λ-矩阵的多项式展开

原文链接

定义. 对于 $m\times n$ 的 $\lambda$-矩阵 $$\mathbf{A}(\lambda )=\left[\begin{array}{ccc}{a}_{11}(\lambda )& ...& a_{1n}(\lambda )\\ ⋮& & ⋮\\ a_{m1}(\lambda )& ...& a_{mn}(\lambda )\end{array}\right]$$

称 L = max ⁡ 1 ≤ i ≤ m 1 ≤ j ≤ n deg ⁡ { a i j ( λ ) } L=\max\limits_{1\leq i\leq m\atop{1\leq j \leq n}}\deg \{a_{ij}(\lambda)\} L=1≤j≤n1≤i≤m​max​deg{ aij​(λ)} 为 $\mathbf{A}(\lambda )$ 的次数, 显然每个元素的次数不超过 $L$.

定理. 对于 $m\times n$ 的 $\lambda$-矩阵 $\mathbf{A}(\lambda )$, 次数为 $L$, 存在唯一的一组常数 $m\times n$ 矩阵 ${\mathbf{A}}_0$, $...$, ${\mathbf{A}}_L$, 使得: $\mathbf{A}(\lambda )={\mathbf{A}}_0+{\mathbf{A}}_1\lambda +...+{\mathbf{A}}_L{\lambda }^L$ (称之为 $\mathbf{A}(\lambda )$ 的多项式展开式).

存在性: 设

$$\mathbf{A}(\lambda )=\left[\begin{array}{ccc}{a}_{11}(\lambda )& ...& a_{1n}(\lambda )\\ ⋮& & ⋮\\ a_{m1}(\lambda )& ...& a_{mn}(\lambda )\end{array}\right]$$

其中 $a_{ij}(\lambda )=a_{ij}^0+a_{ij}^1\lambda +...+a_{ij}^L{\lambda }^L,1\le i\le m,1\le j\le n$, 令 $${\mathbf{A}}_r=\left[\begin{array}{ccc}{a}_{11}^r& ...& a_{1n}^r\\ ⋮& & ⋮\\ a_{m1}^r& ...& a_{mn}^r\end{array}\right],0\le r\le L$$

则可以将 $\mathbf{A}(\lambda )$ 表示为 $\mathbf{A}(\lambda )={\mathbf{A}}_0+{\mathbf{A}}_1\lambda +...+{\mathbf{A}}_L{\lambda }^L$.

唯一性: 若不唯一, 则设存在另外一组 $m\times n$ 矩阵 ${\mathbf{A}}_0^{\prime }$, …, ${\mathbf{A}}_L^{\prime }$, 使得: $\mathbf{A}(\lambda )={\mathbf{A}}_0^{\prime }+{\mathbf{A}}_1^{\prime }\lambda +...+{\mathbf{A}}_L^{\prime }{\lambda }^L$.

$$\mathbf{A}(\lambda )={\mathbf{A}}_0+{\mathbf{A}}_1\lambda +...+{\mathbf{A}}_L{\lambda }^L={\mathbf{A}}_0^{\prime }+{\mathbf{A}}_1^{\prime }\lambda +...+{\mathbf{A}}_L^{\prime }{\lambda }^L$$

$$({\mathbf{A}}_0-{\mathbf{A}}_0^{\prime })+({\mathbf{A}}_1-{\mathbf{A}}_1^{\prime })\lambda +...+({\mathbf{A}}_L-{\mathbf{A}}_L^{\prime }){\lambda }^L=0$$

比较系数可知 ${\mathbf{A}}_0={\mathbf{A}}_0^{\prime }$,…,${\mathbf{A}}_L={\mathbf{A}}_L^{\prime }$. 矛盾.

存在性的过程也提供了展开式的求法.

定理. $\mathbf{A}(\lambda )$ 和 $\mathbf{B}(\lambda )$ 是 $n$ 阶 $\lambda$-矩阵, 记 $\mathrm{deg}\mathbf{A}(\lambda )=L$, $\mathrm{deg}\mathbf{B}(\lambda )=M$, 有: $L,M>0$, 且 $\mathbf{B}(\lambda )$ 的多项式展开式中 ${\lambda }^M$ 项的系数矩阵可逆, 则存在 $n$ 阶 $\lambda$-矩阵 $\mathbf{U}(\lambda )$, $\mathbf{V}(\lambda )$, $\mathrm{deg}\mathbf{V}(\lambda )<M$, 使得 $\mathbf{A}(\lambda )=\mathbf{U}(\lambda )\mathbf{B}(\lambda )+\mathbf{V}(\lambda )$.

证明: 当 $L<M$ 时, 令 $\mathbf{U}(\lambda )=0$, $\mathbf{V}(\lambda )=\mathbf{A}(\lambda )$, 显然 $\mathbf{U}(\lambda )$ 和 $\mathbf{V}(\lambda )$ 即为所求. 接下来用数学归纳法证明当 $L\ge M$ 时结论成立:
当 $L=M=1$ 时, 设 $\mathbf{A}(\lambda )={\mathbf{A}}_0+{\mathbf{A}}_1\lambda$, $\mathbf{B}(\lambda )={\mathbf{B}}_0+{\mathbf{B}}_1\lambda$, 令 $\mathbf{U}(\lambda )={\mathbf{A}}_1{\mathbf{B}}_1^{-1}$, $\mathbf{V}(\lambda )={\mathbf{A}}_0-{\mathbf{A}}_1{\mathbf{B}}_1^{-1}$ 即为所求.
若结论对于 $L=k$ 成立, 当 $L=k+1$ 时: 设 $\mathbf{A}(\lambda )={\mathbf{A}}_0+{\mathbf{A}}_1\lambda +...+{\mathbf{A}}_L{\lambda }^L$, $\mathbf{B}(\lambda )={\mathbf{B}}_0+{\mathbf{B}}_1\lambda +...+{\mathbf{B}}_M{\lambda }^M$, 令 ${\mathbf{A}}^{\prime }(\lambda )=\mathbf{A}(\lambda )-{\mathbf{A}}_L{\mathbf{B}}_M^{-1}{\lambda }^{L-M}\mathbf{B}(\lambda )$, 易验证 ${\mathbf{A}}^{\prime }(\lambda )$ 次数小于 $L$, 根据归纳假设, 存在 $n$ 阶 $\lambda$-矩阵 ${\mathbf{U}}^{\prime }(\lambda )$, ${\mathbf{V}}^{\prime }(\lambda )$, $\mathrm{deg}{\mathbf{V}}^{\prime }(\lambda )<M$, 使得 ${\mathbf{A}}^{\prime }(\lambda )={\mathbf{U}}^{\prime }(\lambda )\mathbf{B}(\lambda )+{\mathbf{V}}^{\prime }(\lambda )$. 进而有 $\mathbf{A}(\lambda )=[{\mathbf{A}}_L{\mathbf{B}}_M^{-1}{\lambda }^{L-M}+{\mathbf{U}}^{\prime }(\lambda )]\mathbf{B}(\lambda )+{\mathbf{V}}^{\prime }(\lambda )$. 定义 $\mathbf{U}(\lambda )={\mathbf{A}}_L{\mathbf{B}}_M^{-1}{\lambda }^{L-M}+{\mathbf{U}}^{\prime }(\lambda )$, $\mathbf{V}(\lambda )={\mathbf{V}}^{\prime }(\lambda )$, 显然 $\mathbf{U}(\lambda )$, $\mathbf{V}(\lambda )$ 即为所求.

同理可证明: $\mathbf{A}(\lambda )$ 和 $\mathbf{B}(\lambda )$ 是 $n$ 阶 $\lambda$-矩阵, 记 $\mathrm{deg}\mathbf{A}(\lambda )=L$, $\mathrm{deg}\mathbf{B}(\lambda )=M$, 有: $L,M>0$, 且 ${\mathbf{B}}_M$ 可逆, 则存在 $n$ 阶 $\lambda$-矩阵 $\mathbf{U}(\lambda )$, $\mathbf{V}(\lambda )$, $\mathrm{deg}\mathbf{V}(\lambda )<M$, 使得 $\mathbf{A}(\lambda )=\mathbf{A}(\lambda )\mathbf{U}(\lambda )+\mathbf{V}(\lambda )$.