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修订公式10.24

Sm1les 6 yıl önce
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      docs/chapter10/chapter10.md

+ 1 - 0
docs/chapter10/chapter10.md

@@ -98,6 +98,7 @@ $$\boldsymbol w_j=\sum_{i=1}^{m} \phi\left(\boldsymbol{x}_{i}\right) \alpha_{i}^
 其中,$\boldsymbol{\alpha}^j=(\alpha_{1}^j;\alpha_{2}^j;...;\alpha_{m}^j)\in \mathbb{R}^{m \times 1} $。所以公式(10.21)可以进一步变换为
 $$\mathbf{Z}\mathbf{Z}^{\mathrm{T}}\mathbf{Z}\boldsymbol{\alpha}^j=\lambda_j\mathbf{Z}\boldsymbol{\alpha}^j $$
 $$\mathbf{Z}\mathbf{Z}^{\mathrm{T}}\mathbf{Z}\boldsymbol{\alpha}^j=\mathbf{Z}\lambda_j\boldsymbol{\alpha}^j $$
+由于主成分分析的目标是求出$\boldsymbol w_j$,也等价于要求出满足上式的$\boldsymbol{\alpha}^j$,显然,此时满足$\mathbf{Z}^{\mathrm{T}}\mathbf{Z}\boldsymbol{\alpha}^j=\lambda_j\boldsymbol{\alpha}^j $的$\boldsymbol{\alpha}^j$一定满足上式,所以问题转化为了求解满足下式的$\boldsymbol{\alpha}^j$:
 $$\mathbf{Z}^{\mathrm{T}}\mathbf{Z}\boldsymbol{\alpha}^j=\lambda_j\boldsymbol{\alpha}^j $$
 令$\mathbf{Z}^{\mathrm{T}}\mathbf{Z}=\mathbf{K}$,那么上式可化为
 $$\mathbf{K}\boldsymbol{\alpha}^j=\lambda_j\boldsymbol{\alpha}^j $$