# Thread: PageRank Question: How does (AR' + E) = (A + E x 1)R'

1. ## PageRank Question: How does (AR' + E) = (A + E x 1)R'

I am trying to understand the classic 1998 PageRank paper (from the Google guys). I'm confused here:

"In matrix notation we have R' = c(AR' + E). Since ||R'||_1 = 1, we can rewrite this as R' = c(A + E x 1)R' where 1 is the vector consisting of all ones"

How did the paper go from "(AR' + E)" to "(A + E x 1)R'"? I don't understand.

R' is a vector of size n, A is an n x n matrix, and E is a vector of size n.

2. Originally Posted by VinceW
I am trying to understand the classic 1998 PageRank paper (from the Google guys). I'm confused here:

"In matrix notation we have R' = c(AR' + E). Since ||R'||_1 = 1, we can rewrite this as R' = c(A + E x 1)R' where 1 is the vector consisting of all ones"

How did the paper go from "(AR' + E)" to "(A + E x 1)R'"? I don't understand.

R' is a vector of size n, A is an n x n matrix, and E is a vector of size n.
The notation E x 1 must mean the n x n matrix in which each column is equal to E. Then the i'th element of (E x 1)R' will be the i'th element of E times the sum of the elements of R'. But ||R'||_1 = 1 (and presumably the elements of R' are non-negative), so the sum of the elements of R' is 1. Therefore (E x 1)R' = E.

3. OK, my coworkers cracked this one.

||R'||_1 = 1 (that's a subscript) means that all the terms of R sum to one rather than the typical euclidean magnitude.

"E x 1" is basic multiplication rather than a cross product (hard to tell which they meant), so (E times 1) times R' = E. E is n x 1 vector and 1 is a 1 x n vector of ones.

Thanks!

4. Wow, thanks! You solved it as well.