做滾動圖的免費(fèi)網(wǎng)站百度競價返點(diǎn)開戶
本質(zhì)上在做什么？
內(nèi)積是距離度量，核函數(shù)相當(dāng)于將低維空間的距離映射到高維空間的距離，并非對特征直接映射。
為什么要求核函數(shù)是對稱且Gram矩陣是半正定？
核函數(shù)對應(yīng)某一特征空間的內(nèi)積，要求①核函數(shù)對稱；②Gram矩陣半正定。
證明內(nèi)積對應(yīng)的Gram矩陣半正定：
α T K α = [ α 1 , α 2 , ? , α n ] [ κ ( x 1 , x 1 ) κ ( x 1 , x 2 ) ? κ ( x 1 , x n ) κ ( x 2 , x 1 ) κ ( x 2 , x 2 ) ? κ ( x 1 , x n ) ? ? ? ? κ ( x n , x 1 ) κ ( x n , x 2 ) ? κ ( x n , x n ) ] [ α 1 α 2 ? α n ] = ∑ i = 1 n ∑ j = 1 n α i κ ( x i , x j ) α j = ∑ i = 1 n ∑ j = 1 n α i α j ? ? ( x i ) , ? ( x j ) ? = ? ∑ i = 1 n α i ? ( x i ) , ∑ j = 1 n α j ? ( x j ) ? = ∥ ∑ i = 1 n α i ? ( x i ) ∥ 2 2 ? 0 \begin{aligned} {{ \bm \alpha}^{\rm T} {\bm K} { \bm \alpha}} &=\begin{bmatrix} {\alpha}_1, {\alpha}_2, \cdots, {\alpha}_n \end{bmatrix} \begin{bmatrix} \kappa \left( {\bm x}_1, {\bm x}_1 \right) &\kappa \left( {\bm x}_1, {\bm x}_2 \right) &\cdots &\kappa \left( {\bm x}_1, {\bm x}_n \right) \\ \kappa \left( {\bm x}_2, {\bm x}_1 \right) &\kappa \left( {\bm x}_2, {\bm x}_2 \right) &\cdots &\kappa \left( {\bm x}_1, {\bm x}_n \right) \\ \vdots &\vdots &\ddots &\vdots \\ \kappa \left( {\bm x}_n, {\bm x}_1 \right) &\kappa \left( {\bm x}_n, {\bm x}_2 \right) &\cdots &\kappa \left( {\bm x}_n, {\bm x}_n \right) \\ \end{bmatrix} \begin{bmatrix} {\alpha}_1 \\ {\alpha}_2 \\ \vdots \\ {\alpha}_n \\ \end{bmatrix} \\ &= \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n} {\alpha}_i \kappa \left( {\bm x}_i, {\bm x}_j \right) {\alpha}_j \\ &= \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n} {\alpha}_i {\alpha}_j \langle \phi \left( {\bm x}_i \right), \phi \left( {\bm x}_j \right) \rangle\\ &= \langle \sum\limits_{i=1}^{n} {\alpha}_i \phi \left( {\bm x}_i \right), \sum\limits_{j=1}^{n} {\alpha}_j \phi \left( {\bm x}_j \right) \rangle \\ &= \lVert \sum\limits_{i=1}^{n} {\alpha}_i \phi \left( {\bm x}_i \right) \rVert^2_2 \\ &\geqslant 0 \end{aligned} αTKα?=[α1?,α2?,?,αn??] ?κ(x1?,x1?)κ(x2?,x1?)?κ(xn?,x1?)?κ(x1?,x2?)κ(x2?,x2?)?κ(xn?,x2?)??????κ(x1?,xn?)κ(x1?,xn?)?κ(xn?,xn?)? ? ?α1?α2??αn?? ?=i=1∑n?j=1∑n?αi?κ(xi?,xj?)αj?=i=1∑n?j=1∑n?αi?αj???(xi?),?(xj?)?=?i=1∑n?αi??(xi?),j=1∑n?αj??(xj?)?=∥i=1∑n?αi??(xi?)∥22??0?