Analysis on Graphs - download pdf or read online

By Alexander Grigoryan

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Example text

14). 14). 16). 17). (b) We need to construct N 1 linearly independent eigenfunctions with the eigenvalue N . As above, set V = f0; 1; :::; N 1g and consider the following N 1 functions fk N 1 for k = 1; 2; :::N 1 : 8 i = 0; < 1; 1; i = k; fk (i) = : 0; otherwise. We have Lfk (i) = fk (i) 1 N 1 X j6=i fk (j) : 44 CHAPTER 2. SPECTRAL PROPERTIES OF THE LAPLACE OPERATOR P If i = 0 then fk (0) = 1 and in the sum j = k, and all others vanish, whence X 1 Lfk (0) = fk (0) N 1 1 1, for fk (j) j6=0 N fk (0) : N 1 N 1 P 1 and in the sum j6=k fk (j) there is exactly one term = 1, for = 1+ If i = k then fk (k) = j = 0, whence fk (j) there is exactly one term = j6=0 1 Lfk (k) = fk (k) = = N 1 1 1 N If i 6= 0; k then fk (i) = 0, while in the sum others are 0, whence Lfk (i) = 0 = 1 P X fk (j) N 1 j6=k N = j6=k N N 1 fk (k) : fk (j) there are terms 1; 1 and all fk (i) : 1 Hence, Lfk = NN 1 fk .

26), that is, by k1 + k2 + ::: + n kn where j 2 [0; n] where each ki = 0 or 1. Hence, each eigenvalue of f0; 1gn is equal to 2j n is the number of 1's in the sequence k1 ; :::; kn : The multiplicity of the eigenvalue 2j is n equal to the number of binary sequences fk1 ; :::; kn g where 1 occurs exactly j times. This number is given by the binomial coe cient nj : Hence, all the eigenvalues of the Laplace where j = 0; 1; :::; n, and the multiplicity of this eigenvalue operator on f0; 1gn are 2j n n is j : Note that the total sum of all multiplicities is n X n j j=0 = 2n ; n that is the number of vertices in f0; 1g as expected.

6. 6 1 =1 1: It log 1" 1 (x) . (V ) 0. Here " must be chosen so that " << minx Eigenvalues of Zm Let us give an example of computation of the eigenvalues k of L: Frequently it is more convenient to compute the eigenvalues k = 1 k of the Markov operator P = id L: Let us compute the eigenvalues of the Markov operator on the cycle graph Cm with simple weight. Recall that Cm = Zm = f0; 1; :::; m 1g and the connections are 0 1 2 ::: m 1 0: The Markov operator is given by P f (k) = 1 (f (k + 1) + f (k 2 1)) where k is regarded as a residue mod m.

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