早稲田大学 創造理工学研究科 経営システム工学専攻 2021年8月実施 线性代数

标签：

• Waseda-University
• Eigenvalues-and-Eigenvectors
• Basis-and-Dimension

Author

思齐塾, 祭音Myyura

Description

$A$ を任意の $n$ 次正方行列 (n-dimension matrix) とし、 $B$ を $n$ 次の正則行列 (non-singular matrix) とするとき、 $A$ の固有値 (eigenvalue) と $B^{-1}AB$ の固有値は一致することを示せ。

题目描述

设 $A$ 为任意 $n$ 阶方阵，$B$ 为 $n$ 阶可逆矩阵。证明 $A$ 与

$$B^{-1}AB$$

的特征值相同。

Kai

Let $\lambda$ be an eigenvalue of $A$ , and let $v$ be the corresponding eigenvector. Then $Av = \lambda v$ . We want to show that $\lambda$ is also an eigenvalue of $B^{-1}AB$ . Let $w = B^{-1}v$ . Since $B$ is invertible, $B^{-1}v \neq 0$ if $v\neq 0$ . Thus $w$ is a non-zero vector. Then

$$(B^{-1}AB)w =B^{-1}AB(B^{-1}v) =B^{-1}Av =\lambda B^{-1}v =\lambda w.$$

This shows that $\lambda$ is an eigenvalue of $B^{-1}AB$ . Suppose $\lambda$ is an eigenvalue of $B^{-1}AB$ . Then there exists $v \neq 0$ such that $B^{-1}ABv = \lambda v$ . Multiply both sides by $B$ from the left. $B(B^{-1}ABv) = B(\lambda v)$ , so $ABv = \lambda Bv$ . Let $w = Bv$ . Since $B$ is invertible and $v \neq 0$ , $Bv \neq 0$ . Thus, $w \neq 0$ . Therefore $Aw = \lambda w$ . This means $\lambda$ is an eigenvalue of $A$ . Therefore, the eigenvalues of $A$ and $B^{-1}AB$ are the same.