Consider the n×n matrix
An=(01…1−10⋱⋮⋮⋱⋱1−1…−10).
It can be shown that the characteristic polynomial of An is given by p(λ)=det. So, when n is even, we have \det(A_n)=(-1)^np(0)=1 and the eigenvalues of A_n appear in reciprocal pairs, i.e. if \lambda is an eigenvalue, so is 1/\lambda.
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