Math Problem Statement

Note that A(Cn) = Qn + Q n = Qn + Qn n −1 is a polynomial in Qn. Thus, the eigenvalues of A(Cn) are obtained by evaluating the same polynomial at each of the eigenvalues of Qn. Thus, by Lemma 3.5, the eigenvalues of A(Cn) are ωk + ωn−k , k = 1,..., n. Note that ωk + ωn−k = ω k

  • ω−k = e 2 πik n + e − 2πik n = 2 cos 2π k n , k = 1,..., n, and the proof is complete.

Solution

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Math Problem Analysis

Mathematical Concepts

Eigenvalues
Circulant Matrices
Polynomials in Matrices

Formulas

Eigenvalues of circulant matrices: λ_k = 2 cos(2πk/n)

Theorems

Lemma 3.5 (Eigenvalues of A(C_n))

Suitable Grade Level

Advanced Undergraduate