题目
1 -1 4-|||-设α1= 2 a2= 3 α3 = -1 试用施密特正交化过程-|||--1 1 0-|||-把这组向量标准正交化.
题目解答
答案
解析
步骤 1：计算 ${\beta }_{1}$
${\beta }_{1}={\alpha }_{1}=\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right.$
步骤 2：计算 ${\beta }_{2}$
${\beta }_{2}={\alpha }_{2}-\dfrac {[ {\alpha }_{2},{\beta }_{1}] }{[ {\beta }_{1},{\beta }_{1}] }{\beta }_{1}$
其中，$[ {\alpha }_{2},{\beta }_{1}] =1*1+3*2+(-1)*(-1)=8$，$[ {\beta }_{1},{\beta }_{1}] =1*1+2*2+(-1)*(-1)=6$
所以，${\beta }_{2}=\left (\begin{matrix} 1\\ 3\\ -1\end{matrix} ) \right. -\dfrac {8}{6}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right. =\dfrac {5}{3}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right.$
步骤 3：计算 ${\beta }_{3}$
${\beta }_{3}={\alpha }_{3}-\dfrac {[ {\alpha }_{3},{\beta }_{1}] }{[ {\beta }_{1},{\beta }_{1}] }{\beta }_{1}-\dfrac {[ {\alpha }_{3},{\beta }_{2}] }{[ {\beta }_{2},{\beta }_{2}] }{\beta }_{2}$
其中，$[ {\alpha }_{3},{\beta }_{1}] =1*1+(-1)*2+0*(-1)=-1$，$[ {\alpha }_{3},{\beta }_{2}] =1*(-1)+(-1)*1+0*1=-2$，$[ {\beta }_{2},{\beta }_{2}] =\dfrac {5}{3}*\dfrac {5}{3}*(-1)*(-1)+\dfrac {5}{3}*\dfrac {5}{3}*1*1+\dfrac {5}{3}*\dfrac {5}{3}*1*1=\dfrac {50}{9}$
所以，${\beta }_{3}=\left (\begin{matrix} 1\\ -1\\ 0\end{matrix} ) \right. -\dfrac {-1}{6}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right. -\dfrac {-2}{\dfrac {50}{9}}\dfrac {5}{3}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right. =\dfrac {1}{\sqrt {2}}\left (\begin{matrix} 1\\ 0\\ 1\end{matrix} ) \right.$
步骤 4：单位化
${\varepsilon }_{1}=\dfrac {{\beta }_{1}}{||{\beta }_{1}||}=\dfrac {1}{\sqrt {6}}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right.$
${\varepsilon }_{2}=\dfrac {{\beta }_{2}}{||{\beta }_{2}||}=\dfrac {1}{\sqrt {3}}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right.$
${\varepsilon }_{3}=\dfrac {{\beta }_{3}}{||{\beta }_{3}||}=\dfrac {1}{\sqrt {2}}\left (\begin{matrix} 1\\ 0\\ 1\end{matrix} ) \right.$
${\beta }_{1}={\alpha }_{1}=\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right.$
步骤 2：计算 ${\beta }_{2}$
${\beta }_{2}={\alpha }_{2}-\dfrac {[ {\alpha }_{2},{\beta }_{1}] }{[ {\beta }_{1},{\beta }_{1}] }{\beta }_{1}$
其中，$[ {\alpha }_{2},{\beta }_{1}] =1*1+3*2+(-1)*(-1)=8$，$[ {\beta }_{1},{\beta }_{1}] =1*1+2*2+(-1)*(-1)=6$
所以，${\beta }_{2}=\left (\begin{matrix} 1\\ 3\\ -1\end{matrix} ) \right. -\dfrac {8}{6}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right. =\dfrac {5}{3}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right.$
步骤 3：计算 ${\beta }_{3}$
${\beta }_{3}={\alpha }_{3}-\dfrac {[ {\alpha }_{3},{\beta }_{1}] }{[ {\beta }_{1},{\beta }_{1}] }{\beta }_{1}-\dfrac {[ {\alpha }_{3},{\beta }_{2}] }{[ {\beta }_{2},{\beta }_{2}] }{\beta }_{2}$
其中，$[ {\alpha }_{3},{\beta }_{1}] =1*1+(-1)*2+0*(-1)=-1$，$[ {\alpha }_{3},{\beta }_{2}] =1*(-1)+(-1)*1+0*1=-2$，$[ {\beta }_{2},{\beta }_{2}] =\dfrac {5}{3}*\dfrac {5}{3}*(-1)*(-1)+\dfrac {5}{3}*\dfrac {5}{3}*1*1+\dfrac {5}{3}*\dfrac {5}{3}*1*1=\dfrac {50}{9}$
所以，${\beta }_{3}=\left (\begin{matrix} 1\\ -1\\ 0\end{matrix} ) \right. -\dfrac {-1}{6}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right. -\dfrac {-2}{\dfrac {50}{9}}\dfrac {5}{3}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right. =\dfrac {1}{\sqrt {2}}\left (\begin{matrix} 1\\ 0\\ 1\end{matrix} ) \right.$
步骤 4：单位化
${\varepsilon }_{1}=\dfrac {{\beta }_{1}}{||{\beta }_{1}||}=\dfrac {1}{\sqrt {6}}\left (\begin{matrix} 1\\ 2\\ -1\end{matrix} ) \right.$
${\varepsilon }_{2}=\dfrac {{\beta }_{2}}{||{\beta }_{2}||}=\dfrac {1}{\sqrt {3}}\left (\begin{matrix} -1\\ 1\\ 1\end{matrix} ) \right.$
${\varepsilon }_{3}=\dfrac {{\beta }_{3}}{||{\beta }_{3}||}=\dfrac {1}{\sqrt {2}}\left (\begin{matrix} 1\\ 0\\ 1\end{matrix} ) \right.$