实对称矩阵特征值的微分

考虑矩阵A的第p个特征值$lambda_p$及属于它的特征向量$v_p$, egin{align} A v_p = lambda_p v_p end{align} 等式两边同时取微分 egin{align} (dA) v_p + A d v_p = (dlambda_p) v_p + lambda_p d v_p, end{align} 等式两边同时乘以$v_p^T$得 egin{align}label{eq:vpt} v_p^T (dA) v_p + v_p^T A d v_p = v_p^T (dlambda_p) v_p + v_p^T lambda_p d v_p = (dlambda_p) (v_p^T v_p) + lambda_p (v_p^T d v_p ), end{align} 根据A的实对称属性,可知第一个等号左边$v_p^T A = (A^T v_p)^T = (A v_p)^T = lambda_p v_p^T$, 带入( ef{eq:vpt})式得 egin{align} v_p^T (dA) v_p + lambda_p (v_p^T d v_p ) = (dlambda_p) (v_p^T v_p) + lambda_p (v_p^T d v_p ), end{align} 根据特征向量$v_p$的长度恒为1,可以计算上式中 $v_p^T d v_p$满足如下关系: egin{align} 0 = d 1 = d |v_p|_2^2 = 2 (v_p^T d v_p), end{align} 因此有 egin{align} v_p^T (dA) v_p = dlambda_p cdot 1 . end{align}

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原文地址:https://www.cnblogs.com/xyq-deeplearning/p/15111307.html