Question

Let A be an n × n real symmetric matrix with its row and column sums both equal to 0. Let λ₁, . . . , λ_n be the eigenvalues of A, with λ_n = 0, and with corresponding eigenvectors v₁,...,v_n (these exist because A is real symmetric). Note that v_n = (1, . . . , 1). Let A[i] be the result of deleting the ith row and column.

Prove that detA[i] = (λ₁···λ_n-1)/n. Thus, the number of spanning trees of G is the product of the nonzero eigenvalues of L, divided by n.

Answer 1

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n-1 times Note that Each poincipa minor LCG) equa) the humber ot sponning tyeesot G Since Ano, the sum ot pro ducト ot eigen ualu , taken n-lata time, equal A-An. ol Hen ce Prov ed