二次项吸收

参考：Manifold regularized matrix completion for multi-label learning with
ADMM

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试着推导下：
$$\begin{gathered} \frac{\rho }{2}\Vert Z-W+\frac{1}{\rho }\Lambda {\Vert }_F^2 \\ =\frac{\rho }{2}tr[(Z-W+\frac{1}{\rho }\Lambda )(Z-W+\frac{1}{\rho }\Lambda {)}^T] \\ =\frac{\rho }{2}tr[(Z-W+\frac{1}{\rho }\Lambda )[(Z-W{)}^T+(\frac{1}{\rho }\Lambda {)}^T] \\ =\frac{\rho }{2}[\Vert Z-W{\Vert }_F^2+tr((\frac{1}{\rho }\Lambda )(Z-W{)}^T)+tr((Z-W)(\frac{1}{\rho }\Lambda {)}^T)+\Vert \frac{1}{\rho }\Lambda {\Vert }_F^2] \\ =\frac{\rho }{2}[\Vert Z-W{\Vert }_F^2+2tr((\frac{1}{\rho }\Lambda )(Z-W{)}^T)+\Vert \frac{1}{\rho }\Lambda {\Vert }_F^2] \end{gathered}$$

貌似不对……

换一个：
$$\begin{gathered} \frac{\rho }{2}\Vert Z-W+\frac{1}{\rho }\Lambda {\Vert }_F^2 \\ =\frac{\rho }{2}\sum (z-w+\frac{1}{\rho }\Lambda {)}^2 \\ =\frac{\rho }{2}\sum [(z-w{)}^2+(\frac{1}{\rho }\Lambda {)}^2+2(z-w)\frac{1}{\rho }\Lambda ] \\ =\frac{\rho }{2}\Vert Z-W{\Vert }_F^2+⟨Z-W,\Lambda ⟩+\frac{1}{2\rho }\Vert \Lambda {\Vert }_F^2 \end{gathered}$$

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之前疑惑，不管怎么算，都多了一项，不知道为啥!!!

后面经 知乎 长驱鬼侠大神解答。是因为：

$因为表达式是对Z求min，所以加\Lambda 的任意函数都成立$，所以可以任意凑项。即将$\frac{1}{2\rho }\Vert \Lambda {\Vert }_F^2$视为无关紧要的拼凑项。

大神V5, 点赞大神！！！