每週問題 July 8, 2013

在代數中,若 a>0,則 a+a^{-1}\ge 2。那麼對應的矩陣性質該如何證明呢?

Let A and B be n\times n Hermitian matrices. We denote A\succcurlyeq B if A-B is positive semidefinite. If A\succ 0, i.e., A is positive definite, show that A+A^{-1}\succcurlyeq 2I.

 
參考解答:

Hermitian 正定矩陣 A 可對角化為 A=U\Lambda U^\ast,其中 \Lambda=\hbox{diag}(\lambda_1,\ldots,\lambda_n) 且每一 \lambda_i>0U 是么正 (unitary) 矩陣,即 U^\ast=U^{-1}。利用 \lambda_i+\lambda_i^{-1}\ge 2 計算二次型:

\displaystyle  \mathbf{x}^\ast(A+A^{-1})\mathbf{x}=\mathbf{x}^\ast U(\Lambda+\Lambda^{-1})U^\ast\mathbf{x}\ge 2\mathbf{x}^\ast UU^\ast\mathbf{x}=2\mathbf{x}^\ast\mathbf{x}

故證明 A+A^{-1}\succcurlyeq 2I

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