「半监督学习」笔记（四）

令
$\begin{aligned} &W = (w_1, \, \cdots, w_m)\\ &F = (f(x_1), \cdots, f(x_m)) = (f_1, \cdots, f_m) \end{aligned}$
则有

$\begin{al$

$igned} FWF^T &= (\sum_{i=1}^m f_i(W)_{1,i},\cdots, \sum_{i=1}^m f_j(W)_{m,i})F^T \\ &= (Fw_1, \cdots, Fw_m)F^T \\ &= \sum_{i=1}^m Fw_if_i^T \\ &= \sum_{i,j=1}^m (W)_{i,j}f_i f_j^T \end{aligned}$

FWFT=(i=1∑mfi(W)1,i,⋯,i=1∑mfj(W)m,i)FT=(Fw1,⋯,Fwm)FT=i=1∑mFwifiT=i,j=1∑m(W)i,jfifjT

故而
$\begin{aligned} Tr(FWF^T) &= \sum_{i,j=1}^m Tr(f_i f_j^T)(W)_{i,j} \\ &= {\displaystyle\sum_{i=1}^{m}\sum_{j=1}^m}f(x_i)^Tf(x_j)(W)_{ij} \end{aligned}$

令 $d_j = {\displaystyle \sum_{j=1}^m (W)_{ij} = ||w_j||_1}$，则记
$D = diag(d_1, \cdots, d_m)$，故而
${\displaystyle\sum_{i=1}^{m}\sum_{j=1}^m}(W)_{ij}||f(x_i)||^2 = Tr(FDF^T)$

因而
$E(f) = Tr(F(D-W)F^T)$