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。

This entry was posted in pow 向量空間, 每週問題 and tagged . Bookmark the permalink.

發表迴響

在下方填入你的資料或按右方圖示以社群網站登入:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / 變更 )

Twitter picture

You are commenting using your Twitter account. Log Out / 變更 )

Facebook照片

You are commenting using your Facebook account. Log Out / 變更 )

Google+ photo

You are commenting using your Google+ account. Log Out / 變更 )

連結到 %s