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
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