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标题: 蓝以中上册线性空间与线性变换 299页习题三22 解答 [打印本页]

作者: castelu 时间: 2016-6-6 17:37
标题: 蓝以中上册线性空间与线性变换 299页习题三22 解答
习题三22：
　　设
$$B=\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)$$
　　在$M_2(K)$中定义变换如下：
$$AX=B^{-1}XB(\forall X \in M_2(K))$$
　　证明$A$是$M_2(K)$内一个线性变换，并找出$\lambda_0 \in K,X_0 \in M_2(K),X_0 \ne 0$，使$AX_0=\lambda_0X_0$。

解：
　　对
$$\forall X,Y \in M_2(K)$$
　　有
$$A(X+Y)=B^{-1}(X+Y)B=B^{-1}XB+B^{-1}YB=AX+AY$$
　　对任意实数$k$，有
$$A(kX)=B^{-1}(kX)B=kB^{-1}XB=kAX$$
　　故$A$是$M_2(K)$内一个线性变换。
　　由于
$$B=\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)$$
　　所以
$$B^*=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)$$
　　并且$|B|=1$
　　故
$$B^{-1}=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)$$
　　在$M_2(K)$中取一组基
$$\epsilon_{11}=\left( {\begin{array}{*{20}{c}}
{1}&{0}\\
{0}&{0}
\end{array}} \right),\epsilon_{12}=\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right),\epsilon_{21}=\left( {\begin{array}{*{20}{c}}
{0}&{0}\\
{1}&{0}
\end{array}} \right),\epsilon_{22}=\left( {\begin{array}{*{20}{c}}
{0}&{0}\\
{0}&{1}
\end{array}} \right)$$
　　就有
$$\begin{eqnarray*}
A\epsilon_{11}&=&B^{-1}\epsilon_{11}B=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{1}&{0}\\
{0}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{2}
\end{array}} \right)=-\epsilon_{11}-\epsilon_{12}+2\epsilon_{21}+2\epsilon_{22}
\end{eqnarray*}$$
$$\begin{eqnarray*}
A\epsilon_{12}&=&B^{-1}\epsilon_{12}B=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{0}&{1}\\
{0}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{2}&{1}\\
{-4}&{-2}
\end{array}} \right)=2\epsilon_{11}+\epsilon_{12}-4\epsilon_{21}-2\epsilon_{22}
\end{eqnarray*}$$
$$\begin{eqnarray*}
A\epsilon_{21}&=&B^{-1}\epsilon_{21}B=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{0}&{0}\\
{1}&{0}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{1}&{1}
\end{array}} \right)=-\epsilon_{11}-\epsilon_{12}+\epsilon_{21}+\epsilon_{22}
\end{eqnarray*}$$
$$\begin{eqnarray*}
A\epsilon_{22}&=&B^{-1}\epsilon_{22}B=\left( {\begin{array}{*{20}{c}}
{1}&{1}\\
{-2}&{-1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{0}&{0}\\
{0}&{1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{-1}&{-1}\\
{2}&{1}
\end{array}} \right)\\
&=&\left( {\begin{array}{*{20}{c}}
{2}&{1}\\
{-2}&{-1}
\end{array}} \right)=2\epsilon_{11}+\epsilon_{12}-2\epsilon_{21}-\epsilon_{22}
\end{eqnarray*}$$
　　所以
$$\left\{ \begin{array}{l}
A\epsilon_{11}=-\epsilon_{11}-\epsilon_{12}+2\epsilon_{21}+2\epsilon_{22}\\
A\epsilon_{12}=2\epsilon_{11}+\epsilon_{12}-4\epsilon_{21}-2\epsilon_{22}\\
A\epsilon_{21}=-\epsilon_{11}-\epsilon_{12}+\epsilon_{21}+\epsilon_{22}\\
A\epsilon_{22}=2\epsilon_{11}+\epsilon_{12}-2\epsilon_{21}-\epsilon_{22}
\end{array} \right.$$
　　$A$在这组基下的矩阵为
$$A=\left( {\begin{array}{*{20}{c}}
{-1}&{2}&{-1}&{2}\\
{-1}&{1}&{-1}&{1}\\
{2}&{-4}&{1}&{-2}\\
{2}&{-2}&{1}&{-1}
\end{array}} \right)$$
　　$A$的特征多项式为
$$f(\lambda)=(\lambda^2-1)^2$$
　　属于特征值$\lambda_0=1$的其中一个特征向量是
$$\left( {\begin{array}{*{20}{c}}
{-1}\\
{-\frac{1}{2}}\\
{1}\\
{0}
\end{array}} \right)$$
　　由于任何数域都包含有理数域作为它的一部分，所以上述所求的特征值和特征向量符合题意
　　故取
$$\lambda_0=1,X_0=\left( {\begin{array}{*{20}{c}}
{-1}&{-\frac{1}{2}}\\
{1}&{0}
\end{array}} \right)$$
　　使
$$AX_0=\lambda_0X_0$$

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