蓝以中上册 向量空间与矩阵 138页 习题五7 解答

castelu

习题五7：
　　设$A$是数域$K$上的$n$阶方阵。证明：
（1）若$A^2=E$，则
$$r(A+E)+r(A-E)=n$$
（1）若$A^2=A$，则
$$r(A)+r(A-E)=n$$

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解：
（1）由于$x+1$与$x-1$互素
$$\exists u(x), v(x) \in P[x]$$
$$u(x)(x+1)+v(x)(x-1)=1$$
　　将$A$代入，得到
$$u(A)(A+E)+v(A)(A-E)=E$$
　　考虑$2n$阶方阵
$$\begin{eqnarray*}
\left( {\begin{array}{*{20}{c}}
{A+E}&{O}\\
{O}&{A-E}
\end{array}} \right)&\to&\left( {\begin{array}{*{20}{c}}
{A+E}&{u(A)(A+E)}\\
{O}&{A-E}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{A+E}&{E}\\
{O}&{A-E}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{O}&{E}\\
{A^2-E}&{O}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{O}&{E}\\
{O}&{O}
\end{array}} \right)
\end{eqnarray*}$$
　　于是
$$r(A+E)+r(A-E)=n$$
（2）由于$x$与$x-1$互素
$$\exists u(x), v(x) \in P[x]$$
$$u(x)x+v(x)(x-1)=1$$
　　将$A$代入，得到
$$u(A)A+v(A)(A-E)=E$$
　　考虑$2n$阶方阵
$$\begin{eqnarray*}
\left( {\begin{array}{*{20}{c}}
{A}&{O}\\
{O}&{A-E}
\end{array}} \right)&\to&\left( {\begin{array}{*{20}{c}}
{A}&{u(A)A}\\
{O}&{A-E}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{A}&{E}\\
{O}&{A-E}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{O}&{E}\\
{A^2-A}&{O}
\end{array}} \right)\\
&\to&\left( {\begin{array}{*{20}{c}}
{O}&{E}\\
{O}&{O}
\end{array}} \right)
\end{eqnarray*}$$
　　于是
$$r(A)+r(A-E)=n$$