每週問題 July 29, 2013

證明 A^\ast A=0 蘊含 A=0

Suppose all the matrix multiplications are well defined. Prove the following statements.
(a) If A^\ast A=0, then A=0.
(b) If BAA^\ast=CAA^\ast, then BA=CA.


(a) 令 A 為一 m\times n 階矩陣。利用 \hbox{rank}(A^\ast A)=\hbox{rank}A,可知 \hbox{rank}A=0,故 A=0。下面證明矩陣秩等式。矩陣 A 的零空間等於 A^{\ast}A 的零空間。注意 N(A)=N(A^{\ast}A) 等價於 N(A)\subseteq N(A^{\ast}A)N(A^{\ast}A)\subseteq N(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},等號兩邊同時左乘 \mathbf{x}^{\ast},可得

\displaystyle  \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

A\mathbf{x}=0。根據 AA^{\ast}A 的秩—零度定理表達式 \dim C(A)+\dim N(A)=n\dim C(A^{\ast}A)+\dim N(A^{\ast}A)=n,即證明 \mathrm{rank}A=\mathrm{rank}(A^{\ast}A)

(b) 寫出

\displaystyle  (BAA^\ast-CAA^\ast)(B-C)^\ast=(BA-CA)(BA-CA)^\ast=0

利用 (a) 立得 BA=CA

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