每週問題 January 6, 2014

這是推導可逆矩陣的練習問題。

Let A and B be n\times n matrices. If I-AB is nonsingular, show that I-BA is nonsingular.

 
參考解答:

解法1:直接計算

\displaystyle\begin{aligned}  (I-BA)(I+B(I-AB)^{-1}A)&=I-BA+(B-BAB)(I-AB)^{-1}A\\  &=I-BA+B(I-AB)(I-AB)^{-1}A\\  &=I-BA+BA=I.\end{aligned}

解法2:I-AB 是可逆矩陣,表示 AB 的特徵值不為 1。因為 ABBA 有相同的特徵值,BA 的特徵值也不為 1,故 I-BA 是可逆矩陣。

This entry was posted in pow 線性方程與矩陣代數, 每週問題 and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s