By Alexander Grigoryan

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2. Let V = f1; 2; 3g with edges 1 2 then Lf (1) = f (1) 3 1, that is, (V; E) = C3 = K3 . We have 1 (f (2) + f (3)) 2 and similar identities for Lf (2) and Lf (3) : The action of L can be written as a matrix multiplication: 0 1 0 10 1 Lf (1) 1 1=2 1=2 f (1) @ Lf (2) A = @ 1=2 1 1=2 A @ f (2) A : Lf (3) 1=2 1=2 1 f (3) 3 The characteristic polynomial of the above 3 3 matrix is 3 2 + 49 its roots, we obtain the following eigenvalues of L: = 0 (simple) and multiplicity 2. 3. Let V = f1; 2; 3g with edges 1 2 3.

Also, we have always N 1 2 and the inequality is strict if the graph is non-bipartite. 3 Convergence to equilibrium Let P be the Markov operator associated with a weighted graph (V; ). We consider it as a linear operator from F to F. Recall that it is related to the Laplace operator L by the identity P = id L. It follows that all the eigenvalues of P have the form 1 where is an eigenvalue of L, and the eigenfunctions of P and L are the same. N 1 Denote k = 1 k so that f k gk=0 is the sequence of all the eigenvalues of P in the decreasing order, counted with multiplicities.

An operator A is called symmetric (or self-adjoint) with respect to this inner product if (Au; v) = (u; Av) for all u; v 2 V. The following theorem collects important results from Linear Algebra about symmetric operators. Theorem. Let A be a symmetric operator in a N -dimensional inner product space V over R: (a) All eigenvalues of A are real. Hence, we can enumerate all the eigenvalues of A in increasing order as 1 ; :::; N where each eigenvalue is counted with multiplicity. (b) (Diagonalization of symmetric operators) There is an orthonormal1 basis fvk gN k=1 in V such that each vk is an eigenvector of A with the eigenvalue k , that is Avk = k vk (equivalently, the matrix of A in the basis fvk g is diag ( 1 ; :::; N )).

### Analysis on Graphs by Alexander Grigoryan

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