Summing over y and using y rx y 1 we have x x vy vx

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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 P sum so some cancellation will occur. Summing over y and using y r(x, y ) = 1 we have X X |vy | < |vx | y x a contradiction. Suppose now that all of the vx 0. Using X vy = 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 the ve...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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