矩阵理论及其应用邱启荣习题3.5题解

(1) P= ( − 1 0 1 − 1 − 1 2 1 1 − 1 ) \begin{pmatrix} -1 & 0&1 \\ -1 & -1&2\\1&1&-1 \end{pmatrix} ​−1−11​0−11​12−1​ ​
A= ( 1 0 1 1 1 0 − 1 2 1 ) \begin{pmatrix} 1 & 0&1 \\ 1 & 1&0\\-1&2&1 \end{pmatrix} ​11−1​012​101​ ​
B=P${}^{-1}$AP= ( − 2 − 3 5 0 0 2 − 2 − 2 5 ) \begin{pmatrix} -2 & -3&5 \\ 0 & 0&2\\-2&-2&5 \end{pmatrix} ​−20−2​−30−2​525​ ​
$\begin{array}{rl}(2)T(\alpha )& =T({\alpha }_1)+6T({\alpha }_2)-T({\alpha }_3)\\ & =(1,0,1)+(6,6,0)+(1,-2,-1)\\ & =(8,4,0)\end{array}$
$\begin{array}{rl}T(\beta )& =T(e_1)-T(e_2)+T(e_3)\\ & =(-2,-3,5)+(0,0,-2)+(-2,-2,5)\\ & =(-4,-5,8)\end{array}$
注意$\alpha$是行向量