多项式相乘的矩阵形式

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多项式相乘的矩阵形式_黄嘉威.pdf (266.95 KB, Downloads: 25)

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1．多项式相乘的矩阵形式
$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$就是一个多项式相乘的简单例子．虽然多项式相乘可以用卷积来计算，可是卷积的性质还是没有矩阵那么好．
$(a x+b)(c x+d)= \begin{cases}a c x^2\\&+(bc+a d)x\\&&+bd\end{cases}=\left(\begin{array}{lll}x^2 & x & 1\end{array}\right)\left(\begin{array}{ll}a & 0 \\ b & a \\ 0 & b\end{array}\right)\left(\begin{array}{l}c \\ d\end{array}\right) .$
如是者不管两个多项式有多长，都能分解成三个矩阵的乘积．$$\left(a x^2+b x+c\right)(d x+e)=\left(\begin{array}{llll}x^3 & x^2 & x & 1\end{array}\right)\left(\begin{array}{ll}a & 0 \\ b & a \\ c & b \\ 0 & c\end{array}\right)\left(\begin{array}{l}d \\ e\end{array}\right)$$以下会用到矩阵的逆和特征多项式来求解问题，但是用到这些性质之前先把矩阵化成方阵．这里首先说明矩阵上消元的方法．这相当于在$x^2=2$时，把$(ax+b)(cx+d)=(ad+bc)x+bd+2ac$也表达成矩阵乘积的问题．
\begin{aligned} & \left(\begin{array}{lll}x^2 & x & 1\end{array}\right)\left(\begin{array}{ll}a & 0 \\ b & a \\ 0 & b\end{array}\right)\left(\begin{array}{l}c \\ d\end{array}\right)=\left(\begin{array}{ll}x & 1\end{array}\right)\left(\begin{array}{cc}b & a \\ 0+2 a & b+2 \times 0\end{array}\right)\left(\begin{array}{l}c \\ d\end{array}\right) \\ & =\left(\begin{array}{ll}x & 1\end{array}\right)\left(\begin{array}{cc}b & a \\ 2 a & b\end{array}\right)\left(\begin{array}{l}c \\ d\end{array}\right)=\left(\begin{array}{ll}x & 1\end{array}\right)\left(\begin{array}{c}b c+a d \\ 2 a c+b d\end{array}\right)\end{aligned}我们看到第二步已经把中间的矩阵化为方阵，最后算出来的与结果一致．

2．分母有理化
将 $\frac{1}{1+2 \sqrt[3]{2}+3 \sqrt[3]{4}}$ 表为 $1,\sqrt[3]{2},\sqrt[3]{4}$ 的线性组合．对于这个问题在[1]中提供了四种解法，当中用了辗转相除法、待定系数法、线性方程组求解．其中解2不能避免多项式相乘的复杂运算．以下用书[1]中的解2求解上述问题．
$a^3=2$，$(1+2a+3a^2)\left(x+y a+z a^2\right)=1$，则 $x+y a+z a^2$ 为问题所需要的解．
\begin{align*}\left(\begin{array}{lllll}1 & a & a^2 & a^3 & a^4\end{array}\right)\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ 0 & 3 & 2 \\ 0 & 0 & 3\end{array}\right)&=\left(\begin{array}{llll}1 & a & a^2 & a^3\end{array}\right)\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 6 \\ 3 & 2 & 1 \\ 0 & 3 & 2\end{array}\right)\\
&=\left(\begin{array}{lll}1 & a & a^2\end{array}\right)\left(\begin{array}{lll}1 & 6 & 4 \\ 2 & 1 & 6 \\ 3 & 2 & 1\end{array}\right)\end{align*}
于是就得到解2中的线性方程组，求解之，问题就解决了．这里就用到了矩阵的逆．
\begin{aligned} & \left(\begin{array}{lll}1 & a & a^2\end{array}\right)\left(\begin{array}{lll}1 & 6 & 4 \\ 2 & 1 & 6 \\ 3 & 2 & 1\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{lll}1 & a & a^2\end{array}\right)\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right), \\ & \left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{lll}1 & 6 & 4 \\ 2 & 1 & 6 \\ 3 & 2 & 1\end{array}\right)^{-1}\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\frac{-1}{89}\left(\begin{array}{c}11 \\ 16 \\ 1\end{array}\right)\end{aligned}
此解法也适用于书[1]中的例7，即$a^3=a-1,(8-6a+a^2)(x+ya+za^2)=1$.
\begin{aligned} & \left(\begin{array}{ccc}8 & 0 & 0 \\ -6 & 8 & 0 \\ 1 & -6 & 8 \\ 0 & 1 & -6 \\ 0 & 0 & 1\end{array}\right) \rightarrow\left(\begin{array}{ccc}8 & 0 & 0 \\ -6 & 8 & -1 \\ 1 & -6 & 9 \\ 0 & 1 & -6\end{array}\right) \rightarrow\left(\begin{array}{ccc}8 & -1 & 6 \\ -6 & 9 & -7 \\ 1 & -6 & 9\end{array}\right), \\ & \left(\begin{array}{ccc}8 & -1 & 6 \\ -6 & 9 & -7 \\ 1 & -6 & 9\end{array}\right)^{-1}\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\frac{1}{427}\left(\begin{array}{l}39 \\ 47 \\ 27\end{array}\right)\\&x+y a+z a^2= \frac{1}{427}\left(39+47 a+27 a^2\right) .\end{aligned}于是对于分母有理化问题就转化成了求多项式相乘方阵的逆的问题．

3．极小多项式
这里先引入一个简单的问题：若$x^2+ax+b=0,u=cx+d$，构造$u$的极小多项式．
假设关于$u$的方程有$u_1=cx_1+d,u_2=cx_2+d$两个解．
\begin{aligned} & u_1=c x_1+d \\&u_2=c x_2+d \\ & u_1+u_2=c\left(x_1+x_2\right)+2 d=2 d-a c \\ & u_1 u_2=\left(c x_1+d\right)\left(c x_2+d\right)=c^2 x_1 x_2+c d\left(x_1+x_2\right)+d^2= b c^2-a c d+d^2\end{aligned}关于$u$的方程为$u^2+(ac-2d)u+bc^2-acd+d^2=0$，方程左边就是极小多项式．

如是者，所有关于$u$的基本对称多项式都要求出来，这样会很麻烦．但事实上，只需要计算$u$在多项式相乘时的方阵，并求出其特征多项式，问题就一下子解决了．
\begin{aligned} & \left(\begin{array}{ll}c & 0 \\ d & c \\ 0 & d\end{array}\right) \rightarrow\left(\begin{array}{cc}d-a c & c \\ -b c & d\end{array}\right) \\ & \left|\begin{array}{cc}\lambda-(d-a c) & -c \\ b c & \lambda-d\end{array}\right|=\lambda^2+(2 d-a c) \lambda+d^2-a c d+b c^2 .\end{aligned}以下证明：设$x$为多项式方程的根，$u(x)$在多项式相乘时的方阵为$U$，$U$的特征多项式为$f(λ)$，则$f(u(x))=0$.

对$\sum_{r=0}^{n-1} k_{2,r} x^r=u(x) \sum_{r=0}^{n-1} k_{1,r} x^r$进行矩阵分解:
\begin{gather*}\left(\begin{array}{llll}x^{n-1} & x^{n-2} & \cdots & 1\end{array}\right)\left(\begin{array}{lllll}k_{2,n-1} & k_{2,n-2} & \cdots & k_{2,1}\end{array}\right)^T\\= \left(\begin{array}{llll}x^{n-1} & x^{n-2} & \cdots & 1\end{array}\right) U\left(\begin{array}{llll}k_{1, n-1} & k_{1,n-2} & \cdots & k_{1,1}\end{array}\right)^T\end{gather*}
$u(x)= u(x) \times 1,[u(x)]^2=u(x) \times u(x),[u(x)]^m=u(x) \times [u(x)]^{m-1}$，如此类推：\[
[u(x)]^m=\left(\begin{array}{llll}
x^{n-1} & x^{n-2} & \cdots & 1
\end{array}\right) U^m\left(\begin{array}{llll}
0 & 0 & \cdots & 1
\end{array}\right)^T\]
设 $U$ 的特征多项式为 $f(\lambda)=\sum_{m=0}^n c_m \lambda^m$, 则有\begin{aligned}
& \sum_{m=0}^n c_m U^m=0 \\
& f(U)=f(u(x))=\sum_{m=0}^n c_m[u(x)]^m\\&=\left(\begin{array}{llll}
x^{n-1} & x^{n-2} & \cdots & 1
\end{array}\right)
\left(\sum_{m=0}^n c_m U^m\right)\left(\begin{array}{llll}
0 & 0 & \cdots & 1
\end{array}\right)^T=0
\end{aligned}问题得证，于是通过特征多项式就能构造出一个关于$u$的方程．这个方程不一定是极小的，但一般都是极小的．

再以书[1]中习题5-3第4题为例：设$a$是$x^3-2x+2$的根，求$a^2-1$的极小多项式．
\begin{aligned} & \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right) \rightarrow\left(\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 1 \\ -2 & -1 & 0 \\ 0 & 0 & -1\end{array}\right) \rightarrow\left(\begin{array}{ccc}1 & 0 & 1 \\ -2 & 1 & 0 \\ 0 & -2 & -1\end{array}\right), \\ & \left|\begin{array}{ccc}\lambda-1 & 0 & -1 \\ 2 & \lambda-1 & 0 \\ 0 & 2 & \lambda+1\end{array}\right|=\lambda^3-\lambda^2-\lambda-3\end{aligned}结果就是\[\left(a^2-1\right)^3-\left(a^2-1\right)^2-\left(a^2-1\right)-3=0 .\]
于是对于极小多项式问题就转化成了求多项式相乘方阵的特征多项式的问题．

References
[1] 韩士安,林磊.近世代数[M].北京:科学出版社,2009.