# 问题：

Show that the transpose, complex conjugation, and adjoint operations distribute over the tensor product,$(A \otimes B)^{*}=A^{*} \otimes B^{*} ;(A \otimes B)^{T}=A^{T} \otimes B^{T} ;(A \otimes B)^{\dagger}=A^{\dagger} \otimes B^{\dagger}$；

## 解答

$(A \otimes B)^{*}=\left[\begin{array}{ccc}a_{11} B & a_{12} B & \ldots \\ a_{21} B & a_{22} B & \ldots \\ \vdots & & \ddots\end{array}\right]^{*} =\left[\begin{array}{ccc}a_{11}^{*} B^{*} & a_{21}^{*} B^{*} & \ldots \\ a_{12}^{*} B^{*} & a_{22}^{*} B^{*} & \ldots \\ \vdots & & \ddots\end{array}\right]=A^{*} \otimes B^{*}$;

$(A \otimes B)^{T}=\left[\begin{array}{ccc}a_{11} B & a_{12} B & \ldots \\ a_{21} B & a_{22} B & \ldots \\ \vdots & & \ddots\end{array}\right]^{T}=\left[\begin{array}{ccc}a_{11}^{T} B^{T} & a_{12}^{T} B^{T} & \ldots \\ a_{21}^{T} B^{T} & a_{22}^{T} B^{T} & \ldots \\ \vdots & & \ddots\end{array}\right]=A^{T} \otimes B^T$;

$(A \otimes B)^{\dagger}=((A \otimes B)^{T})^*=(A^{T} \otimes B^{T})^*=(A^{T})^* \otimes (A^{T})^*=A^{\dagger} \otimes B^{\dagger}$;

#### 参考

[1]www.qtumist.com