每週問題 March 3, 2014

這是關於二個半正定矩陣積的跡數與二次型問題。

Let A and B be real symmetric matrices and let A and B be positive semidefinite. Prove the following statements.
(a) If \text{trace}(AB)=0, then AB=BA=0.
(b) If \mathbf{x}^TAB\mathbf{x}=0 for all real \mathbf{x}, then AB=BA=0.

 
參考解答:

(a) 為方便說明,我們以 A\succcurlyeq 0 表示 A 是一實對稱正定矩陣。令 \sqrt{A} 表示 A 的平方根,滿足 A=\sqrt{A}\sqrt{A}\sqrt{A}\succcurlyeq 0。使用跡數的循環不變性,

\displaystyle\begin{aligned} 0=\text{trace}(AB)&=\text{trace}(\sqrt{A}\sqrt{A}\sqrt{B}\sqrt{B})=\text{trace}(\sqrt{B}\sqrt{A}\sqrt{A}\sqrt{B})\\ &=\text{trace}((\sqrt{A}\sqrt{B})^T(\sqrt{A}\sqrt{B}))=\Vert \sqrt{A}\sqrt{B}\Vert^2_F, \end{aligned}

其中 \Vert\cdot\Vert_F 表示 Frobenius 範數。上式表明 \sqrt{A}\sqrt{B}=0。左乘 \sqrt{A},右乘 \sqrt{B},即得 AB=0,而且 BA=B^TA^T=(AB)^T=0

(b) 假設每一 \mathbf{x} 使得 \mathbf{x}^TAB\mathbf{x}=0。令 \mathbf{e}_i 為標準單位向量,即第 i 元為 1,其餘元為 0。計算二次型,如下:

\displaystyle 0=\mathbf{e}_i^TAB\mathbf{e}_i=(AB)_{ii}

\displaystyle 0=(\mathbf{e}_i+\mathbf{e}_j)^TAB(\mathbf{e}_i+\mathbf{e}_j)=(AB)_{ij}+(AB)_{ji},~~i\neq j

所以,AB 是反對稱 (anti-symmetric) 矩陣,即 AB=-(AB)^T=-BA。等號兩邊取跡數並使用循環不變性,可得

\displaystyle \text{trace}(AB)=\text{trace}(-BA)=-\text{trace}(BA)=-\text{trace}(AB)

\text{trace}(AB)=0。根據 (a),推得 AB=BA=0

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