蓝以中下册 幂零线性变换的Jordan标准型 93页 习题二13 解答

castelu

习题二13：
　　给定数域$K$上的$m$阶方阵$A$和$n$阶方阵$B$，满足$A^2=0$，$B^2=0$。又设
$$C,D \in M_{m,n}(K)$$
　　令
$$F=\left( {\begin{array}{*{20}{c}}
{A}&{C}\\
{0}&{B}
\end{array}} \right),G=\left( {\begin{array}{*{20}{c}}
{A}&{D}\\
{0}&{B}
\end{array}} \right)$$
　　如果
$$r(F)=r(G)=r(A)+r(B),r(AC+CB)=r(AD+DB)$$
　　证明$F,G$在$K$内相似。

---

解：
　　因为
$$A^2=0,B^2=0$$
　　故$A,B$均为幂零矩阵，其若尔当形中若尔当块最高阶为二阶。现设它们的若尔当形为$A_1,B_1$，即
$$T_1^{-1}AT_1=A_1=\left( {\begin{array}{*{20}{c}}
{J_1}&{0}\\
{0}&{0}
\end{array}} \right),J_1=\left( {\begin{array}{*{20}{c}}
{D_1}&{}&{}&{}\\
{}&{D_1}&{}&{}\\
{}&{}&{\ddots}&{}\\
{}&{}&{}&{D_1}
\end{array}} \right)_{2r \times 2r}$$
$$T_2^{-1}BT_2=B_1=\left( {\begin{array}{*{20}{c}}
{J_2}&{0}\\
{0}&{0}
\end{array}} \right),J_2=\left( {\begin{array}{*{20}{c}}
{D_1}&{}&{}&{}\\
{}&{D_1}&{}&{}\\
{}&{}&{\ddots}&{}\\
{}&{}&{}&{D_1}
\end{array}} \right)_{2s \times 2s}$$
　　其中
$$D_1=\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right)$$
　　于是有
$$F_1=\left( {\begin{array}{*{20}{c}}
{T_1^{-1}}&{0}\\
{0}&{T_2^{-1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{A}&{C}\\
{0}&{B}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{T_1}&{0}\\
{0}&{T_2}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{A_1}&{T_1^{-1}CT_2}\\
{0}&{B_1}
\end{array}} \right)$$
$$G_1=\left( {\begin{array}{*{20}{c}}
{T_1^{-1}}&{0}\\
{0}&{T_2^{-1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{A}&{D}\\
{0}&{B}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{T_1}&{0}\\
{0}&{T_2}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{A_1}&{T_1^{-1}DT_2}\\
{0}&{B_1}
\end{array}} \right)$$
　　$F_1$与$F$相似，$G_1$与$G$相似，只需证$F_1$与$G_1$相似
　　矩阵左乘、右乘满秩方阵其秩不变，故有
$$r(F_1)=r(F)=r(A)+r(B)=r(A_1)+r(B_1)=r+s$$
$$r(G_1)=r(G)=r(A)+r(B)=r(A_1)+r(B_1)=r+s$$
　　又设
$$T_1^{-1}CT_2=\overline C,T_1^{-1}DT_2=\overline D$$
　　那么
$$\begin{eqnarray*}
A_1\overline C+\overline CB_1&=&(T_1^{-1}AT_1)(T_1^{-1}CT_2)+(T_1^{-1}CT_2)(T_2^{-1}BT_2)\\
&=&T_1^{-1}(AC+CB)T_2
\end{eqnarray*}$$
$$\begin{eqnarray*}
A_1\overline D+\overline DB_1&=&(T_1^{-1}AT_1)(T_1^{-1}DT_2)+(T_1^{-1}DT_2)(T_2^{-1}BT_2)\\
&=&T_1^{-1}(AD+DB)T_2
\end{eqnarray*}$$
　　故有
$$\begin{eqnarray*}
r(A_1\overline C+\overline CB_1)&=&r(AC+CB)=r(AD+DB)\\
&=&r(A_1\overline D+\overline DB_1)
\end{eqnarray*}$$
　　因此，我们不妨直接设
$$A=\left( {\begin{array}{*{20}{c}}
{J_1}&{0}\\
{0}&{0}
\end{array}} \right),B=\left( {\begin{array}{*{20}{c}}
{J_2}&{0}\\
{0}&{0}
\end{array}} \right)$$
　　再将$C$分块如下
$$C=\left( {\begin{array}{*{20}{c}}
{C_1}&{C_2}\\
{C_3}&{C_4}
\end{array}} \right),C_1=\left( {\begin{array}{*{20}{c}}
{c_{11}}&{c_{12}}&{\cdots}&{c_{1,2s}}\\
{c_{21}}&{c_{22}}&{\cdots}&{c_{2,2s}}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{c_{2r,1}}&{c_{2r,2}}&{\cdots}&{c_{2r,2s}}
\end{array}} \right)$$
　　此时$F$表示成
$$F=\left( {\begin{array}{*{20}{c}}
{J_1}&{0}&{C_1}&{C_2}\\
{0}&{0}&{C_3}&{C_4}\\
{0}&{0}&{J_2}&{0}\\
{0}&{0}&{0}&{0}
\end{array}} \right)$$
　　当$A=0$或$B=0$时，我们有
$$ACB=ADB=0$$
　　下面设
$$A \ne 0,B \ne 0$$
　　在$J_1$的第$1,3,\cdots,2r-1$行有一个$1$，利用它们经初等列变换把其右边所有元素消为$0$
　　$J_2$的第$2,4,\cdots,2s$列有一个$1$，利用它们经初等行变换把其上边所有元素消为$0$。如此可把$C_1$化为
$$\left( {\begin{array}{*{20}{c}}
{0}&{0}&{0}&{0}&{\cdots}&{0}&{0}\\
{c_{21}}&{0}&{c_{23}}&{0}&{\cdots}&{c_{2,2s-1}}&{0}\\
{0}&{0}&{0}&{0}&{\cdots}&{0}&{0}\\
{c_{41}}&{0}&{c_{43}}&{0}&{\cdots}&{c_{4,2s-1}}&{0}\\
{\vdots}&{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}&{\vdots}\\
{0}&{0}&{0}&{0}&{\cdots}&{0}&{0}\\
{c_{2r,1}}&{0}&{c_{2r,3}}&{0}&{\cdots}&{c_{2r,2s-1}}&{0}
\end{array}} \right)$$
　　如果$c_{21} \ne 0$，利用它经初等行、列变换可把$F$的第$2$行所有其他元素化为$0$，第$m+1$列所有其他元素化为$0$
　　如此则立得
$$r(F)>r(J_1)+r(J_2)=r(A)+r(B)$$
　　与题设矛盾
　　故必
$$c_{21}=0$$
　　同理，必有
$$c_{2i,2j-1}=0(i=1,2,\cdots,r;j=1,2,\cdots,s)$$
　　于是$C_1$可进一步写成如下分块形式
$$C_1=\left( {\begin{array}{*{20}{c}}
{M_{11}}&{M_{12}}&{\cdots}&{M_{1s}}\\
{M_{21}}&{M_{22}}&{\cdots}&{M_{2s}}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{M_{r1}}&{M_{r2}}&{\cdots}&{M_{rs}}
\end{array}} \right),M_{ij}=\left( {\begin{array}{*{20}{c}}
{*}&{*}\\
{0}&{*}
\end{array}} \right)$$
　　现在
$$\begin{eqnarray*}
D_1M_{ij}D_1&=&\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{m_1}&{m_2}\\
{0}&{m_3}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{0}&{m_3}\\
{0}&{0}
\end{array}} \right)
\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
{0}&{0}\\
{0}&{0}
\end{array}} \right)
\end{eqnarray*}$$
　　于是
$$\begin{eqnarray*}J_1C_1J_2&=&\left( {\begin{array}{*{20}{c}}
{D_1}&{}&{}&{}\\
{}&{D_1}&{}&{}\\
{}&{}&{\ddots}&{}\\
{}&{}&{}&{D_1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{M_{11}}&{M_{12}}&{\cdots}&{M_{1s}}\\
{M_{21}}&{M_{22}}&{\cdots}&{M_{2s}}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{M_{r1}}&{M_{r2}}&{\cdots}&{M_{rs}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{D_1}&{}&{}&{}\\
{}&{D_1}&{}&{}\\
{}&{}&{\ddots}&{}\\
{}&{}&{}&{D_1}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{D_1M_{11}D_1}&{D_1M_{12}D_1}&{\cdots}&{D_1M_{1s}D_1}\\
{D_1M_{21}D_1}&{D_1M_{22}D_1}&{\cdots}&{D_1M_{2s}D_1}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{D_1M_{r1}D_1}&{D_1M_{r2}D_1}&{\cdots}&{D_1M_{rs}D_1}
\end{array}} \right)=0
\end{eqnarray*}$$
　　利用这结果，我们有
$$ACB=\left( {\begin{array}{*{20}{c}}
{J_1}&{0}\\
{0}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{C_1}&{C_2}\\
{C_3}&{C_4}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{J_2}&{0}\\
{0}&{0}
\end{array}} \right)=
\left( {\begin{array}{*{20}{c}}
{J_1C_1J_2}&{0}\\
{0}&{0}
\end{array}} \right)=0$$
　　同理，由
$$r(G)=r(A)+r(B)$$
　　也推出
$$ADB=0$$
　　现在
$$F^2=\left( {\begin{array}{*{20}{c}}
{0}&{AC+CB}\\
{0}&{0}
\end{array}} \right),G^2=\left( {\begin{array}{*{20}{c}}
{0}&{AD+DB}\\
{0}&{0}
\end{array}} \right)$$
$$F^3=\left( {\begin{array}{*{20}{c}}
{0}&{ACB}\\
{0}&{0}
\end{array}} \right)=0,G^3=\left( {\begin{array}{*{20}{c}}
{0}&{ADB}\\
{0}&{0}
\end{array}} \right)=0$$
　　故$F,G$均为幂零矩阵，有唯一特征值$\lambda_0=0$
　　而
$$r(F)=r(G),r(F^2)=r(AC+CB)=r(AD+DB)=r(G^2),r(F^3)=r(G^3)=0$$
　　$F,G$的若尔当形中
　　一阶若尔当块个数为
$$r(F^0)+r(F^2)-2r(F)=r(G^0)+r(G^2)-r(G)$$
　　二阶若尔当块个数为
$$r(F)+r(F^3)-2r(F^2)=r(G)+r(G^3)-2r(G^2)$$
　　三阶若尔当块个数为
$$r(F^2)+r(F^4)-2r(F^3)=r(G^2)+r(G^4)-2r(G^3)$$
　　四阶以上若尔当块个数均为$0$
　　因此，$F,G$的若尔当形相同，即$F$与$G$相似
　　现考查
$$F=\left( {\begin{array}{*{20}{c}}
{J}&{0}\\
{0}&{J}
\end{array}} \right),G=\left( {\begin{array}{*{20}{c}}
{J}&{E}\\
{0}&{J}
\end{array}} \right)$$
　　其中
$$J=\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right),E=\left( {\begin{array}{*{20}{c}}
{1}&{0}\\
{0}&{1}
\end{array}} \right)$$
　　$F$已为若尔当形。而
$$G^2=\left( {\begin{array}{*{20}{c}}
{0}&{2J}\\
{0}&{0}
\end{array}} \right),G^3=0$$
　　$G$的若尔当形中一阶若尔当块个数为
$$r(G^0)+r(G^2)-2r(G)=4+1-2 \times 2=1$$
　　而二阶若尔当块个数为
$$r(G)+r(G^3)-2r(G^2)=2+0-2 \times 1=0$$
　　三阶若尔当块个数为
$$r(G^2)+r(G^4)-2r(G^3)=1+0-0=1$$
　　故$G$的若尔当形为
$$\left( {\begin{array}{*{20}{c}}
{0}&{0}&{0}&{0}\\
{0}&{0}&{1}&{0}\\
{0}&{0}&{0}&{1}\\
{0}&{0}&{0}&{0}
\end{array}} \right)$$
　　与$F$的若尔当形不同，从而$F$与$G$不相似。其原因是，现在
$$r(F)=r(A)+r(B)=r(J)+r(J)=2=r(G)$$
　　但
$$AC+CB=0$$
　　而
$$AD+DB=2J \ne 0$$