在上一篇部落格中，我們知道目標函式變為 $arg \min \limits_{f \in \R^6} f^TLf$，即找到一個$f$，使得 $f^TLf$ 取得最小值

這篇部落格將通過求導的方式取得目標函式的最小值。

1. 求$f^TLf$的導數

目標函式的未知量為$f$，那麼 $f^TLf$ 的導數可以表示為$\displaystyle \frac{\partial}{\partial f} f^TLf$。

這裡為了方便證明，使用一個二維向量作為例子，推廣到更高維空間也是一樣的，即假設 $f^T = [f_1 \space \space f_2]$

，$L = \left[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}\right]$

故：

$\begin{array}{cc}{\displaystyle {\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }f}{f}^{T}Lf}}& {\displaystyle ={\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }f}\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{c}{f}_1\\ f_2\end{array}\right]}}\\ {\displaystyle ={\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }f}\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]\left[\begin{array}{c}{a}_{11}{f}_1+a_{12}{f}_2\\ a_{21}{f}_1+a_{22}{f}_2\end{array}\right]}}\\ {\displaystyle ={\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }f}{a}_{11}{f}_1^2+a_{12}{f}_1{f}_2+a_{21}{f}_1{f}_2+a_{22}{f}_2^2}}\\ {\displaystyle ={\displaystyle [\begin{array}{c}{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{f}_1}{a}_{11}{f}_1^2+a_{12}{f}_1{f}_2+a_{21}{f}_1{f}_2+a_{22}{f}_2^2}\\ {\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{f}_2}{a}_{11}{f}_1^2+a_{}}\end{array}}}\end{array}$

12f1f2+a21f1f2+a22f22]=[2a11f1+a12f2+a21f2a12f1+a21f1+2a22f2]=[[a11a12a21a22]+[a11a21a12a22]][f1f2]=(L+LT)f\begin{aligned}\displaystyle \frac{\partial}{\partial f} f^TLf &= \displaystyle \frac{\partial}{\partial f} \left[ \begin{matrix} f_1 & f_2 \end{matrix}\right] \left[\begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{matrix}\right] \left[\begin{matrix}f_{1}\\ f_{2}\end{matrix}\right] \\ & = \displaystyle \frac{\partial}{\partial f} \left[ \begin{matrix} f_1 & f_2 \end{matrix}\right] \left[\begin{matrix}a_{11}f_1 + a_{12}f_2 \\ a_{21}f_1 + a_{22}f_2 \end{matrix}\right] \\ & = \displaystyle \frac{\partial}{\partial f} a_{11}f_1^2 + a_{12}f_1f_2 + a_{21}f_1f_2 + a_{22}f_2^2 \\ & = \displaystyle \left[ \begin{matrix}\displaystyle \frac{\partial}{\partial f_1} a_{11}f_1^2 + a_{12}f_1f_2 + a_{21}f_1f_2 + a_{22}f_2^2 \\ \displaystyle \frac{\partial}{\partial f_2} a_{11}f_1^2 + a_{12}f_1f_2 + a_{21}f_1f_2 + a_{22}f_2^2\end{matrix} \right] \\ & = \left[ \begin{matrix}2a_{11}f_1 + a_{12}f_2 + a_{21}f_2 \\ a_{12}f_1 + a_{21}f_1 + 2a_{22}f_2 \end{matrix} \right] \\ & = \displaystyle \left[ \left[ \begin{matrix}a_{11} & a_{12} \\ a_{21} & a{22}\end{matrix}\right] + \left[ \begin{matrix}a_{11} & a_{21} \\ a_{12} & a{22}\end{matrix}\right]\right]\left[ \begin{matrix}f_{1} \\ f_{2} \end{matrix}\right] \\ & = \displaystyle (L+L^T)f \end{aligned}