λ-矩阵的多项式展开
原文链接
定义. 对于 m × n m \times n m×n 的 λ \lambda λ-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 = 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≤mmaxdeg{aij(λ)} 为 A ( λ ) \mathbf{A}(\lambda) A(λ) 的次数, 显然每个元素的次数不超过 L L L.
定理. 对于 m × n m \times n m×n 的 λ \lambda λ-矩阵 A ( λ ) \mathbf{A}(\lambda) A(λ), 次数为 L L L, 存在唯一的一组常数 m × n m \times n m×n 矩阵 A 0 \mathbf{A}_0 A0, . . . ... ..., A L \mathbf{A}_{L} AL, 使得: A ( λ ) = A 0 + A 1 λ + . . . + A L λ L \mathbf{A}(\lambda)=\mathbf{A}_{0}+\mathbf{A}_{1}\lambda+...+\mathbf{A}_{L}\lambda^{L} A(λ)=A0+A1λ+...+ALλL (称之为 A ( λ ) \mathbf{A}(\lambda) A(λ) 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 i j ( λ ) = a i j 0 + a i j 1 λ + . . . + a i j L λ L , 1 ≤ i ≤ m , 1 ≤ j ≤ n a_{ij}(\lambda)=a_{ij}^{0}+a_{ij}^{1}\lambda + ... + a_{ij}^{L}\lambda^{L}, \ 1 \leq i \leq m, \ 1 \leq j \leq n aij(λ)=aij0+aij1λ+...+aijLλL, 1≤i≤m, 1≤j≤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 A ( λ ) \mathbf{A}(\lambda) A(λ) 表示为 A ( λ ) = A 0 + A 1 λ + . . . + A L λ L \mathbf{A}(\lambda)=\mathbf{A}_{0}+\mathbf{A}_{1}\lambda+...+\mathbf{A}_{L}\lambda^{L} A(λ)=A0+A1λ+...+ALλL.
唯一性: 若不唯一, 则设存在另外一组 m × n m \times n m×n 矩阵 A 0 ′ \mathbf{A}'_0 A0′, …, A L ′ \mathbf{A}'_{L} AL′, 使得: A ( λ ) = A 0 ′ + A 1 ′ λ + . . . + A L ′ λ L \mathbf{A}(\lambda)=\mathbf{A}'_{0}+\mathbf{A}'_{1}\lambda+...+\mathbf{A}'_{L}\lambda^{L} A(λ)=A0′+A1′λ+...+AL′λL.
A ( λ ) = A 0 + A 1 λ + . . . + A L λ L = A 0 ′ + A 1 ′ λ + . . . + A L ′ λ L \mathbf{A}(\lambda)=\mathbf{A}_{0}+\mathbf{A}_{1}\lambda+...+\mathbf{A}_{L}\lambda^{L}=\mathbf{A}'_{0}+\mathbf{A}'_{1}\lambda+...+\mathbf{A}'_{L}\lambda^{L} A(λ)=A0+A1λ+...+ALλL=A0′+A1′λ+...+AL′λL
( A 0 − A 0 ′ ) + ( A 1 − A 1 ′ ) λ + . . . + ( A L − A L ′ ) λ L = 0 (\mathbf{A}_{0}-\mathbf{A}'_{0})+(\mathbf{A}_{1}-\mathbf{A}'_{1})\lambda+...+(\mathbf{A}_{L}-\mathbf{A}'_{L})\lambda^L=\mathbf{0} (A0−A0′)+(A1−A1′)λ+...+(AL−AL′)λL=0
比较系数可知 A 0 = A 0 ′ \mathbf{A}_{0}=\mathbf{A}'_{0} A0=A0′,…, A L = A L ′ \mathbf{A}_{L}=\mathbf{A}'_{L} AL=AL′. 矛盾.
存在性的过程也提供了展开式的求法.
定理. A ( λ ) \mathbf{A}(\lambda) A(λ) 和 B ( λ ) \mathbf{B}(\lambda) B(λ) 是 n n n 阶 λ \lambda λ-矩阵, 记 deg A ( λ ) = L \deg \mathbf{A}(\lambda)=L degA(λ)=L, deg B ( λ ) = M \deg \mathbf{B}(\lambda)=M degB(λ)=M, 有: L , M > 0 L, M \gt 0 L,M>0, 且 B ( λ ) \mathbf{B}(\lambda) B(λ) 的多项式展开式中 λ M \lambda^{M} λM 项的系数矩阵可逆, 则存在 n n n 阶 λ \lambda λ-矩阵 U ( λ ) \mathbf{U}(\lambda) U(λ), V ( λ ) \mathbf{V}(\lambda) V(λ), deg V ( λ ) < M \deg\mathbf{V}(\lambda)<M degV(λ)<M, 使得 A ( λ ) = U ( λ ) B ( λ ) + V ( λ ) \mathbf{A}(\lambda)=\mathbf{U}(\lambda)\mathbf{B}(\lambda)+\mathbf{V}(\lambda) A(λ)=U(λ)B(λ)+V(λ).
证明: 当 L < M L<M L<M 时, 令 U ( λ ) = 0 \mathbf{U}(\lambda)=\mathbf{0} U(λ)=0, V ( λ ) = A ( λ ) \mathbf{V}(\lambda)=\mathbf{A}(\lambda) V(λ)=A(λ), 显然 U ( λ ) \mathbf{U}(\lambda) U(λ) 和 V ( λ ) \mathbf{V}(\lambda) V(λ) 即为所求. 接下来用数学归纳法证明当 L ≥ M L\geq M L≥M 时结论成立:
当 L = M = 1 L=M=1 L=M=1 时, 设 A ( λ ) = A 0 + A 1 λ \mathbf{A}(\lambda)=\mathbf{A}_0+\mathbf{A}_1 \lambda A(λ)=A0+A1λ, B ( λ ) = B 0 + B 1 λ \mathbf{B}(\lambda)=\mathbf{B}_0+\mathbf{B}_1 \lambda B(λ)=B0+B1λ, 令 U ( λ ) = A 1 B 1 − 1 \mathbf{U}(\lambda)=\mathbf{A}_1\mathbf{B}_{1}^{-1} U(λ)=A1B1−1, V ( λ ) = A 0 − A 1 B 1 − 1 \mathbf{V}(\lambda)=\mathbf{A}_0-\mathbf A_1\mathbf B_{1}^{-1} V(λ)=A0−A1B1−1 即为所求.
若结论对于 L = k L=k L=k 成立, 当 L = k + 1 L=k+1 L=k+1 时: 设 A ( λ ) = A 0 + A 1 λ + . . . + A L λ L \mathbf{A}(\lambda)=\mathbf{A}_{0}+\mathbf{A}_{1}\lambda+...+\mathbf{A}_{L}\lambda^{L} A(λ)=A0+A1λ+...+ALλL, B ( λ ) = B 0 + B 1 λ + . . . + B M λ M \mathbf{B}(\lambda)=\mathbf{B}_{0}+\mathbf{B}_{1}\lambda+...+\mathbf{B}_{M}\lambda^{M} B(λ)=B0+B1λ+...+BMλM, 令 A ′ ( λ ) = A ( λ ) − A L B M − 1 λ L − M B ( λ ) \mathbf {A}'(\lambda) = \mathbf{A}(\lambda)-\mathbf{A}_{L}\mathbf{B}^{-1}_{M}\lambda^{L-M}\mathbf{B}(\lambda) A′(λ)=A(λ)−ALBM−1λL−MB(λ), 易验证 A ′ ( λ ) \mathbf{A}'(\lambda) A′(λ) 次数小于 L L L, 根据归纳假设, 存在 n n n 阶 λ \lambda λ-矩阵 U ′ ( λ ) \mathbf{U}'(\lambda) U′(λ), V ′ ( λ ) \mathbf{V}'(\lambda) V′(λ), deg V ′ ( λ ) < M \deg\mathbf{V}'(\lambda)<M degV′(λ)<M, 使得 A ′ ( λ ) = U ′ ( λ ) B ( λ ) + V ′ ( λ ) \mathbf{A}'(\lambda)=\mathbf{U}'(\lambda)\mathbf{B}(\lambda)+\mathbf{V}'(\lambda) A′(λ)=U′(λ)B(λ)+V′(λ). 进而有 A ( λ ) = [ A L B M − 1 λ L − M + U ′ ( λ ) ] B ( λ ) + V ′ ( λ ) \mathbf{A}(\lambda)=[\mathbf{A}_{L}\mathbf{B}^{-1}_{M}\lambda^{L-M}+\mathbf{U}'(\lambda)]\mathbf{B}(\lambda)+\mathbf{V}'(\lambda) A(λ)=[ALBM−1λL−M+U′(λ)]B(λ)+V′(λ). 定义 U ( λ ) = A L B M − 1 λ L − M + U ′ ( λ ) \mathbf{U}(\lambda)=\mathbf{A}_{L}\mathbf{B}^{-1}_{M}\lambda^{L-M}+\mathbf{U}'(\lambda) U(λ)=ALBM−1λL−M+U′(λ), V ( λ ) = V ′ ( λ ) \mathbf{V}(\lambda)=\mathbf{V}'(\lambda) V(λ)=V′(λ), 显然 U ( λ ) \mathbf{U}(\lambda) U(λ), V ( λ ) \mathbf{V}(\lambda) V(λ) 即为所求.
同理可证明: A ( λ ) \mathbf{A}(\lambda) A(λ) 和 B ( λ ) \mathbf{B}(\lambda) B(λ) 是 n n n 阶 λ \lambda λ-矩阵, 记 deg A ( λ ) = L \deg \mathbf{A}(\lambda)=L degA(λ)=L, deg B ( λ ) = M \deg \mathbf{B}(\lambda)=M degB(λ)=M, 有: L , M > 0 L, M \gt 0 L,M>0, 且 B M \mathbf{B}_{M} BM 可逆, 则存在 n n n 阶 λ \lambda λ-矩阵 U ( λ ) \mathbf{U}(\lambda) U(λ), V ( λ ) \mathbf{V}(\lambda) V(λ), deg V ( λ ) < M \deg\mathbf{V}(\lambda)<M degV(λ)<M, 使得 A ( λ ) = A ( λ ) U ( λ ) + V ( λ ) \mathbf{A}(\lambda)=\mathbf{A}(\lambda)\mathbf{U}(\lambda)+\mathbf{V}(\lambda) A(λ)=A(λ)U(λ)+V(λ).