每週問題 March 6, 2017

證明 Hermitian 矩陣的秩與跡數不等式。

Let A be an n\times n nonzero Hermitian matrix. Prove that

\displaystyle \hbox{rank}A\ge\frac{(\hbox{trace}A)^2}{\hbox{trace}(A^2)}.

 
參考解答:

任一 n\times n 階 Hermitian 矩陣 A 可么正對角化為 A=U\Lambda U^\ast,其中 U^\ast=U^{-1}\Lambda=\hbox{diag}(\lambda_1,\ldots,\lambda_n)。因為 A 是非零 Hermitian 矩陣,至少存在一個非零特徵值。在不失一般性的原則下,令 \lambda_1,\ldots,\lambda_rA 的非零特徵值 (包含相重特徵值),其中 r=\hbox{rank}\Lambda=\hbox{rank}A。使用算術平均數小於或等於平方平均數 (方均根),

\displaystyle (\hbox{trace}A)^2=(\lambda_1+\cdots+\lambda_r)^2\le r(\lambda_1^2+\cdots+\lambda_r^2)=r\,\hbox{trace}(A^2)

因此得證。

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