Circulant Matrix

设循环矩阵

A = a_{0} a_{n - 1} a_{n - 2} ⋮ a_{1} a_{1} a_{0} a_{n - 1} ⋮ a_{2} \dots \dots \dots ⋱ \dots a_{n - 2} a_{n - 3} a_{n - 4} ⋮ a_{n - 1} a_{n - 1} a_{n - 2} a_{n - 3} ⋮ a_{0} \in M_{n} (R)

设 $ϵ_{k} = e^{\frac{2 kπ i}{n}} k = 0, \dots, n - 1$ , 令

f = a_{0} + a_{1} x + \dots + a_{n - 2} x^{n - 2} + a_{n - 1} x^{n - 1} \in C [x]

于是利用 $ϵ_{k}^{n} = 1$ 有

f (ϵ_{k}) ϵ_{k} f (ϵ_{k}) ϵ_{k}^{n - 1} f (ϵ_{k}) = a_{0} + a_{1} ϵ_{k} + \dots + a_{n - 2} ϵ_{k}^{n - 2} + a_{n - 1} ϵ_{k}^{n - 1} = a_{n - 1} + a_{0} ϵ_{k} + \dots + a_{n - 3} ϵ_{k}^{n - 2} + a_{n - 2} ϵ_{k}^{n - 1} ⋮ = a_{1} + a_{2} ϵ_{k} + \dots + a_{n - 1} ϵ_{k}^{n - 2} + a_{0} ϵ_{k}^{n - 1}

可以写成矩阵形式

f (ϵ_{k}) = 1 ϵ_{k} ⋮ ϵ_{k}^{n - 1} = A 1 ϵ_{k} ⋮ ϵ_{k}^{n - 1}

设

V = 1 ϵ_{0} ⋮ ϵ_{0}^{n - 1} 1 ϵ_{1} ⋮ ϵ_{1}^{n - 1} \dots \dots ⋱ \dots 1 ϵ_{n} ⋮ ϵ_{n - 1}^{n - 1}

则 $V diag (f (ϵ_{0}), \dots, f (ϵ_{n - 1})) = A V$ . 由Vandermonde行列式知 $V$ 可逆，于是

A = V diag (f (ϵ_{0}), \dots, f (ϵ_{n - 1})) V^{- 1}

两边取行列式有

det A = f (ϵ_{0}) \dots f (ϵ_{n - 1})

Lin's Notes Garden