Question

12．已知矩阵A=$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$，B=$[\begin{array}{l}{0}&{1}\\{1}&{0}\end{array}]$，求满足条件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$．

Answer 1

分析 利用矩阵乘法公式求出AB=$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$，由f（λ）=|λE-AB|=$|\begin{array}{l}{λ-2}&{-1}\\{-1}&{λ-2}\end{array}|$=（λ-2）²-1=0，求出矩阵AB的特征根，由此能求出满足条件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量．

解答 解：∵矩阵A=$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$，B=$[\begin{array}{l}{0}&{1}\\{1}&{0}\end{array}]$，
∴AB=$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$$[\begin{array}{l}{0}&{1}\\{1}&{0}\end{array}]$=$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$，
∴f（λ）=|λE-AB|=$|\begin{array}{l}{λ-2}&{-1}\\{-1}&{λ-2}\end{array}|$=（λ-2）²-1=0，
解得矩阵AB的特征根为λ₁=1，λ₂=3，
设λ=1对应的特征向量$\overrightarrow{α}$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
∵（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$，∴$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，即$[\begin{array}{l}{2x+y}\\{x+2y}\end{array}]=[\begin{array}{l}{x}\\{y}\end{array}]$，
解得x=-y，
∴满足条件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$=$[\begin{array}{l}{1}\\{-1}\end{array}]$．
∴设λ=3对应的特征向量$\overrightarrow{α}$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
∵（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$，∴$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=3$[\begin{array}{l}{x}\\{y}\end{array}]$，即$[\begin{array}{l}{2x+y}\\{x+2y}\end{array}]=[\begin{array}{l}{3x}\\{3y}\end{array}]$，
解得x=y，
∴满足条件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$=$[\begin{array}{l}{1}\\{1}\end{array}]$．

点评 本题考查矩阵的特征向量的求法，考查矩阵的特征向量、特征值等基础知识，考查推理论证能力、运算求解能力，考查化归与转化思想、函数与方程思想，是基础题．