每週問題 October 6, 2014

本週問題是證明 Hermitian 正定矩陣的積必可對角化。

Let A and B be n\times n Hermitian matrices. Show that if A or B is positive definite, then AB is diagonalizable.

 
參考解答:

假設 A 是正定矩陣,推知 A 的特徵值皆為正數,故 A 是一可逆矩陣。因此,AB 相似於 \sqrt{A}^{-1}AB\sqrt{A}=\sqrt{A}B\sqrt{A},其中 \sqrt{A}A 的平方根,也是 Hermitian。由於 \sqrt{A}B\sqrt{A} 是 Hermitian 矩陣,故可么正對角化 (unitarily diagonalizable) 為 \sqrt{A}B\sqrt{A}=UDU^\ast,其中 U 是一么正矩陣,滿足 U^\ast=U^{-1}D 是實對角矩陣。據此,AB 亦可對角化,如下:

\displaystyle AB=\sqrt{A}\left(\sqrt{A}B\sqrt{A}\right)\sqrt{A}^{-1}=\sqrt{A}(UDU^\ast)\sqrt{A}^{-1}=\left(\sqrt{A}U\right)D\left(\sqrt{A}U\right)^{-1}

按照相同方式亦可證明若 B 是正定矩陣,則 AB 可對角化。

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