每週問題 November 14, 2016

證明一個常見於多變量統計學的矩陣 I_m-A(A^{\ast}A)^{-1}A^\ast 是半正定。

Let m\ge n and let A be a complex m\times n matrix of rank n. Show that the Hermitian matrix B=I_m-A(A^{\ast}A)^{-1}A^\ast is positive semidefinite.

 
參考解答:

因為 \hbox{rank}(A^\ast A)=\hbox{rank}A=nn\times n 階 Gramian 矩陣 A^\ast A 是可逆的。直接計算可得 B=B^\ast=B^2B 是一個 Hermitian 冪等 (idempotent) 矩陣,也就是正交投影矩陣。因此,B 的特徵值為 01,即得證。事實上,冪等矩陣 B 滿足 \hbox{rank}B=\hbox{trace}B,而

\begin{aligned} \hbox{trace}B&=\hbox{trace}(I_m-A(A^{\ast}A)^{-1}A^\ast)\\ &=\hbox{trace}{I_m}-\hbox{trace}(A(A^{\ast}A)^{-1}A^\ast)\\ &=\hbox{trace}I_m-\hbox{trace}((A^{\ast}A)^{-1}A^\ast A)\\ &=\hbox{trace}I_m-\hbox{trace}I_n\\ &=m-n, \end{aligned}

故知 Bm-n 個特徵值 1n 個特徵值 0

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