每週問題 August 18, 2014

這是關於在複數系和實數系的矩陣相似性問題。

Let A and B be n\times n real matrices. Show that if B=P^{-1}AP, where P is a complex matrix, then there exists a real matrix T such that B=T^{-1}AT. That is, if two real square matrices are similar over \mathbb{C}, then they must be similar over \mathbb{R}.

 
參考解答:

P=C+iD,其中 CD 是實矩陣,i=\sqrt{-1}。因為 AB 是實矩陣,AP=PB,即 AC+iAD=CB+iDB,意味 AC=CBAD=DB。寫出 D=D'-I,並令 T=C+tD=(C+tD')-tI,其中 t 是實數。若 \lambda_1,\ldots,\lambda_nC+tD' 的特徵值,則 T 的特徵值為 \lambda_j-tj=1,\ldots,n。我們一定可以找到 t\in\mathbb{R} 使得 t\neq \lambda_jj=1,\ldots,n,故證明存在一實可逆矩陣 T 滿足 AT=TB

This entry was posted in pow 特徵分析, 每週問題 and tagged . Bookmark the permalink.

1 Response to 每週問題 August 18, 2014

  1. Meiyue Shao says:

    比较一般的结论是:矩阵相似与否与域的选取无关。这里的技术只适用于实数域和复数域,不过可以用于证明另一个结论:两个实矩阵若酉相似则实正交相似(Specht定理)。

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