λ-矩阵的多项式展开

原文链接

定义. 对于 $m\times n$ 的 $\lambda$-矩阵 A ( λ ) = [ a 11 ( λ ) . . . a 1 n ( λ ) ⋮ ⋮ a m 1 ( λ ) . . . a m n ( λ ) ] \mathbf{A}(\lambda)=\begin{bmatrix} a_{11}(\lambda) & ... & a_{1n}(\lambda)\\ \vdots & & \vdots \\ a_{m1}(\lambda) & ... & a_{mn}(\lambda) \end{bmatrix} A(λ)= ​a11​(λ)⋮am1​(λ)​......​a1n​(λ)⋮amn​(λ)​ ​

称 $L=\underset{\genfrac{}{}{0ex}{}{1\le i\le m}{1\le j\le n}}{\max }\mathrm{deg}\{a_{ij}(\lambda )\}$ 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

A ( λ ) = [ a 11 ( λ ) . . . a 1 n ( λ ) ⋮ ⋮ a m 1 ( λ ) . . . a m n ( λ ) ] \mathbf{A}(\lambda)=\begin{bmatrix} a_{11}(\lambda) & ... & a_{1n}(\lambda)\\ \vdots & & \vdots \\ a_{m1}(\lambda) & ... & a_{mn}(\lambda) \end{bmatrix} A(λ)= ​a11​(λ)⋮am1​(λ)​......​a1n​(λ)⋮amn​(λ)​ ​

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 A r = [ a 11 r . . . a 1 n r ⋮ ⋮ a m 1 r . . . a m n r ] ,   0 ≤ r ≤ L \mathbf{A}_{r}=\begin{bmatrix} a_{11}^{r} & ... & a_{1n}^{r}\\ \vdots & & \vdots \\ a_{m1}^{r} & ... & a_{mn}^{r} \end{bmatrix},\ 0\leq r\leq L Ar​= ​a11r​⋮am1r​​......​a1nr​⋮amnr​​ ​, 0≤r≤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 $\mathbf{A}(\lambda )=\mathbf{A}(\lambda )\mathbf{U}(\lambda )+\mathbf{V}(\lambda )$.