每週問題 October 24, 2016

對於任一複矩陣 A\hbox{rank}(A^\ast A)=\hbox{rank}(A^TA) 總是成立嗎?

Let A be an m\times n complex matrix. Prove or disprove the following statements.

(a) \hbox{rank}(A^\ast A)=\hbox{rank}A.
(b) \hbox{rank}(A^TA)=\hbox{rank}A.

 
參考解答:

(a) 正確。我們先證明 m\times n 矩陣 An\times n 矩陣 A^\ast A 有相同的零空間 (nullspace),記為 N(A)=N(A^\ast A)。若 A\mathbf{x}=\mathbf{0},則 A^{\ast}A\mathbf{x}=A^{\ast}\mathbf{0}=\mathbf{0}。若 A^{\ast}A\mathbf{x}=\mathbf{0},則 \mathbf{x}^{\ast}A^{\ast}A\mathbf{x}=\left(A\mathbf{x}\right)^{\ast}\left(A\mathbf{x}\right)=\Vert A\mathbf{x}\Vert^2=0,故 A\mathbf{x}=\mathbf{0}。使用秩—零度定理 \hbox{rank}A+\dim N(A)=n\hbox{rank}(A^\ast A)+\dim N(A^{\ast}A)=n,證明 \mathrm{rank}A=\mathrm{rank}(A^{\ast}A)

(b) 錯誤。例如,A=\left[\!\!\begin{array}{cr} 1&-i\\ i&1 \end{array}\!\!\right],可得 \hbox{rank}A=1,但 \hbox{rank}(A^TA)=0。當 A 是一個實矩陣時,\hbox{rank}(A^TA)=\hbox{rank}A 方成立。

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