标准形方法I:等价标准形

标签：begin right end 等价 标准 pmatrix array 方法 left

在高等代数课程里，我们学习过矩阵的等价、相似与合同三种等价关系，在每个等价类里我们都可以选择一个“相对简单”的代表元，称为等价标准形. 若待解决的问题在某一等价关系（相抵、相似于或合同）下不变，或与等价类的代表元的选取无关，则对一般矩阵\(A\)的问题可以转化为选取\(A\)的（相抵、相似 或合同）标准形来解决，从而为解决相关问题带来极大的方便. 例如，任给\(n\)阶方阵\(A\), 求\(A\)的中心化子\(\mathcal{C}(A)=\{X\in \mbox{F}^{n\times n}\mid AX=XA\}\)的维数. 最直接的做法就是设\(X=(x_{ij})_{n\times n}\),由\(AX=XA\)可得到含\(n^2\)个未知量、\(n^2\)个方程的线性方程组，这个方程组解空间的维数等于\(\dim\mathcal{C}(A)\). 当\(n\)比较大时，解这个线性方程组是相当困难的. 注意到若\(B\)相似于\(A\), 即存在可逆矩阵\(P\)使得\(P^{-1}AP=B\), 则

\[(P^{-1}AP)(P^{-1}XP)=(P^{-1}XP)(P^{-1}AP), \;\mbox{即}\; B(P^{-1}XP)=(P^{-1}XP)B. \]

所以\(\mathcal{C}(B)=\{P^{-1}XP\mid X\in\mathcal{C}(A)\}\), 从而\(\dim\mathcal{C}(A)=\dim \mathcal{C}(B)\), 即矩阵中心化子的维数在矩阵相似下不变. 由Jordan标准形定理知，任意复方阵都相似于Jordan形矩阵，所以我们可以取\(A\)的Joran标准形矩阵\(J\), 而计算\(\mathcal{C}(J)\)显然比直接计算\(\mathcal{C}(A)\)更简单. 注意到对于这个问题，\(A\)的相似类中任一矩阵的中心化子的维数都是相等的，所以我们可以取\(A\)的相似类中“最简单”的那个代表元来计算，从而使问题得到简化，这就是标准形方法.

本节首先讨论利用矩阵的等价（或相抵）标准形方法解决问题. 设\(A\)是\(n\)阶方阵, \({\rm r}(A)=r\). 则存在可逆矩阵\(P,Q\) 使得\(PAQ=\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}\), 称为矩阵\(A\)的等价标准形. 由此可得矩阵\(A,B\)等价当且仅当\({\rm r}(A)=\r(B)\). 所以很多涉及矩阵等价或与矩阵秩相关的问题时，可以选取矩阵的等价标准形或将\(A\)分解为\(A=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}\)来寻求思路解决问题.

例 1 [汕头大学2005] 设\(A\)是\(n\)阶方阵, \({\rm r}(A)=r\). 证明存在秩为\(n-r\)的非零矩阵\(B\)和\(C\), 使得\(AB=O, CA=O\).

证明 因为\({\rm r}(A)=r\), 故存在可逆矩阵\(P,Q\)使得

\[PAQ=\left(\begin{array}{cc} E_r&0\\ 0&0\end{array}\right). \]

令\(B=Q\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}\), \(C=\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}P\). 则\(\r(B)=\r(C)=n-r\), 且

\[\begin{array}{l} AB=P^{-1}(PAQ)\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}=P^{-1} \begin{pmatrix}E_r&O\\ O&O\end{pmatrix}\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}=O,\\ CA=\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}(PAQ)Q^{-1}=\begin{pmatrix}O&O\\ O&E_{n-r}\end{pmatrix}\begin{pmatrix}E_r&O\\ O&O\end{pmatrix}Q^{-1}=O. \end{array} \]

练习 [厦门大学2007] 设\(A\)是\(n\)阶方阵且\(|A|=0\), 求证：存在\(n\)阶非零方阵\(B\)使得\(AB=BA=O\).

例2 [南开大学2004] 设\(A,B\)分别是数域\(\mbox{F}\)上的\(m\times s\)与\(s\times n\)阶矩阵，\(AB=C\). 证明：若\(\r(A)=r\), 则存在秩为\(\min\{s-r,n\}\)的\(s\times n\)矩阵\(D\), 使得对任意的\(n\)阶矩阵\(Q\), 都有\(A(DQ+B)=C\).

证明 由题设，\(A(DQ+B)=C\)当且仅当\(ADQ=O\). 由\(Q\)的任意性知只需证存在秩为\(\min\{s-r,n\}\)的\(s\times n\)矩阵\(D\), 使得\(AD=O\).

(方法一) 由题设，存在可逆矩阵\(P,R\)使得

\[A=P\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}R. \]

取\(D=R^{-1}\begin{pmatrix} O_{r\times n}\\ X_{(s-r)\times n}\end{pmatrix}\), 其中\(\r(X)=\min\{s-r,n\}\). 则\(\r(D)=\min\{s-r,n\}\)满足所求.

(方法二) 因为\(\r(A)=r\), 所以设\(Ax=0\)的基础解系为\(\eta_1,\eta_2,\cdots,\eta_{s-r}\). 令

\[D=\left\{\begin{array}{ll} (\eta_1,\eta_2,\cdots,\eta_n),& n\leqslant s-r;\\ (\eta_1,\eta_2,\cdots,\eta_{s-r},0,\cdots,0), & n>s-r. \end{array}\right. \]

则\(\r(D)=\min\{s-r,n\}\)满足所求.

例3 设\(A\)是\(m\times n\)阶方阵. 若\(r(A)=m\), 则\(A\)称为行满秩矩阵; 若\(r(A)=n\), 则\(A\)称为列满秩矩阵. 证明

(1) \(A\)列满秩\(\Leftrightarrow\) 存在可逆矩阵\(P\)使得\(A=P\left(\begin{array}{c} E_n\\ O\end{array}\right)\).

(2) (中国科技大学2015) \(A\)行满秩\(\Leftrightarrow\) 存在可逆矩阵\(Q\)使得\(A=\left(\begin{array}{cc} E_m&O\end{array}\right)Q\).

证明 (1) \(\Rightarrow\): 因为\(r(A)=n\), 所以
存在初等矩阵\(P_1,P_2,\ldots,P_s\)使得$$P_s\cdots P_2P_1A=\left(\begin{array}{c} E_n\ O\end{array}\right).$$

令\(P=P_1^{-1}P_2^{-1}\cdots P_s^{-1}\), 则
\(A=P\left(\begin{array}{c} E_n\\ O\end{array}\right)\).

\(\Leftarrow\): 由\(P\)可逆知\(r(A)=\r(P\left(\begin{array}{c} E_n\\ O\end{array}\right)) =r\left(\begin{array}{c} E_n\\ O\end{array}\right)=n\).

练习 [浙江大学1999] 若\(A\)是行满秩矩阵，则存在矩阵\(B\), 使得\(AB=E\).

例4 [满秩分解] 设\(A\)是\(m\times n\)阶方阵. \(\r(A)=r\). 则

(1) 存在\(m\times r\)阶
列满秩矩阵\(H\)与\(r\times n\)阶行满秩矩阵\(L\)使得\(A=HL\);

(2) 如果\(A=HL=H_1L_1\), 其中\(H,H_1\)是列满秩矩阵, \(L,L_1\)是行满秩矩阵, 则存在\(r\)阶可逆矩阵\(P\)使得\(H=H_1P\), \(L=P^{-1}L_1\).

证明 (1) 因为\({\rm r}(A)=r\), 故存在可逆矩阵\(R,S\) 使得

\[RAS=\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right). \]

故

\[\begin{array}{rl} A=&R^{-1}\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)S^{-1}\\ =& \left(\begin{array}{cc} R_1&R_2\\ R_3&R_4\end{array}\right) \left(\begin{array}{c} E_r\\ O\end{array}\right) \cdot \left(\begin{array}{cc} E_r&O\end{array}\right) \left(\begin{array}{cc} S_1&S_2\\ S_3&S_4\end{array}\right)\\ =& \left(\begin{array}{c} R_1\\ R_3\end{array}\right) \left(\begin{array}{cc} S_1&S_2\end{array}\right) \stackrel{\Delta}{=}HL. \end{array} \]

其中\(\r(H)=\r(R^{-1}\left(\begin{array}{c} E_r\\ O\end{array}\right))=\r \left(\begin{array}{c} E_r\\ O\end{array}\right)=r\), 类似地,

\[\r(L)=\r(\left(\begin{array}{cc} E_r&O\end{array}\right)S^{-1})=\r \left(\begin{array}{cc} E_r&O\end{array}\right)=r. \]

(2) 由于\(L\)是行满秩矩阵, 所以由上题(2)知存在可逆矩阵\(Q\)使得\(L=\left(\begin{array}{cc} E_r&O\end{array}\right)Q\). 令\(N=Q^{-1}\left(\begin{array}{c} E_r\\ O\end{array}\right)\), 则\(LN=E_r\).
所以

\[H=HE_r=H(LN)=(HL)N=(H_1L_1)N=H_1(L_1N)=H_1P. \]

由于\(H\)是列满秩矩阵, 所以存在矩阵\(r\times m\)行满秩矩阵\(M\) 使得\(MH=E_r\). 故

\[E_r=MH=MH_1P, \mbox{即}P^{-1}=MH_1. \]

于是

\[L=(MH)L=(MH_1)L_1=P^{-1}L_1. \]

注 [北京理工大学2005]除(1)外，还要求证明(3) \(Ax=0\)与\(Lx=0\)同解.

例5 [特征多项式降阶公式]
设\(A,B\)分别是\(m\times n, n\times m\)阶矩阵，\(m\geqslant n\). 则

\[|\lambda E_m-AB|= \lambda^{m-n}|\lambda En-BA|. \]

证明 设\({\rm r}(A)=r\), 则存在\(m,n\)阶可逆矩阵\(P,Q\)使得

\[PAQ=\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}. \]

令\(Q^{-1}BP^{-1}=\begin{pmatrix} B_1&B_2\\ B_3&B_4\end{pmatrix}\), 其中\(B_1\)为\(r\)阶矩阵. 则

\[PABP^{-1}=\begin{pmatrix} B_1&B_2\\ O&O\end{pmatrix},\quad Q^{-1}BAQ=\begin{pmatrix} B_1&O\\ B_3&O\end{pmatrix}. \]

所以

\[|\lambda E_m-AB|=\begin{vmatrix} \lambda E_r-B_1&-B_2\\ O&\lambda E_{m-r}\end{vmatrix} =\lambda^{m-r}|\lambda E_r-B_1|, \]

\[|\lambda E_n-BA|=\begin{vmatrix} \lambda E_r-B_1&O\\ -B_3&\lambda E_{n-r}\end{vmatrix} =\lambda^{n-r}|\lambda E_r-B_1|. \]

故\(|\lambda E_m-AB|= \lambda^{m-n}|\lambda En-BA|. \)

例6设\(A,B\)是\(n\)阶方阵, \({\rm r}(ABA)={\rm r}(B)\). 求证：\(AB\sim BA\).

证明 设存在可逆矩阵\(P,Q\)使得

\[A=P\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)Q,\quad B=Q^{-1}\left(\begin{array}{cc} B_{11}&B_{12}\\ B_{21}&B_{22}\end{array}\right)P^{-1}. \]

所以

\[AB=P\left(\begin{array}{cc} B_{11}&B_{12}\\ O&O\end{array}\right)P^{-1},\; BA=Q^{-1}\left(\begin{array}{cc} B_{11}&O\\ B_{21}&O\end{array}\right)Q, \]

\[ABA=P\left(\begin{array}{cc} B_{11}&O\\ O&O\end{array}\right)Q. \]

所以\({\rm r}(B_{11})={\rm r}(ABA)={\rm r}(B)\), 因此存在可逆矩阵\(X,Y\)使得

\[B_{21}=XB_{11}, B_{12}=B_{11}Y. \]

所以

\[\begin{array}{l} AB=P\left(\begin{array}{cc} E_r&-Y\\ O&E_r\end{array}\right) \left(\begin{array}{cc} B_{11}&O\\ O&O\end{array}\right) \left(\begin{array}{cc} E_r&Y\\ O&E_r\end{array}\right)P^{-1},\\ BA=Q^{-1}\left(\begin{array}{cc} E_r&O\\ X&E_r\end{array}\right)\left(\begin{array}{cc} B_{11}&O\\ O&O\end{array}\right) \left(\begin{array}{cc} E_r&O\\ -X&E_r\end{array}\right)Q. \end{array} \]

因此\(AB\)与\(BA\)相似.

下面的例子可以看作矩阵方程\(AX-YA=O\)有秩为\(r\)的解.

例7 设\(A\in\mbox{F}^{n\times n}\), \({\rm r}(A)=r\). 求证：存在秩为\(r\)的\(n\)阶方阵\(B,C\)使得\(AB=CA\).

证明 由\({\rm r}(A)=r\)知存在可逆矩阵\(P,Q\)使得\(PAQ=\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}\), 则\(A=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}\). 令

\[B=Q\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}, \]

则

\[AB=AQ\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}Q\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}. \]

令\(C=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}P\), 则

\[CA=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}P P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}=AB. \]

例8 设\(A,B\)分别是复数域\(\mathbb{C}\)上的\(n\)阶与\(m\)阶矩阵, \(0<r\leqslant \min\{n,m\}\). 证明：如果\(AX-XB=O\)有秩为\(r\) 的矩阵解, 则\(A\) 与\(B\)至少有\(r\)个公共特征值(重根按重数计算).

证明 设\(C\)是\(AX-XB=O\)的矩阵解, 且\({\rm r}(C)=r\). 则存在可逆矩阵\(P\in \mathbb{C}^{n\times n}\), \(Q\in \mathbb{C}^{m\times m}\), 使得

\[C=P\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)Q. \]

由于\(AC=CB\), 所以

\[AP\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)Q= P\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)QB,\]

从而

\[P^{-1}AP\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)=\left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)QBQ^{-1}. \]

将\(P^{-1}AP, QBQ^{-1}\)写成分块矩阵的形式

\[P^{-1}AP=\begin{pmatrix} A_{11}&A_{12} A_{21}&A_{22}\end{pmatrix}, \quad QBQ^{-1}=\begin{pmatrix} B_{11}&B_{12} B_{21}&B_{22}\end{pmatrix}. \]

故

\[\left(\begin{array}{cc} A_{11}&O\\ A_{21}&O\end{array}\right) =\left(\begin{array}{cc} B_{11}&B_{12}\\ O&O\end{array}\right). \]

由此得\(A_{11}=B_{11}, A_{21}=O, B_{12}=O\).
所以

\[P^{-1}AP=\left(\begin{array}{cc} A_{11}&A_{12}\\ O&A_{22}\end{array}\right),\quad QBQ^{-1}=\left(\begin{array}{cc} A_{11}&O\\ B_{12}&B_{22}\end{array}\right). \]

所以\(A,B\)的特征多项式分别为

\[f_A(\lambda)=|\lambda E_r-A_{11}|\cdot |\lambda E-A_{22}|,\quad f_B(\lambda)=|\lambda E_r-A_{11}|\cdot |\lambda E-B_{22}|. \]

它们有公因式\(|\lambda E_r-A_{11}|\), 因而\(A\) 与\(B\)至少有\(r\) 个公共特征值(重根按重数计算).

例8证明：矩阵方程\(AXA=A\)对任意\(m\times n\)矩阵\(A\)都有解.

证明 设\({\rm r}(A)=r\), 则存在\(m\)阶可逆矩阵\(P\)和\(n\)阶可逆矩阵\(Q\)使得

\[PAQ=\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}. \]

令\(Q^{-1}XP^{-1}=\begin{pmatrix} X_{11}&X_{12}\\ X_{21}&X_{22} \end{pmatrix}\), 则由\(AXA=A\)可得

\[(PAQ)(Q^{-1}XP^{-1})(PAQ)=PAQ, \]

即

\[\begin{pmatrix} E_r&O\\ O&O \end{pmatrix} \begin{pmatrix} X_{11}&X_{12}\\ X_{21}&X_{22} \end{pmatrix}\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}=\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}. \]

故\(\begin{pmatrix} X_{11}&O\\ O&O \end{pmatrix}=\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}\), 于是
\(X_{11}=E_r\). 由此可得\(X=Q\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}P\)即为\(AXA=A\)的解.

例9 [Roth定理] 设\(A,B,C\)分别为\(m\times n\), \(n\times m\)与\(n\times n\) 矩阵, 则\({\rm r}\left(\begin{array}{cc} A&O\\O&B\end{array}\right) ={\rm r}\left(\begin{array}{cc} A&O\\ C&B\end{array}\right) \)的充要条件为存在矩阵\(X,Y\)使得\(XA-BY=C\).

证明
\(\Longleftarrow\) 由\(\left(\begin{array}{cc} E&O\\ -X&E\end{array}\right) \left(\begin{array}{cc} A&O\\C&B\end{array}\right) \left(\begin{array}{cc} E&O\\ Y&E\end{array}\right) =\left(\begin{array}{cc} A&O\\O&B\end{array}\right) \)即得.

\(\Longrightarrow\) 设\(P_1AQ_1=\left(\begin{array}{cc} E_r&O\\O&O\end{array}\right)\stackrel{\Delta}{=}H_1,\; P_2BQ_2=\left(\begin{array}{cc} E_s&O\\O&O\end{array}\right)\stackrel{\Delta}{=}H_2. \) 则
\begin{equation}
\left(\begin{array}{cc} P_1&O\O&P_2\end{array}\right)
\left(\begin{array}{cc} A&O\O&B\end{array}\right)
\left(\begin{array}{cc} Q_1&O\O&Q_2\end{array}\right)
=\left(\begin{array}{cccc} E_r&&&\ &O&&\ &&E_s&\ &&&O\end{array}\right),
\end{equation}
\begin{equation}
\begin{array}{ll}
&\left(\begin{array}{cc} P_1&O\O&P_2\end{array}\right)
\left(\begin{array}{cc} A&O\ C&B\end{array}\right)
\left(\begin{array}{cc} Q_1&O\O&Q_2\end{array}\right)
=\left(\begin{array}{cc} P_1AQ_1&O\ P_2CQ_1&P_2BQ_2\end{array}\right)\
=&
\left(\begin{array}{cccc} E_r&O&&\ O&O&&\ D_1&D_2&E_s&O\ D_3&D_4&O&O\end{array}\right)
=\left(\begin{array}{cc} H_1&O\ D&H_2\end{array}\right).
\end{array}
\end{equation}
从(2.5)中消去\(D_1\), 即

\[\left(\begin{array}{cccc} E_r&&&\\ O&E&&\\ -D_1&O&E_s&O\\ O&O&O&E\end{array}\right) \left(\begin{array}{cccc} E_r&O&&\\ O&O&&\\ D_1&D_2&E_s&O\\ D_3&D_4&O&O\end{array}\right) = \left(\begin{array}{cccc} E_r&O&&\\ O&O&&\\ O&D_2&E_s&O\\ D_3&D_4&O&O\end{array}\right). \]

此过程可简记为

\[\left(\begin{array}{cc} -D_1&O\\ O&O\end{array}\right) \left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)+ \left(\begin{array}{cc} D_1&D_2\\ D_3&D_4\end{array}\right) = \left(\begin{array}{cc} O&D_2\\ D_3&D_4\end{array}\right). \]

继续消去\(D_3,D_2\)的过程可简记为

\[\left(\begin{array}{cc} O&O\\ -D_3&O\end{array}\right) \left(\begin{array}{cc} E_r&O\\ O&O\end{array}\right)+ \left(\begin{array}{cc} O&D_2\\ D_3&D_4\end{array}\right) = \left(\begin{array}{cc} O&D_2\\ O&D_4\end{array}\right). \]

\[\left(\begin{array}{cc} E_s&O\\ O&O\end{array}\right)\left(\begin{array}{cc} O&-D_2\\ O&O\end{array}\right)+ \left(\begin{array}{cc} O&D_2\\ O&D_4\end{array}\right) = \left(\begin{array}{cc} O&O\\ O&D_4\end{array}\right). \]

由题设, (2.4)与(2.5)式右端的秩相等, 所以\(D_4=O\). 从而该消去过程相当于存在矩阵\(U=\left(\begin{array}{cc} O&O\\ -D_3&O\end{array}\right)+\left(\begin{array}{cc} -D_1&O\\ O&O\end{array}\right),V=\left(\begin{array}{cc} O&-D_2\\ O&O\end{array}\right)\)使得

\[UH_1+H_2V+D=O. \]

即

\[U(P_1AQ_1)+(P_2BQ_2)V=-P_2CQ_1, \]

或等价地,

\[(-P_2^{-1}UP_1)A-B(Q_2VQ_1^{-1})=C. \]

令\(X=-P_2^{-1}UP_1\), \(Y=Q_2VQ_1^{-1}\), 即证.

例10 设\(A,B\)分别为\(m\times n\)与\(n\times m\)矩阵, 则\({\rm r}(A)+{\rm r}(B)={\rm r}(AB)+n\) 的充要条件为存在矩阵\(X,Y\)使得
\(XA-BY=E_n\).

证明 由Sylvester不等式的证明过程知只需证明

\[{\rm r}\left(\begin{array}{cc} A&O\\O&B\end{array}\right) ={\rm r}\left(\begin{array}{cc} A&O\\ E&B\end{array}\right) \Longleftrightarrow \exists X,Y, \mbox{s.t.} XA-BY=E_n. \]

这由例9即得.

标签：begin,right,end,等价,标准,pmatrix,array,方法,left
From： https://www.cnblogs.com/xuyg/p/18350205