例 2.56

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• 第 2 章解答题 13

例 2.56

计算下列循环矩阵 $A$ 的行列式的值：

$$A=\left(\begin{array}{ccccc}{a}_1& a_2& a_3& \cdots & a_n\\ a_n& a_1& a_2& \cdots & a_{n-1}\\ a_{n-1}& a_n& a_1& \cdots & a_{n-2}\\ ⋮& ⋮& ⋮& & ⋮\\ a_2& a_3& a_4& \cdots & a_1\end{array}\right).$$

解答

解 作多项式 $f(x)=a_1+a_{2}x+a_3{x}^2+\cdots +a_n{x}^{n-1}$，令 ${\epsilon }_1,{\epsilon }_2,\cdots ,{\epsilon }_n$ 是 $1$ 的所有 $n$ 次方根。又令

$$V=\left(\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {\epsilon }_1& {\epsilon }_2& {\epsilon }_3& \cdots & {\epsilon }_n\\ {\epsilon }_1^2& {\epsilon }_2^2& {\epsilon }_3^2& \cdots & {\epsilon }_n^2\\ ⋮& ⋮& ⋮& & ⋮\\ {\epsilon }_1^{n-1}& {\epsilon }_2^{n-1}& {\epsilon }_3^{n-1}& \cdots & {\epsilon }_n^{n-1}\end{array}\right),\Lambda =\left(\begin{array}{ccccc}f({\epsilon }_1)& 0& 0& \cdots & 0\\ 0& f({\epsilon }_2)& 0& \cdots & 0\\ 0& 0& f({\epsilon }_3)& \cdots & 0\\ ⋮& ⋮& ⋮& & ⋮\\ 0& 0& 0& \cdots & f({\epsilon }_n)\end{array}\right),$$

则

$$AV=\left(\begin{array}{ccccc}f({\epsilon }_1)& f({\epsilon }_2)& f({\epsilon }_3)& \cdots & f({\epsilon }_n)\\ {\epsilon }_{1}f({\epsilon }_1)& {\epsilon }_{2}f({\epsilon }_2)& {\epsilon }_{3}f({\epsilon }_3)& \cdots & {\epsilon }_{n}f({\epsilon }_n)\\ {\epsilon }_1^{2}f({\epsilon }_1)& {\epsilon }_2^{2}f({\epsilon }_2)& {\epsilon }_3^{2}f({\epsilon }_3)& \cdots & {\epsilon }_n^{2}f({\epsilon }_n)\\ ⋮& ⋮& ⋮& & ⋮\\ {\epsilon }_1^{n-1}f({\epsilon }_1)& {\epsilon }_2^{n-1}f({\epsilon }_2)& {\epsilon }_3^{n-1}f({\epsilon }_3)& \cdots & {\epsilon }_n^{n-1}f({\epsilon }_n)\end{array}\right)=V\Lambda .$$

因此

$$|A||V|=|AV|=|V\Lambda |=|V||\Lambda |.$$

因为 ${\epsilon }_i$ 互不相同，所以 $|V|\ne 0$，从而

$$|A|=|\Lambda |=f({\epsilon }_1)f({\epsilon }_2)\cdots f({\epsilon }_n).$$