在阅读该页内容之前,我们向量子计算的开创者费曼和Deutsch致敬,同时向三位量子信息学的奠基人Charles H. Bennett, David Deutsch, Peter Shor表示敬意.
问题：
Show that$ (A^†)^ † = A$.
说明$ (A^†)^ † = A$.
解答
令$$A=\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right),$$
$$A^†=\left(\begin{array}{cccc}
a_{11}^* & a_{21}^* & \cdots & a_{n1}^* \\
a_{21}^* & a_{22}^* & \cdots & a_{2n}^* \\
\vdots & \vdots & & \vdots \\
a_{1n}^* & a_{2n}^* & \cdots & a_{nn}^*
\end{array}\right),$$
$$\left(A^{\dagger}\right)^{\dagger}=\left(\begin{array}{cccc}
\left(a_{11}^{*}\right)^{*} & \left(a_{12}^{*}\right)^{*} & \cdots & \left(a_{1 n}^{*}\right)^{*} \\
\left(a_{21}^{*}\right)^{*} & \left(a_{22}^{*}\right)^{*} & \cdots & \left(a_{2 n}^{*}\right)^{*} \\
\vdots & \vdots & & \vdots \\
\left(a_{n 1}^{*}\right)^{*} & \left(a_{n 2}^{*}\right)^{*} & \cdots & \left(a_{n n}^{*}\right)^{*}
\end{array}\right).$$
故$ (A^†)^ † = A$.
参考
[1]www.qtumist.com
参与者
作者：HKL, W65
贡献者:Dingyan, Wjw,Wxw
