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f(x) is a strictly increasing function of x, for all x 2 IR nf274g. Hence, for all x&lt;1: f(x) &lt;f(1) &lt;f(b)=)b&gt;1=c, which completes the proof. The proof is due in part to Ronny Roth ([27]). We bring an additional lemma, which can be used for an alternative proof of proposition 2. Lemma 4 Let A =[ai;j] be a non-negative, irreducible n n matrix with an (i; j)-switch property. Let be an eigenvalue of A, and let w =(w1;:::;wn) T beacorresponding eigenvector. Then: 6 = ai;i, ai;j =) wi = wj. 2 Proof: Since w is an eigenvector which corresponds to [Aw] i = a i;iw i + a i;jw j + [Aw] j = a j;iw i + a j;jw j + nX l6=i;j nX l6=i;j a i;lw l = a j;lw l = Subtracting the second equation from the rst, we get:,wehave: w i w j (a i;i, a j;i)w i +(a i;j, a j;j)w j = (w i, w j) Since a i;i, a j;i = a j;j, a i;j, we get: (a i;i, a i;j)(w i, w j) = (w i,w j) Hence, 6 = a i;i, a i;j =) w i, w j =0. 2 87 By [24], for any square irreducible n n matrix B: (B) min nX 1 i n j=1 In our case, this lower bound yields (Ak) 1662. Exploiting the structure of Ak and its principal eigenvector, we arrive at the following 3 3 equation set

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Year: 2011
OAI identifier: oai:CiteSeerX.psu:10.1.1.184.8697
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