangling node) the importance score of m/n (Exercise 9), and the matrix M is substochastic for any m < since the matrix A is substochastic. Therefore the modified formula yields nonzero importance scores for dangling links (if m > ) but does not resolve the issue of dangling nodes. In the remainder of this article, we only consider webs with no dangling nodes. The equation x = Mx can also be cast as x = ( m)ax + ms, (.) where s is a column vector with all entries /n. Note that Sx = s if i x i =. We will prove below that V (M) is always one-dimensional, but first let s look at a couple of examples. Example : For the web of four pages in Figure. with matrix A given by (.), the new formula gives (with m =.5) M = , and yields importance scores x.68, x.4, x.88, and x 4.. This yields the same ranking of pages as the earlier computation, but the scores are slightly different. Example shows more explicitly the advantages of using M in place of A. Example : As a second example, for the web of Figure. with m =.5 we obtain the matrix M = (.)..... The space V (M) is indeed one-dimensional, with normalized eigenvector components of x =., x =., x =.85, x 4 =.85, and x 5 =.. The modification, using M instead of A, allows us to compare pages in different subwebs. Each entry M ij of M defined by equation (.) is strictly positive, which motivates the following definition. Definition.. A matrix M is positive if M ij > for all i and j. This is the key property that guarantees dim(v (M)) =, which we prove in the next section.

7 THE \$5,,, EIGENVECTOR 7.. Analysis of the matrix M. Note that Proposition shows that V (M) is nonempty since M is stochastic. The goal of this section is to show that V (M) is in fact one-dimensional. This is a consequence of the following two propositions. Proposition. If M is positive and column-stochastic, then any eigenvector in V (M) has all positive or all negative components. Proof. We use proof by contradiction. First note that in the standard triangle inequality i y i i y i (with all y i real) the inequality is strict when the y i are of mixed sign. Suppose x V (M) contains elements of mixed sign. From x = Mx we have x i = n j= M ijx j and the summands M ij x j are of mixed sign (since M ij > ). As a result we have a strict inequality x i = M ij x j < j= M ij x j. (.4) Sum both sides of inequality (.4) from i = to i = n, and swap the i and j summations. Then use the fact that M is column-stochastic ( i M ij = for all j) to find x i < i= i= j= M ij x j = j= ( n ) M ij x j = j= i= x j, a contradiction. Hence x cannot contain both positive and negative elements. If x i for all i (and not all x i are zero) then x i > follows immediately from x i = n j= M ijx j and M ij >. Similarly x i for all i implies that each x i <. The following proposition will also be useful for analyzing dim(v (M)): Proposition. Let v and w be linearly independent vectors in lr m, m. Then, for some values of s and t that are not both zero, the vector x = sv + tw has both positive and negative components. Proof. Linear independence implies neither v nor w is zero. Let d = i v i. If d = then v must contain P components of mixed sign, and taking s = and t = yields the conclusion. i If d set s = wi d, t =, and x = sv + tw. Since v and w are independent x. However, i x i =. We conclude that x has both positive and negative components. We can now prove that using M in place of A yields an unambiguous ranking for any web with no dangling nodes. Lemma.. If M is positive and column-stochastic then V (M) has dimension. Proof. We again use proof by contradiction. Suppose there are two linearly independent eigenvectors v and w in the subspace V (M). For any real numbers s and t that are not both zero, the nonzero vector x = sv + tw must be in V (M), and so have components that are all negative or all positive. But by Proposition, for some choice of s and t the vector x must contain components of mixed sign, a contradiction. We conclude that V (M) cannot contain two linearly independent vectors, and so has dimension one. Lemma. provides the punchline for our analysis of the ranking algorithm using the matrix M (for < m < ). The space V (M) is one-dimensional, and moreover, the relevant eigenvectors have entirely positive or negative components. We are thus guaranteed the existence of a unique eigenvector x V (M) with positive components such that i x i =. Exercise 7. Prove that if A is an n n column-stochastic matrix and m, then M = ( m)a + ms is also a column-stochastic matrix. Exercise 8. Show that the product of two column-stochastic matrices is also column-stochastic. Exercise 9. Show that a page with no backlinks is given importance score m n by formula (.). Exercise. Suppose that A is the link matrix for a strongly connected web of n pages (any page can be reached from any other page by following a finite number of links). Show that dim(v (A)) = as follows. Let (A k ) ij denote the (i, j)-entry of A k. Note that page i can be reached from page j in one step if and only A ij > (since A ij > means there s a link from j to i!) Show that (A ) ij > if and only if page i can be reached from page j in exactly two steps. Hint: (A ) ij = k A ika kj ; all A ij are non-negative, so (A ) ij > implies that for some k both A ik and A kj are positive. j=

8 8 K. BRYAN AND T. LEISE Show more generally that (A p ) ij > if and only if page i can be reached from page j in EXACTLY p steps. Argue that (I + A + A + + A p ) ij > if and only if page i can be reached from page j in p or fewer steps (note p = is a legitimate choice any page can be reached from itself in zero steps!) Explain why I + A + A + + A n is a positive matrix if the web is strongly connected. Use the last part (and Exercise 8) so show that B = n (I + A + A + + A n ) is positive and column-stochastic (and hence by Lemma., dim(v (B)) = ). Show that if x V (A) then x V (B). Why does this imply that dim(v (A)) =? Exercise. Consider again the web in Figure., with the addition of a page 5 that links to page, where page also links to page 5. Calculate the new ranking by finding the eigenvector of M (corresponding to λ = ) that has positive components summing to one. Use m =.5. Exercise. Add a sixth page that links to every page of the web in the previous exercise, but to which no other page links. Rank the pages using A, then using M with m =.5, and compare the results. Exercise. Construct a web consisting of two or more subwebs and determine the ranking given by formula (.). At present the web contains at least eight billion pages how does one compute an eigenvector for an eight billion by eight billion matrix? One reasonable approach is an iterative procedure called the power method (along with modifications) that we will now examine for the special case at hand. It is worth noting that there is much additional analysis one can do, and many improved methods for the computation of PageRank. The reference [7] provides a typical example and additional references. 4. Computing the Importance Score Eigenvector. The rough idea behind the power method for computing an eigenvector of a matrix M is this: One starts with a typical vector x, then generates the sequence x k = Mx k (so x k = M k x ) and lets k approaches infinity. The vector x k is, to good approximation, an eigenvector for the dominant (largest magnitude) eigenvalue of M. However, depending on the magnitude of this eigenvalue, the vector x k may also grow without bound or decay to the zero vector. One thus typically rescales at each iteration, say by Mx k Mx k computing x k =, where can be any vector norm. The method generally requires that the corresponding eigenspace be one-dimensional, a condition that is satisfied in the case when M is defined by equation (.). To use the power method on the matrices M that arise from the web ranking problem we would generally need to know that any other eigenvalues λ of M satisfy λ <. This assures that the power method will converge to the eigenvector we want. Actually, the following proposition provides what we need, with no reference to any other eigenvalues of M! Definition 4.. The -norm of a vector v is v = i v i. Proposition 4. Let M be a positive column-stochastic n n matrix and let V denote the subspace of lr n consisting of vectors v such that j v j =. Then Mv V for any v V, and Mv c v for any v V, where c = max j n min i n M ij <. straightforward: Let w = Mv, so that w i = n j= M ijv j and w i = i= i= j= M ij v j = ( n ) v j M ij = Hence w = Mv V. To prove the bound in the proposition note that w = e i w i = e i M ij v j, i= j= i= i= j= Proof. To see that Mv V is v j =. See [5] for a general introduction to the power method and the use of spectral decomposition to find the rate of convergence of the vectors x k = M k x. j=

9 THE \$5,,, EIGENVECTOR 9 where e i = sgn(w i ). Note that the e i are not all of one sign, since i w i = (unless w in which case the bound clearly holds). Reverse the double sum to obtain w = ( n ) v j e i M ij = j= i= a j v j, (4.) where a j = n i= e im ij. Since the e i are of mixed sign and i M ij = with < M ij <, it is easy to see that We can thus bound j= < + min i n M ij a j min i n M ij <. a j min i n M ij <. Let c = max j n min i n M ij. Observe that c < and a j c for all j. From equation (4.) we have w = a j v j = a j v j a j v j c v j = c v, j= j= which proves the proposition. Proposition 4 sets the stage for the following proposition. Proposition 5. Every positive column-stochastic matrix M has a unique vector q with positive components such that Mq = q with q =. The vector q can be computed as q = lim k M k x for any initial guess x with positive components such that x =. Proof. From Proposition the matrix M has as an eigenvalue and by Lemma. the subspace V (M) is one-dimensional. Also, all non-zero vectors in V (M) have entirely positive or negative components. It is clear that there is a unique vector q V (M) with positive components such that i q i =. Let x be any vector in lr n with positive components such that x =. We can write x = q + v where v V (V as in Proposition 4). We find that M k x = M k q + M k v = q + M k v. As a result j= j= M k x q = M k v. (4.) A straightforward induction and Proposition 4 shows that M k v c k v for c < (c as in Proposition 4) and so lim k M k v =. From equation (4.) we conclude that lim k M k x = q. Example: Let M be the matrix defined by equation (.) for the web of Figure.. We take x = [.4,.,.8,.8,.9] T as an initial guess; recall that we had q = [.,.,.85,.85,.] T. The table below shows the value of M k x q for several values of k, as well as the ratio M k x q / M k x q. Compare this ratio to c from Proposition 4, which in this case is.94. k M k x q M k x q M k x q It is clear that the bound M k x q c k x q is rather pessimistic (note.85 is the value m, and.85 turns out to be the second largest eigenvalue for M). One can show that in general the power method will converge asymptotically according to Mx k q λ x q, where λ

11 THE \$5,,, EIGENVECTOR [4] S. Brin and L. Page, The anatomy of a large-scale hypertextual web search engine, http : //www db.stanford.edu/ backrub/google.html (accessed August, 5). [5] A. Farahat, T. Lofaro, J. C. Miller, G. Rae, and L.A. Ward, Authority Rankings from HITS, PageRank, and SALSA: Existence, Uniqueness, and Effect of Initialization, SIAM J. Sci. Comput., 7 (6), pp. 8-. [6] A. Hill, Google Inside Out, Maximum PC, April 4, pp [7] S. Kamvar, T. Haveliwala, and G. Golub, Adaptive methods for the computation of PageRank, Linear Algebra Appl., 86 (4), pp [8] A. N. Langville and C. D. Meyer, A survey of eigenvector methods of web information retrieval, SIAM Review, 47 (5), pp [9] A. N. Langville and C. D. Meyer, Deeper inside PageRank, Internet Math., (5), pp [] C. D. Meyer, Matrix Analysis and Applied Linear Algebra, SIAM, Philadelphia,. [] Cleve Moler, The world s largest matrix computation, http : // notes /clevescorner/oct cleve.html (accessed August, 5). [] Mostafa, J., Seeking better web searches, Sci. Amer., 9 (5), pp [] Sara Robinson, The Ongoing search for efficient web search algorithms, SIAM News, 7 (Nov 4). [4] Ian Rogers, The Google Pagerank algorithm and how it works, http : // (accessed August, 5). [5] W. J. Stewart, An Introduction to the Numerical Solution of Markov Chains, Princeton University Press, Princeton, 994.

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