对称和反对称多项式矩阵的合同标准形

Congruent Canonical Forms of Symmetric and Skew-Symmetric Polynomial Matrices

摘要: 本文主要研究一元多项式环上对称矩阵和反对称矩阵的合同标准形。利用多项式和初等矩阵的性质，推广传统的理论到多项式矩阵上。通过行列式不为零的矩阵，可将一元多项式环上对称矩阵合同到三对角矩阵。由于反对称多项式矩阵的对角线元素都为零，进一步可以证明反对称多项式矩阵合同到三对角矩阵，且元素有整除性质。

Abstract: This paper studies the congruent canonical forms of symmetric and skew-symmetric matrices over a univariate polynomial ring. Using the properties of polynomials and elementary matrices, the classical theory will be extended to polynomial matrices. The symmetric matrix over a univariate polynomial ring can be congruent to a tridiagonal matrix through a regular matrix. Since the diagonal elements of the skew-symmetric polynomial matrix are zero, it can be further proved that the skew-symmetric polynomial matrix is congruent to a tridiagonal matrix whose elements have the property of division.

1. 引言

二次型理论起源于18世纪对二次曲面分类问题的研究，是整个数学领域的一个重要研究课题。1852年，Sylvester提出了实系数二次型的惯性定理 [1] ，指出实系数二次型通过选取合适的基，可以表示成

$x_1^2+x_2^2+\cdots +x_p^2-x_{p+1}^2-\cdots -x_{p+q}^2$

实际上，Jacobi也独立的证明了该定理。用矩阵语言描述惯性定理是，任意的对称矩阵 $A\in {ℝ}^{n\times n}$，都存在可逆矩阵 $C\in {ℝ}^{n\times n}$ 使得

$C^{\text{T}}AC=\left[\begin{array}{ccc}{I}_p& 0& 0\\ 0& -I_q& 0\\ 0& 0& 0\end{array}\right]$

其中 $I_m$ 表示m阶单位矩阵 [2] 。最早对二次型的研究主要集中在整系数二次型，费马、欧拉、拉格朗日、高斯等著名数学家在研究整系数二次型时，都做出了重要贡献。Jacobi在 [3] 中指出整系数二次型通过线性替换可以化简成一类简洁的形式，以现在矩阵语言的观点看：任意的对称矩阵 $B\in {\mathbb{Z}}^{n\times n}$，都存在 $C\in {\mathbb{Z}}^{n\times n}$ 且 $\det C\ne 0$ 使得 $C^{\text{T}}AC$ 是三对角矩阵。整数环 $\mathbb{Z}$ 上的特殊对称矩阵有更好的标准形，可参阅文献 [4] 。

整数环和多项式环有相似的性质，受文献 [3] 的启发，本文主要使用矩阵方法讨论一元多项式环上对称矩阵和反对称矩阵的合同标准形。在第二部分首先给出一个证明主要结果所需要的引理，该引理也提供了化简多项式矩阵到三对角矩阵的运算方法。对于传统多项式矩阵 $A\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$，都存在可逆矩阵 $P\left(\lambda \right),Q\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 使得 $P\left(\lambda \right)A\left(\lambda \right)Q\left(\lambda \right)$ 是对角矩阵，即Smith标准形。本文将证明反对称多项式矩阵合同到三对角矩阵且元素有类似的整除性质。

2. 主要结果

在下面行文中 $F\left[\lambda \right]$ 表示域F上的一元多项式环。为了证明两个主要结论，首先给出一个引理。

引理1设 $X\left(\lambda \right)=\left[\begin{array}{c}{x}_1\left(\lambda \right)\\ x_2\left(\lambda \right)\\ ⋮\\ x_n\left(\lambda \right)\end{array}\right]\in F{\left[\lambda \right]}^n$，则存在矩阵 $P\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 且 $\det P\left(\lambda \right)\ne 0$ 使得

$P\left(\lambda \right)X\left(\lambda \right)=\left[\begin{array}{c}d\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right]$， $d\left(\lambda \right)\in F\left[\lambda \right].$

证明. 由于 $x_1\left(\lambda \right),x_2\left(\lambda \right),\cdots ,x_n\left(\lambda \right)\in F\left[\lambda \right]$，则存在 $u_1\left(\lambda \right),u_2\left(\lambda \right),\cdots ,u_n\left(\lambda \right)\in F\left[\lambda \right]$ 使得

$u_1\left(\lambda \right)x_1\left(\lambda \right)+u_2\left(\lambda \right)x_2\left(\lambda \right)+\cdots +u_n\left(\lambda \right)x_n\left(\lambda \right)=d\left(\lambda \right)$，

其中 $d\left(\lambda \right)$ 是多项式 $x_1\left(\lambda \right),x_2\left(\lambda \right),\cdots ,x_n\left(\lambda \right)$ 的首项系数为1的最大公因式，即

$d\left(\lambda \right)=\left(x_1\left(\lambda \right),x_2\left(\lambda \right),\cdots ,x_n(\; \lambda \; )\right)$

此时必然存在 $k\in \left\{1,2,\cdots ,n\right\}$ 使得 $u_k\ne 0$，令 $P_{ij}$ 为单位矩阵 $I_n$ 交换了第i行和第j行所得矩阵，则有

$P_{1k}\left(\lambda \right)X\left(\lambda \right)=\left[\begin{array}{c}{x}_k\left(\lambda \right)\\ x_2\left(\lambda \right)\\ ⋮\\ x_n\left(\lambda \right)\end{array}\right].$

再取矩阵

$P_2\left(\lambda \right)=\left[\begin{array}{cccc}{u}_k\left(\lambda \right)& u_2\left(\lambda \right)& \cdots & u_n\left(\lambda \right)\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right]$

则有

$P_2\left(\lambda \right)P_{1k}X\left(\lambda \right)=\left[\begin{array}{c}d\left(\lambda \right)\\ x_2\left(\lambda \right)\\ ⋮\\ x_n\left(\lambda \right)\end{array}\right]$， $d\left(\lambda \right)=\left(x_1\left(\lambda \right),x_2\left(\lambda \right),\cdots ,x_n\left(\lambda \right)\right)\in F\left[\lambda \right].$

取矩阵

$P_3\left(\lambda \right)=\left[\begin{array}{cccc}1& 0& \cdots & 0\\ -\frac{x_2\left(\lambda \right)}{d\left(\lambda \right)}& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ -\frac{x_n\left(\lambda \right)}{d\left(\lambda \right)}& 0& \cdots & 1\end{array}\right]$

由 $d\left(\lambda \right)=\left(x_1\left(\lambda \right),x_2\left(\lambda \right),\cdots ,x_n\left(\lambda \right)\right)\in F\left[\lambda \right]$ 可知 $-\frac{x_2\left(\lambda \right)}{d\left(\lambda \right)},\cdots ,-\frac{x_n\left(\lambda \right)}{d\left(\lambda \right)}\in F\left[\lambda \right]$，因此 $P_3\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$。所以，令 $P\left(\lambda \right)=P_3\left(\lambda \right)P_2\left(\lambda \right)P_{1k}$，则满足 $P\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$， $\det P_3\left(\lambda \right)\ne 0$ 且

$P\left(\lambda \right)X\left(\lambda \right)=\left[\begin{array}{c}d\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right]$， $d\left(\lambda \right)\in F\left[\lambda \right].$

利用上面的引理，可以证明下面的定理，该定理刻画了对称多项式矩阵的合同标准型。

定理2设 $A\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 且 ${\left(A\left(\lambda \right)\right)}^{\text{T}}=A\left(\lambda \right)$，则存在矩阵 $C\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 满足 $\det C\left(\lambda \right)\ne 0$ 使得

${\left(C\left(\lambda \right)\right)}^{\text{T}}A\left(\lambda \right)C\left(\lambda \right)=\left[\begin{array}{ccccc}{d}_{11}\left(\lambda \right)& d_{12}\left(\lambda \right)& 0& \cdots & 0\\ d_{12}\left(\lambda \right)& d_{22}\left(\lambda \right)& d_{23}\left(\lambda \right)& \cdots & 0\\ 0& d_{23}\left(\lambda \right)& d_{33}\left(\lambda \right)& \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & d_{n-1,n}\left(\lambda \right)\\ 0& 0& \cdots & d_{n-1,n}\left(\lambda \right)& d_{nn}\left(\lambda \right)\end{array}\right].$

证明. 设

$A\left(\lambda \right)=\left[\begin{array}{cc}{a}_{11}\left(\lambda \right)& A_{12}\left(\lambda \right)\\ {\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& A_{22}\left(\lambda \right)\end{array}\right]\in F{\left[\lambda \right]}^{n\times n}$

其中 $a_{11}\left(\lambda \right)\in F\left[\lambda \right]$， ${\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-1}$。

由引理1可知，存在 $P_1\left(\lambda \right)\in F{\left[\lambda \right]}^{\left(n-1\right)\times \left(n-1\right)}$ 满足 $\det P_1\left(\lambda \right)\ne 0$ 且使得

$P_1\left(\lambda \right){\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}=\left[\begin{array}{c}{d}_{12}\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right]$

因此

$\begin{array}{l}\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]A\left(\lambda \right)\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}{a}_{11}\left(\lambda \right)& A_{12}\left(\lambda \right)\\ {\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& A_{22}\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cc}{a}_{11}\left(\lambda \right)& A_{12}\left(\lambda \right){\left(P_1\left(\lambda \right)\right)}^{\text{T}}\\ P_1\left(\lambda \right){\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& P_1\left(\lambda \right)A_{22}\left(\lambda \right){\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{ccc}{a}_{11}\left(\lambda \right)& d_{12}\left(\lambda \right)& 0\\ d_{12}\left(\lambda \right)& d_{22}\left(\lambda \right)& B_{12}\left(\lambda \right)\\ 0& {\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& B_{22}\left(\lambda \right)\end{array}\right]\end{array}$，

其中 ${\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-2}$ 且 ${\left(B_{22}\left(\lambda \right)\right)}^{\text{T}}=B_{22}\left(\lambda \right)$。

同理，由引理1可知，存在 $P_2\left(\lambda \right)\in F{\left[\lambda \right]}^{\left(n-2\right)\times \left(n-2\right)}$ 满足 $\det P_2\left(\lambda \right)\ne 0$ 且使得

$P_2\left(\lambda \right){\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}=\left[\begin{array}{c}{d}_{23}\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right]$

因此有

$\begin{array}{l}\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]A\left(\lambda \right)\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\\ =\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\left[\begin{array}{ccc}{a}_{11}\left(\lambda \right)& d_{12}\left(\lambda \right)& 0\\ d_{12}\left(\lambda \right)& d_{22}\left(\lambda \right)& B_{12}\left(\lambda \right)\\ 0& {\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& B_{22}\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\\ =\left[\begin{array}{ccc}{a}_{11}\left(\lambda \right)& d_{12}\left(\lambda \right)& 0\\ d_{12}\left(\lambda \right)& d_{22}\left(\lambda \right)& B_{12}\left(\lambda \right){\left(P_2\left(\lambda \right)\right)}^{\text{T}}\\ 0& P_2\left(\lambda \right){\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& P_2\left(\lambda \right)B_{22}\left(\lambda \right){\left(P_2\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cccc}{a}_{11}\left(\lambda \right)& d_{12}\left(\lambda \right)& 0& 0\\ d_{12}\left(\lambda \right)& d_{22}\left(\lambda \right)& d_{23}\left(\lambda \right)& 0\\ 0& d_{23}\left(\lambda \right)& d_{33}\left(\lambda \right)& C_{12}\left(\lambda \right)\\ 0& 0& {\left(C_{12}\left(\lambda \right)\right)}^{\text{T}}& C_{22}\left(\lambda \right)\end{array}\right]\end{array}$，

其中 ${\left(C_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-3}$ 且 ${\left(C_{22}\left(\lambda \right)\right)}^{\text{T}}=C_{22}\left(\lambda \right)$。

重复上述过程，定理得证。

下面讨论反对称多项式矩阵的合同标准型，形式上比对称矩阵要简洁，并且保留了矩阵多项式的Smith标准形的一些性质。

定理3设 $A\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 且 ${\left(A\left(\lambda \right)\right)}^{\text{T}}=-A\left(\lambda \right)$，则存在矩阵 $C\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 满足 $\det C\left(\lambda \right)\ne 0$ 且使得

${\left(C\left(\lambda \right)\right)}^{\text{T}}A\left(\lambda \right)C\left(\lambda \right)=\left[\begin{array}{ccccc}0& -d_1\left(\lambda \right)& 0& \cdots & 0\\ d_1\left(\lambda \right)& 0& -d_2\left(\lambda \right)& \cdots & 0\\ 0& d_2\left(\lambda \right)& 0& \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & -d_{n-1}\left(\lambda \right)\\ 0& 0& \cdots & d_{n-1}\left(\lambda \right)& 0\end{array}\right]$，

且 $d_i\left(\lambda \right)|d_{i+1}\left(\lambda \right)$， $i=1,2,\cdots ,n-1$。

证明. 设

$A\left(\lambda \right)=\left[\begin{array}{cc}0& -A_{12}\left(\lambda \right)\\ {\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& A_{22}\left(\lambda \right)\end{array}\right]\in F{\left[\lambda \right]}^{n\times n}$

其中 $a_{11}\left(\lambda \right)\in F\left[\lambda \right]$， ${\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-1}$， ${\left(A_{22}\left(\lambda \right)\right)}^{\text{T}}=-A_{22}\left(\lambda \right)$。

由引理可知，存在 $P_1\left(\lambda \right)\in F{\left[\lambda \right]}^{\left(n-1\right)\times \left(n-1\right)}$ 满足 $\det P_1\left(\lambda \right)\ne 0$ 且使得

$P_1\left(\lambda \right){\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}=\left[\begin{array}{c}{d}_1\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right].$

因此

$\begin{array}{l}\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]A\left(\lambda \right)\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}0& -A_{12}\left(\lambda \right)\\ {\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& A_{22}\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cc}0& -A_{12}\left(\lambda \right){\left(P_1\left(\lambda \right)\right)}^{\text{T}}\\ P_1\left(\lambda \right){\left(A_{12}\left(\lambda \right)\right)}^{\text{T}}& P_1\left(\lambda \right)A_{22}\left(\lambda \right){\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{ccc}0& -d_1\left(\lambda \right)& 0\\ d_1\left(\lambda \right)& 0& -B_{12}\left(\lambda \right)\\ 0& {\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& B_{22}\left(\lambda \right)\end{array}\right]\end{array}$，

其中 ${\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-2}$ 且 ${\left(B_{22}\left(\lambda \right)\right)}^{\text{T}}=-B_{22}\left(\lambda \right)$。

同理，由引理可知，存在 $P_2\left(\lambda \right)\in F{\left[\lambda \right]}^{\left(n-2\right)\times \left(n-2\right)}$ 满足 $\det P_2\left(\lambda \right)\ne 0$ 且使得

$P_2\left(\lambda \right){\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}=\left[\begin{array}{c}{d}_2\left(\lambda \right)\\ 0\\ ⋮\\ 0\end{array}\right].$

因此

$\begin{array}{l}\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& P_1\left(\lambda \right)\end{array}\right]A\left(\lambda \right)\left[\begin{array}{cc}1& 0\\ 0& {\left(P_1\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\\ =\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\left[\begin{array}{ccc}0& -d_1\left(\lambda \right)& 0\\ d_1\left(\lambda \right)& 0& -B_{12}\left(\lambda \right)\\ 0& {\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& B_{22}\left(\lambda \right)\end{array}\right]\left[\begin{array}{cc}{I}_2& 0\\ 0& P_2\left(\lambda \right)\end{array}\right]\\ =\left[\begin{array}{ccc}0& -d_1\left(\lambda \right)& 0\\ d_1\left(\lambda \right)& 0& -B_{12}\left(\lambda \right){\left(P_2\left(\lambda \right)\right)}^{\text{T}}\\ 0& P_2\left(\lambda \right){\left(B_{12}\left(\lambda \right)\right)}^{\text{T}}& P_2\left(\lambda \right)B_{22}\left(\lambda \right){\left(P_2\left(\lambda \right)\right)}^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cccc}0& -d_1\left(\lambda \right)& 0& 0\\ d_1\left(\lambda \right)& 0& -d_2\left(\lambda \right)& 0\\ 0& d_2\left(\lambda \right)& 0& -C_{12}\left(\lambda \right)\\ 0& 0& {\left(C_{12}\left(\lambda \right)\right)}^{\text{T}}& C_{22}\left(\lambda \right)\end{array}\right]\end{array}$，

其中 ${\left(C_{12}\left(\lambda \right)\right)}^{\text{T}}\in F{\left[\lambda \right]}^{n-3}$ 且 ${\left(C_{22}\left(\lambda \right)\right)}^{\text{T}}=-C_{22}\left(\lambda \right)$。

重复上述过程可知存在矩阵 $C\left(\lambda \right)\in F{\left[\lambda \right]}^{n\times n}$ 满足

$\det C\left(\lambda \right)=det\left(P_1\left(\lambda \right)P_2\left(\lambda \right)\cdots P_{n-2}\left(\lambda \right)\right)\ne 0$

使得

${\left(C\left(\lambda \right)\right)}^{\text{T}}A\left(\lambda \right)C\left(\lambda \right)=\left[\begin{array}{ccccc}0& -d_1\left(\lambda \right)& 0& \cdots & 0\\ d_1\left(\lambda \right)& 0& -d_2\left(\lambda \right)& \cdots & 0\\ 0& d_2\left(\lambda \right)& 0& \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & -d_{n-1}\left(\lambda \right)\\ 0& 0& \cdots & d_{n-1}\left(\lambda \right)& 0\end{array}\right].$

如果存在 $d_i\left(\lambda \right)\cancel{|}{d}_k\left(\lambda \right)$， $i<k$，则不妨设是 $d_1\left(\lambda \right)\cancel{|}{d}_2\left(\lambda \right)$。此时存在 $u\left(\lambda \right),v\left(\lambda \right)\in F\left[\lambda \right]$ 使得

$u\left(\lambda \right)d_1\left(\lambda \right)+v\left(\lambda \right)d_2\left(\lambda \right)=d\left(\lambda \right)=\left(d_1\left(\lambda \right),d_2(\; \lambda \; )\right)$

此时令

$Q\left(\lambda \right)=\left[\begin{array}{cccc}-u\left(\lambda \right)& 0& v\left(\lambda \right)& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& I_{n-3}\end{array}\right]$，

则满足

$Q\left(\lambda \right){\left(C\left(\lambda \right)\right)}^{\text{T}}A\left(\lambda \right)C\left(\lambda \right){\left(Q\left(\lambda \right)\right)}^{\text{T}}=\left[\begin{array}{ccccc}0& -d\left(\lambda \right)& 0& \cdots & 0\\ d\left(\lambda \right)& 0& -d_2\left(\lambda \right)& \cdots & 0\\ 0& d_2\left(\lambda \right)& 0& \ddots & ⋮\\ ⋮& ⋮& \ddots & \ddots & -d_{n-1}\left(\lambda \right)\\ 0& 0& \cdots & d_{n-1}\left(\lambda \right)& 0\end{array}\right]$

其中 $d\left(\lambda \right)|d_2\left(\lambda \right)$。

因此可以通过上述调整，使得存在整出关系 $d_i\left(\lambda \right)|d_{i+1}\left(\lambda \right)$， $i=1,2,\cdots ,n-1$。

定理得证。

3. 结论

本文主要讨论了多项式环上的对称矩阵和反对称矩阵的合同标准形，使用矩阵方法证明了：1) 一元多项式环上的对称矩阵合同到三对角矩阵；2) 一元多项式环上反对称矩阵合同到三对角矩阵，且元素满足一定的整除性。多项式理论中的Smith标准形有简洁的形式和重要的应用，但缺少了对称性。本文中初步探讨了两类保持对称性的标准形，并且指出反对称多项式矩阵还保留了整除性质。本文中使用的矩阵 $C\left(\lambda \right)$ 满足行列式不为零，是否可以替换成可逆的多项式矩阵值得进一步研究。

基金项目

本论文受临沂大学大学生创新创业训练计划项目资助(X202110452281)，临沂大学教学改革和研究项目资助(JG2021M04)，临沂大学“课程思政”教学师范课程项目(K2021SZ099)。

参考文献