在高等代數(shù)課程里，我們學(xué)習(xí)過矩陣的等價(jià)、相似與合同三種等價(jià)關(guān)系，在每個(gè)等價(jià)類里我們都可以選擇一個(gè)“相對(duì)簡(jiǎn)單”的代表元，稱為等價(jià)標(biāo)準(zhǔn)形. 若待解決的問題在某一等價(jià)關(guān)系（相抵、相似于或合同）下不變，或與等價(jià)類的代表元的選取無關(guān)，則對(duì)一般矩陣\(A\)的問題可以轉(zhuǎn)化為選取\(A\)的（相抵、相似 或合同）標(biāo)準(zhǔn)形來解決，從而為解決相關(guān)問題帶來極大的方便. 例如，任給\(n\)階方陣\(A\), 求\(A\)的中心化子\(\mathcal{C}(A)=\{X\in \mbox{F}^{n\times n}\mid AX=XA\}\)的維數(shù). 最直接的做法就是設(shè)\(X=(x_{ij})_{n\times n}\),由\(AX=XA\)可得到含\(n^2\)個(gè)未知量、\(n^2\)個(gè)方程的線性方程組，這個(gè)方程組解空間的維數(shù)等于\(\dim\mathcal{C}(A)\). 當(dāng)\(n\)比較大時(shí)，解這個(gè)線性方程組是相當(dāng)困難的. 注意到若\(B\)相似于\(A\), 即存在可逆矩陣\(P\)使得\(P^{-1}AP=B\), 則

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

所以\(\mathcal{C}(B)=\{P^{-1}XP\mid X\in\mathcal{C}(A)\}\), 從而\(\dim\mathcal{C}(A)=\dim \mathcal{C}(B)\), 即矩陣中心化子的維數(shù)在矩陣相似下不變. 由Jordan標(biāo)準(zhǔn)形定理知，任意復(fù)方陣都相似于Jordan形矩陣，所以我們可以取\(A\)的Joran標(biāo)準(zhǔn)形矩陣\(J\), 而計(jì)算\(\mathcal{C}(J)\)顯然比直接計(jì)算\(\mathcal{C}(A)\)更簡(jiǎn)單. 注意到對(duì)于這個(gè)問題，\(A\)的相似類中任一矩陣的中心化子的維數(shù)都是相等的，所以我們可以取\(A\)的相似類中“最簡(jiǎn)單”的那個(gè)代表元來計(jì)算，從而使問題得到簡(jiǎn)化，這就是標(biāo)準(zhǔn)形方法.

本節(jié)首先討論利用矩陣的等價(jià)（或相抵）標(biāo)準(zhǔn)形方法解決問題. 設(shè)\(A\)是\(n\)階方陣, \({\rm r}(A)=r\). 則存在可逆矩陣\(P,Q\) 使得\(PAQ=\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}\), 稱為矩陣\(A\)的等價(jià)標(biāo)準(zhǔn)形. 由此可得矩陣\(A,B\)等價(jià)當(dāng)且僅當(dāng)\({\rm r}(A)=\r(B)\). 所以很多涉及矩陣等價(jià)或與矩陣秩相關(guān)的問題時(shí)，可以選取矩陣的等價(jià)標(biāo)準(zhǔn)形或?qū)?span id="w0obha2h00" class="math inline">\(A\)分解為\(A=P^{-1}\begin{pmatrix} E_r&O\\ O&O\end{pmatrix}Q^{-1}\)來尋求思路解決問題.

例 1 [汕頭大學(xué)2005] 設(shè)\(A\)是\(n\)階方陣, \({\rm r}(A)=r\). 證明存在秩為\(n-r\)的非零矩陣\(B\)和\(C\), 使得\(AB=O, CA=O\).

證明 因?yàn)?span id="w0obha2h00" class="math inline">\({\rm r}(A)=r\), 故存在可逆矩陣\(P,Q\)使得

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

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

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

練習(xí) [廈門大學(xué)2007] 設(shè)\(A\)是\(n\)階方陣且\(|A|=0\), 求證：存在\(n\)階非零方陣\(B\)使得\(AB=BA=O\).

例2 [南開大學(xué)2004] 設(shè)\(A,B\)分別是數(shù)域\(\mbox{F}\)上的\(m\times s\)與\(s\times n\)階矩陣，\(AB=C\). 證明：若\(\r(A)=r\), 則存在秩為\(\min\{s-r,n\}\)的\(s\times n\)矩陣\(D\), 使得對(duì)任意的\(n\)階矩陣\(Q\), 都有\(A(DQ+B)=C\).

證明 由題設(shè)，\(A(DQ+B)=C\)當(dāng)且僅當(dāng)\(ADQ=O\). 由\(Q\)的任意性知只需證存在秩為\(\min\{s-r,n\}\)的\(s\times n\)矩陣\(D\), 使得\(AD=O\).

(方法一) 由題設(shè)，存在可逆矩陣\(P,R\)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\(\r(X)=\min\{s-r,n\}\). 則\(\r(D)=\min\{s-r,n\}\)滿足所求.

(方法二) 因?yàn)?span id="w0obha2h00" class="math inline">\(\r(A)=r\), 所以設(shè)\(Ax=0\)的基礎(chǔ)解系為\(\eta_1,\eta_2,\cdots,\eta_{s-r}\). 令

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

則\(\r(D)=\min\{s-r,n\}\)滿足所求.

例3 設(shè)\(A\)是\(m\times n\)階方陣. 若\(r(A)=m\), 則\(A\)稱為行滿秩矩陣; 若\(r(A)=n\), 則\(A\)稱為列滿秩矩陣. 證明

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

(2) (中國科技大學(xué)2015) \(A\)行滿秩\(\Leftrightarrow\) 存在可逆矩陣\(Q\)使得\(A=\left(\begin{array}{cc} E_m&O\end{array}\right)Q\).

證明 (1) \(\Rightarrow\): 因?yàn)?span id="w0obha2h00" class="math inline">\(r(A)=n\), 所以
存在初等矩陣\(P_1,P_2,\ldots,P_s\)使得$$P_s\cdots P_2P_1A=\left(\begin{array}{c} E_n\ O\end{array}\right).$$

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

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

練習(xí) [浙江大學(xué)1999] 若\(A\)是行滿秩矩陣，則存在矩陣\(B\), 使得\(AB=E\).

例4 [滿秩分解] 設(shè)\(A\)是\(m\times n\)階方陣. \(\r(A)=r\). 則

(1) 存在\(m\times r\)階
列滿秩矩陣\(H\)與\(r\times n\)階行滿秩矩陣\(L\)使得\(A=HL\);

(2) 如果\(A=HL=H_1L_1\), 其中\(H,H_1\)是列滿秩矩陣, \(L,L_1\)是行滿秩矩陣, 則存在\(r\)階可逆矩陣\(P\)使得\(H=H_1P\), \(L=P^{-1}L_1\).

證明 (1) 因?yàn)?span id="w0obha2h00" class="math inline">\({\rm r}(A)=r\), 故存在可逆矩陣\(R,S\) 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\(\r(H)=\r(R^{-1}\left(\begin{array}{c} E_r\\ O\end{array}\right))=\r \left(\begin{array}{c} E_r\\ O\end{array}\right)=r\), 類似地,

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

(2) 由于\(L\)是行滿秩矩陣, 所以由上題(2)知存在可逆矩陣\(Q\)使得\(L=\left(\begin{array}{cc} E_r&O\end{array}\right)Q\). 令\(N=Q^{-1}\left(\begin{array}{c} E_r\\ O\end{array}\right)\), 則\(LN=E_r\).
所以

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

由于\(H\)是列滿秩矩陣, 所以存在矩陣\(r\times m\)行滿秩矩陣\(M\) 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 [北京理工大學(xué)2005]除(1)外，還要求證明(3) \(Ax=0\)與\(Lx=0\)同解.

例5 [特征多項(xiàng)式降階公式]
設(shè)\(A,B\)分別是\(m\times n, n\times m\)階矩陣，\(m\geqslant n\). 則

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

證明 設(shè)\({\rm r}(A)=r\), 則存在\(m,n\)階可逆矩陣\(P,Q\)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\(B_1\)為\(r\)階矩陣. 則

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

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設(shè)\(A,B\)是\(n\)階方陣, \({\rm r}(ABA)={\rm r}(B)\). 求證：\(AB\sim BA\).

證明 設(shè)存在可逆矩陣\(P,Q\)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\({\rm r}(B_{11})={\rm r}(ABA)={\rm r}(B)\), 因此存在可逆矩陣\(X,Y\)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\(AB\)與\(BA\)相似.

下面的例子可以看作矩陣方程\(AX-YA=O\)有秩為\(r\)的解.

例7 設(shè)\(A\in\mbox{F}^{n\times n}\), \({\rm r}(A)=r\). 求證：存在秩為\(r\)的\(n\)階方陣\(B,C\)使得\(AB=CA\).

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

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

則

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

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

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

例8 設(shè)\(A,B\)分別是復(fù)數(shù)域\(\mathbb{C}\)上的\(n\)階與\(m\)階矩陣, \(0<r\leqslant \min\{n,m\}\). 證明：如果\(AX-XB=O\)有秩為\(r\) 的矩陣解, 則\(A\) 與\(B\)至少有\(r\)個(gè)公共特征值(重根按重?cái)?shù)計(jì)算).

證明 設(shè)\(C\)是\(AX-XB=O\)的矩陣解, 且\({\rm r}(C)=r\). 則存在可逆矩陣\(P\in \mathbb{C}^{n\times n}\), \(Q\in \mathbb{C}^{m\times m}\), 使得

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

由于\(AC=CB\), 所以

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

從而

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

將\(P^{-1}AP, QBQ^{-1}\)寫成分塊矩陣的形式

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

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\(A,B\)的特征多項(xiàng)式分別為

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

它們有公因式\(|\lambda E_r-A_{11}|\), 因而\(A\) 與\(B\)至少有\(r\) 個(gè)公共特征值(重根按重?cái)?shù)計(jì)算).

例8證明：矩陣方程\(AXA=A\)對(duì)任意\(m\times n\)矩陣\(A\)都有解.

證明 設(shè)\({\rm r}(A)=r\), 則存在\(m\)階可逆矩陣\(P\)和\(n\)階可逆矩陣\(Q\)使得

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

令\(Q^{-1}XP^{-1}=\begin{pmatrix} X_{11}&X_{12}\\ X_{21}&X_{22} \end{pmatrix}\), 則由\(AXA=A\)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\(X=Q\begin{pmatrix} E_r&O\\ O&O \end{pmatrix}P\)即為\(AXA=A\)的解.

例9 [Roth定理] 設(shè)\(A,B,C\)分別為\(m\times n\), \(n\times m\)與\(n\times n\) 矩陣, 則\({\rm r}\left(\begin{array}{cc} A&O\\O&B\end{array}\right) ={\rm r}\left(\begin{array}{cc} A&O\\ C&B\end{array}\right) \)的充要條件為存在矩陣\(X,Y\)使得\(XA-BY=C\).

證明
\(\Longleftarrow\) 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.

\(\Longrightarrow\) 設(shè)\(P_1AQ_1=\left(\begin{array}{cc} E_r&O\\O&O\end{array}\right)\stackrel{\Delta}{=}H_1,\; P_2BQ_2=\left(\begin{array}{cc} E_s&O\\O&O\end{array}\right)\stackrel{\Delta}{=}H_2. \) 則
\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}
從(2.5)中消去\(D_1\), 即

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

此過程可簡(jiǎn)記為

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

繼續(xù)消去\(D_3,D_2\)的過程可簡(jiǎn)記為

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

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

由題設(shè), (2.4)與(2.5)式右端的秩相等, 所以\(D_4=O\). 從而該消去過程相當(dāng)于存在矩陣\(U=\left(\begin{array}{cc} O&O\\ -D_3&O\end{array}\right)+\left(\begin{array}{cc} -D_1&O\\ O&O\end{array}\right),V=\left(\begin{array}{cc} O&-D_2\\ O&O\end{array}\right)\)使得

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

即

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

或等價(jià)地,

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

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

例10 設(shè)\(A,B\)分別為\(m\times n\)與\(n\times m\)矩陣, 則\({\rm r}(A)+{\rm r}(B)={\rm r}(AB)+n\) 的充要條件為存在矩陣\(X,Y\)使得
\(XA-BY=E_n\).

證明 由Sylvester不等式的證明過程知只需證明

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

這由例9即得.