## 每週問題 February 1, 2016

Let $A_1$ and $A_2$ be $n\times n$ normal matrices. If $A_1B=BA_2$, show that $A_1^\ast B=BA_2^\ast$.

$A_1=U_1D_1U_1^\ast,~~A_2=U_2D_2U_2^\ast$

$U_1D_1U_1^\ast B=BU_2D_2U_2^\ast$

$D_1U_1^\ast BU_2=U_1^\ast BU_2D_2$

$D_1^\ast U_1^\ast BU_2=U_1^\ast BU_2D_2^\ast$