## 每週問題 February 15, 2016

Prove the following statements.

(a) If $A$ is an $m\times n$ complex matrix, then $\Vert A\Vert_F^2=\hbox{trace}(A^\ast A)\ge 0$, and $\Vert A\Vert_F=0$ if and only if $A=0$.

(b) If $A_1,\ldots,A_k$ are $m\times n$ complex matrices and $\sum_{i=1}^kA_i^\ast A_i=0$, then $A_1=\cdots=A_k=0$.

(c) If $A^\ast A=B^\ast B-BB^\ast$, then $A=0$.

(d) If $A^\ast$ commutes with $A$ and if $A$ commutes with $B$, then $A^\ast$ commutes with $B$. Prove this statement by showing that

$\Vert A^\ast B-BA^\ast\Vert^2_F-\Vert AB-BA\Vert^2_F=\hbox{trace}\left((A^\ast A-AA^\ast)(B^\ast B-BB^\ast)\right)$.

(a) 因為 $A^\ast A$ 是 Hermitian 半正定，寫出正交對角化形式 $A^\ast A=U\,\hbox{diag}(\mu_1,\ldots,\mu_n)\,U^\ast$，其中 $U$ 是么正 (unitary) 矩陣，即 $U^\ast=U^{-1}$，且每一 $\mu_i\ge 0$。因此，$\hbox{trace}(A^\ast A)=\mu_1+\cdots+\mu_n\ge 0$，又 $\hbox{trace}(A^\ast A)=0$ 等價於 $\mu_1=\cdots=\mu_n=0$，即 $A^\ast A=0$。再者，$\hbox{rank}(A^\ast A)=\hbox{rank}A$，推得 $A^\ast A=0$ 等價於 $A=0$

(b) 跡數是線性函數，

$\displaystyle \sum_{i=1}^k\Vert A_i\Vert_F^2=\sum_{i=1}^k\hbox{trace}(A_i^\ast A_i)=\hbox{trace}\left(\sum_{i=1}^kA_i^\ast A_i\right)=0$

$\Vert A_1\Vert_F=\cdots=\Vert A_k\Vert_F=0$。由 (a)，$A_1=\cdots=A_k=0$

(c) 使用給定條件，

$\Vert A\Vert_F^2=\hbox{trace}(A^\ast A)=\hbox{trace}(B^\ast B-BB^\ast)=\hbox{trace}(B^\ast B)-\hbox{trace}(BB^\ast)=0$

$A=0$

(d) 使用跡數循環不變性，

\begin{aligned} \Vert A^\ast B-BA^\ast\Vert^2_F-\Vert AB-BA\Vert^2_F &=\hbox{trace}\left((A^\ast B-BA^\ast)^\ast(A^\ast B-BA^\ast)\right)\\ &~~-\hbox{trace}\left((AB-BA)^\ast(AB-BA)\right)\\ &=\hbox{trace}(B^\ast AA^\ast B-B^\ast ABA^\ast-AB^\ast A^\ast B+AB^\ast BA^\ast)\\ &~~-\hbox{trace}(B^\ast A^\ast AB-B^\ast A^\ast BA-A^\ast B^\ast AB+A^\ast B^\ast BA)\\ &=\hbox{trace}(AA^\ast BB^\ast-ABA^\ast B^\ast-AB^\ast A^\ast B+A^\ast AB^\ast B)\\ &~~-\hbox{trace}(A^\ast ABB^\ast-AB^\ast A^\ast B-ABA^\ast B^\ast +AA^\ast B^\ast B)\\ &=\hbox{trace}\left((A^\ast A-AA^\ast)(B^\ast B-BB^\ast)\right). \end{aligned}

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