每週問題 February 18, 2013

這是關於實對稱可交換矩陣的特徵值問題。

Let A and B be n\times n real symmetric matrices. If AB is symmetric, show that every eigenvalue \lambda of AB can be written as \lambda=\alpha\beta, where \alpha is an eigenvalue of A and \beta is an eigenvalue of B.

 
參考解答:

若實對稱矩陣 AB 的乘積 AB 是對稱矩陣,則 AB=(AB)^T=B^TA^T=BA。因為實對稱矩陣可正交對角化,且 AB=BA 意味 AB 可同時對角化,即知存在正交矩陣 QQ^T=Q^{-1},使得 Q^TAQ=\mathrm{diag}(\alpha_1,\ldots,\alpha_n)Q^TBQ=\mathrm{diag}(\beta_1,\ldots,\beta_n),其中 \alpha_i\beta_i 分別是 AB 的特徵值。所以,

Q^TABQ=(Q^TAQ)(Q^TBQ)=\mathrm{diag}(\alpha_1\beta_1,\ldots,\alpha_n\beta_n)

故證明 AB 的特徵值為 \alpha_i\beta_ii=1,\ldots,n

PowSol-Feb-18-13

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