$\S 1$ Definition of Circular Matrix and Polynomial Representation
Let $K$ be the number field. Take any $n$ numbers $a_1, a_2, \cdots, a_n$ in $K$, the following matrices are called $n$ order circulant matrices on $K$:
$$A=\begin{pmatrix} a_1 & a_2 & a_3 & \cdots & a_n \\ a_n & a_1 & a_2 & \cdots & a_{n-1} \\ a_{n-1} & a_n & a_1 & \cdots & a_{n-2} \\ \vdots & \vdots & \vdots & & \vdots \\ a_2 & a_3 & a_4 & \cdots & a_1 \\ \end{pmatrix}.\qquad(1)$$
Taking $a_2=1$, $a_1=a_3=\cdots=a_n=0$, the following basic cyclic matrix can be obtained:
$$J=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 0 & \cdots & 0 \\ \end{pmatrix}.\qquad(2)$$
From Example 2.1 of the Fudan High Generation White Paper, $J^k=\begin{pmatrix} 0 & I_{nk} \\ I_k & 0 \\ \end{pmatrix}\,(1\leq k\leq n)$ , so $$A=a_1I_n+a_2J+a_3J^2+\cdots+a_nJ^{n-1}.\qquad(3)$$ let $g(x)=a_1+a_2x+a_3x^2+\cdots+ a_nx^{n-1}$, then $g(x)$ is a polynomial whose degree does not exceed $n-1$ on $K$, so that $A=g(J)$, this is the cyclic matrix about the basic cyclic matrix The polynomial representation of dimensional linear space, its set of basis is $\{I_,J,\cdots,J^{n-1}\}$. Then take any circulant matrix $B=h(J)$, where $h(x)$ is a polynomial whose degree does not exceed $n-1$ on $K$, then using polynomial multiplication and $J^n=I_n$ we can see that $AB=g(J)h(J)$ is still a circulant matrix (refer to the white paper Example 2.12). Therefore, $C_n(K)$ is an $n$-dimensional commutative algebra on $K$, isomorphic to $K[x]/(x^n-1)$.
$\S 2$ Properties of Circular Matrices
The following will study the properties of eigenvalues, eigenvectors, and diagonalizability of circulant matrices in turn, from which information such as the determinant, rank, and non-isotropy of circulant matrices can be obtained. These contents are included in the white paper's example 2.52, example 6.9, example 6.32 and the corollary of Example 6.39.
It is easy to calculate the characteristic polynomial $|\lambda I_n-J|=\lambda^n-1$ of the underlying cyclic matrix $J$, so that $J$ has $n$ different eigenvalues in the complex number domain, that is, $n $ subunit root $\omega_k=\cos\dfrac{2k\pi}{n}+i\sin\dfrac{2k\pi}{n}\,(0\leq k\leq n-1)$, so $J$ can be diagonalized in the complex number field. After calculation, the eigenvector of the eigenvalue $\omega_k$ is $\alpha_k=(1,\omega_k,\omega_k^2,\cdots,\omega_k^{n- 1})'$. Block these eigenvectors into a matrix by column: $$P=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & \omega_1 & \omega_2 & \cdots & \omega_{n-1} \\ 1 & \omega_1^2 & \omega_2^2 & \cdots & \omega_{n-1}^2 \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & \omega_1^{n-1} & \omega_2^{n-1} & \cdots & \omega_{n-1}^{n-1} \\ \end{pmatrix},\quad(4)$ $ Then from the properties of Vander Monde determinant or eigenvalue eigenvector, $P$ is non-isomorphic, and satisfies $$P^{-1}JP=\mathrm{diag}\{1,\omega_1,\omega_2, \cdots,\omega_{n-1}\},\quad(5)$$ thus $$P^{-1}AP=P^{-1}g(J)P=g(P^{-1 }JP)=\mathrm{diag}\{g(1),g(\omega_1),g(\omega_2),\cdots,g(\omega_{n-1})\}.\quad(6)$$ From equation (6), the cyclic matrix $A=g(J )$ The eigenvalues are $g(1),g(\omega_1),g(\omega_2),\cdots,g(\omega_{n-1})$, the corresponding eigenvectors are $\alpha_0,\alpha_1 ,\alpha_2,\cdots,\alpha_{n-1}$, and $A$ is diagonalizable. In particular, $|A|=g(1)g(\omega_1)g(\omega_2)\cdots g (\omega_{n-1})$, $r(A)=\sharp\{0\leq k\leq n-1\mid g(\omega_k)\neq 0\}$, $A$ is not equivalent And only if $g(\omega_k)\neq 0\,(0\leq k\leq n-1)$, that is, if and only if $(\lambda^n-1,g(\lambda))=1 $.
Theorem 1 There is a natural algebraic isomorphism between the totality of $n$ complex cyclic matrices $C_n(\mathbb{C})$ and the $n$ complex diagonal matrix total $D_n(\mathbb{C})$$ \xi$.
Prove that the entire complex diagonal matrix of order $n$ becomes an algebra on the complex number field under the addition, multiplication and multiplication of matrices. We define the mapping $\xi$ by formula (6), that is, $\xi: C_n(\mathbb {C})\to D_n(\mathbb{C})$ is defined as $\xi(A)=P^{-1}AP=\mathrm{diag}\{g(1),g(\omega_1), g(\omega_2),\cdots,g(\omega_{n-1})\}$. It is easy to verify that $\xi$ holds matrix addition, number multiplication and multiplication, and thus is an algebraic homomorphism. For any $ \Lambda=\mathrm{diag}\{\lambda_0,\lambda_1,\cdots,\lambda_{n-1}\}$, using the Lagrange interpolation formula, it can be known that there is a polynomial $h( \lambda)$, so that $h(\omega_k)=\lambda_k\,(0\leq k\leq n-1)$. Let $B=h(J)$, then $\xi(B)=\Lambda $, that is, $\xi$ is a surjective. And $\dim C_n(\mathbb{C})=\dim D_n(\mathbb{C})=n$, so $\xi$ is a linear isomorphism, thus is an algebraic isomorphism. $\Box$
Corollary 2 A necessary and sufficient condition for a complex matrix $B$ of order $n$ to be diagonalizable is that $B$ is similar to a circulant matrix.
Prove that the isomorphism $\xi$ in Theorem 1 is realized by similarity transformation and the conclusion is obtained. $\Box$
Corollary 3 If $A\in C_n(K)$ is a non-isotropic matrix, then $A^{-1}$ is also a circulant matrix.
Proof 1 From the conclusion of Theorem 1, $A^{-1}=\xi^{-1}\bigg(\xi(A)^{-1}\bigg)$ is also a cyclic matrix.
Proof 2 According to the Cayley-Hamilton theorem, there is a polynomial $h(\lambda)\in K[\lambda]$ such that $A^{-1}=h(A)$ (refer to Example 6.61 of the white paper). Let $A=g(J)$, then $A^{-1}=h(A)=h(g(J))$ is still a polynomial of $J$, thus a circulant matrix.
Proof Three Set $A=g(J)$, then it can be known from $A$ that $(\lambda^n-1,g(\lambda))=1$. According to the properties of coprime polynomials, there is $ u(\lambda),v(\lambda)$, so that $(\lambda^n-1)u(\lambda)+g(\lambda)v(\lambda)=1$. Let $\lambda=J$ , Substitute into the above formula to get $g(J)v(J)=I_n$, so $A^{-1}=v(J)$ is also a cyclic matrix. $\Box$
Corollary 4 The entire set of $GC_n(K)$ non-heterocyclic matrices of order $GC_n(K)$ on $K$ becomes an Abel group under matrix multiplication.
$\S 3$ Application of Circular Matrices
Below we give an application of circulant matrices.
Proposition 5 has $n^2\,(n\geq 2)$ different numbers in $K$, then there is a full permutation, denoted as $a_1,\cdots,a_{n^2}$, such that $ $\begin{vmatrix} a_1 & a_2 & \cdots & a_n \\ a_{n+1} & a_{n+2} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n^2-n+1} & a_{n^2-n+2} & \cdots & a_{n^2} \\ \end{vmatrix}\neq 0.$$
Prove induction on $n$. When $n=2$, take $a_1,a_2$ first, so that $a_1+a_2\neq 0$, so $\begin{vmatrix} a_1 & a_2 \\ a_2 & a_1 \\ \end{vmatrix}=(a_1-a_2)(a_1+a_2)\neq 0$, so $B=\{(a_1,a_2),(a_2,a_1)\}$ is a set of $K^2$ Basis. Note that $(a_3,a_4)\neq 0$, so by the basis expansion theorem, a basis vector must be selected from the basis $B$, which may be set to $(a_1,a_2)$, so that $\{( a_1,a_2),(a_3,a_4)\}$ becomes a basis of $K^2$, so $\begin{vmatrix} a_1 & a_2 \\ a_3 & a_4 \\ \end{vmatrix}\neq 0$ . Let $n-1$ hold the conclusion, and now prove the case of $n$.
proof one First get $a_1,a_2,\cdots,a_n$, so that $a_1+a_2\omega_k+\cdots+a_n\omega_k^{n-1}\neq 0$ is true for $0\leq k\leq n-1$. This must be done. For example, if $a_2,\cdots,a_n$ are selected first, then there are only $n$ of $a_1$ that do not meet the above conditions, so that $a_1$ that satisfies the above conditions can be obtained. It can be seen from the properties that the circulant matrix $A$ in Eq. (1) is non-isotropic, in particular, the $n$ row vectors $\{\beta_1,\beta_2,\cdots,\beta_n\}$ of $A$ are $ A set of bases for K^n$. By induction hypothesis, the full permutation of $(n-1)^2$ numbers can be selected from the remaining $n^2-n$ numbers, such that $$\begin{ vmatrix} a_{n+2} & \cdots & a_{2n} \\ \vdots & & \vdots \\ a_{n^2-n+2} & \cdots & a_{n^2} \\ \end {vmatrix}\neq 0,$$ arbitrarily choose $a_{n+1},\cdots,a_{n^2-n+1}$ behind, can make $n-1$ row vector $(a_{ n+1},a_{n+2},\cdots,a_{2n})$, $\cdots$, $(a_{n^2-n+1},a_{n^2-n+2} ,\cdots,a_{n^2})$ is linearly independent (refer to Exercise 3.4.9 in Fudan Gaodai Textbook). Therefore, from the basis expansion theorem, it must be possible from the basis $\{\beta_1,\beta_2,\cdots,\ Select a basis vector from beta_n\}$, let’s set it as $\beta_1$, so that $\{(a_1,a_2,\cdots,a_n)$, $(a_{n+1},a_{n+2} ,\cdots,a_{2n})$, $\cdots$, $(a_{n^2-n+1},a_{n^2-n+2},\cdots,a_{n^2})\}$ constitutes a set of basis of $K^n$, so the conclusion is proved.
The second method of proof uses the method of proof by contradiction. Assuming that for all permutations of $n^2$, the corresponding determinants are equal to zero, we will deduce the contradiction. First take $a_1,a_2,\cdots,a_n$, so that $a_1+a_2+ \cdots+a_n\neq 0$, and then by the induction hypothesis, may wish to assume that in the obtained determinant, the algebraic cofactor of $a_1$ is $A_1\neq 0$. Let the algebraic cofactor of the remaining elements $a_i$ be $A_i \,(2\leq i\leq n)$, so $a_1A_1+a_2A_2+\cdots+a_nA_n=0$. In the obtained determinant, swap $a_1$ and $a_i\,(2 \leq i\leq n)$, the remaining $n^2-2$ elements remain unchanged, then there are $a_iA_1+\cdots+a_1A_i+\cdots+a_nA_n=0$. From this, we can get $(a_1-a_i)( A_1-A_i)=0$, but $a_1\neq a_i$, thus $A_1=A_i\,(2\leq i\leq n)$. Finally, $0=a_1A_1+a_2A_2+\cdots+a_nA_n=(a_1+a_2+ \cdots+a_n)A_1\neq 0$, contradicting. $\Box$
Note that Proposition 1 of Proposition 5 is constructive, and it can be used to give an estimate of the total number of full permutations that satisfy the conditions. Proposition 2 of Proposition 5 is provided by Zhu Minzhe, a Grade 16 undergraduate student of Fudan University of Mathematics and Science.
The main conclusions of this paper can also be extended to the domain of feature zero or the domain of feature $p>0$ (requires $p\nmid n$) and its splitting domain. In addition, the solution to problem 13 in Chapter 2 of the white paper also gives Generalization of $b-$ circulant matrices. Interested readers can learn and verify these conclusions by themselves.