Question

12．已知矩陣A=$[\begin{array}{l}{1}&{2}\\{2}&{1}\end{array}]$，B=$[\begin{array}{l}{0}&{1}\\{1}&{0}\end{array}]$，求滿足條件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$．

Answer 1

分析 利用矩陣乘法公式求出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，求出矩陣AB的特征根，由此能求出滿足條件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量．

解答 解：∵矩陣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}]$$[\begin{array}{l}{0}&{1}\\{1}&{0}\end{array}]$=$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$，
∴f（λ）=|λE-AB|=$|\begin{array}{l}{λ-2}&{-1}\\{-1}&{λ-2}\end{array}|$=（λ-2）²-1=0，
解得矩陣AB的特征根為λ₁=1，λ₂=3，
設(shè)λ=1對應的特征向量$\overrightarrow{α}$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
∵（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$，∴$[\begin{array}{l}{2}&{1}\\{1}&{2}\end{array}]$$[\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}]$，
解得x=-y，
∴滿足條件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$=$[\begin{array}{l}{1}\\{-1}\end{array}]$．
∴設(shè)λ=3對應的特征向量$\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}]$，
解得x=y，
∴滿足條件（AB）$\overrightarrow{a}$=λ$\overrightarrow{a}$特征向量$\overrightarrow{a}$=$[\begin{array}{l}{1}\\{1}\end{array}]$．

點評 本題考查矩陣的特征向量的求法，考查矩陣的特征向量、特征值等基礎(chǔ)知識，考查推理論證能力、運算求解能力，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想，是基礎(chǔ)題．