This preview shows page 1. Sign up to view the full content.
Unformatted text preview: = q k 1 and note that vr = v . Since p irreducible, for any x 6= y there is a
path from x to y . Since the shortest such path will not visit any state more
than once, we can always get from x to y in k 1 steps, and it follows that
r(x, y ) > 0.
The next step is to prove that all the vx have the same sign. Suppose not.
In this case since r(x, y ) > 0 we have
|vy | = X vx r(x, y ) < x X
x |vx |r(x, y ) To check the second inequality note that there are terms of both signs in the
sum so some cancellation will occur. Summing over y and using y r(x, y ) = 1
|vy | <
y x a contradiction.
Suppose now that all of the vx 0. Using
vx r(x, y )
x we conclude that vy > 0 for all y . This proves the existence of a positive
solution. To prove uniqueness, note that if p I has rank k 2 then by linear
algebra there are two perpendicular solutions, v and w, but the last argument
implies that we can choose the sign so that vx , wx > 0 for all x. In this case
View Full Document
This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
- Spring '10
- The Land