Therefore we must have n k 2 n 2 which yields n k 2 k

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Therefore, we must have n ( k 2 ) = ( n 2 ) , which yields n = k 2 - k + 1. Proof (continued). Step 5: We claim that A 2 = J + ( k - 1) I, where A is the adjacency matrix of G , I is the n × n identity matrix, and J is the n × n all-1 matrix (i.e. all entries of J are equal to 1). Let the rows and columns of A be indexed by the vertices v 1 , v 2 , . . . , v n of G in this order. Since G is k -regular, for each i , there are exactly k walks of length two from v i to v i , each of which travels from v i to one of its neighbours and then returns to v i . On the other hand, for distinct i and j , since v i and v j have exactly one common neighbour, say z , the walk v i , z, v j is the only walk from v i to v j with length two (and the walk v j , z, v i is the only walk from v j to v i with length two). Since ( A 2 ) ij is the number of walks of length two in G from v i to v j , it follows that ( A 2 ) ii = k for each i and ( A 2 ) ij = 1 for each pair i 6 = j . In other words, A 2 = J + ( k - 1) I . Proof (continued). In the rest proof we will use the following result from linear algebra: Step 6: Let M be an n × n matrix with eigenvalues λ 1 , . . . , λ n . Let f ( x ) = m i =0 a i x i be a polynomial of indeterminant x . Then f ( λ 1 ) , . . . , f ( λ n ) are eigenvalues of f ( M ) = m i =0 a i M i . (Exercise: Prove this result by your- self.) Proof (continued). Step 7: We claim that G has eigenvalues ± k - 1, with total multiplicity n - 1, and k , with multiplicity 1. In fact, one can show that J has eigenvalues 0 and n with multiplicities n - 1 and 1 respectively. (Exercise: Prove this result by yourself.) Since A 2 = J + ( k - 1) I is a polynomial of J , by Step 6, A 2 has eigenvalues 0 + ( k - 1) = k - 1 and n + ( k - 1) = k 2 with multiplicities n - 1 and 1 respectively. (Note that n = k 2 - k + 1 as proved in Step 4.) Thus, by the result in Step 6, G has eigenvalues ± k - 1, with total multiplicity n - 1, and k , with multiplicity 1. (If λ is an eigenvalue of G , then by Step 6, λ 2 is an eigenvalue of A 2 and hence λ 2 = k - 1 or k 2 , implying that λ = ± k - 1 or ± k . As G is k -regular, k must be an eigenvalue of G , and so - k cannot be an eigenvalue of G as the total multiplicity is equal to n .) Proof (continued). Step 8: We now derive a contradiction and thus com- plete the proof.
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By Step 7, the sum of the eigenvalues of G is equal to t 1 k - 1 + t 2 ( - k - 1) + k for some integers t 1 , t 2 0 with t 1 + t 2 = n - 1. Since the sum of the eigenvalues of G must be equal to 0, we have t k - 1 = k , where t = t 2 - t 1 is an integer. That is, k/ k - 1 is an integer, but this hap- pens only when k = 2 (exercise: prove this statement by yourself). Hence n = k 2 - k + 1 = 2 2 - 2 + 1 = 3. That is, G is a friendship graph with exactly three vertices. Since any two of these three vertices have exactly one common neighbour, G must be the complete graph K 3 (which is the same as the cycle C 3 ). Therefore, Δ( G ) = 2 = n - 1, but this contradicts the assumption that Δ( G ) < n - 1. This completes the proof. 5 Eulerian graphs Problem 1 (K¨ onigsberg Bridge Problem) . Can you walk around K¨ onigsberg, crossing each of the seven bridges exactly once, and return to the starting point?
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