价值250亿美元的特征向量——Google背后的线性代数

原文标题:THE nk, 1\le k\le nxkkx_kx_j\gt x_kjkx_j=0jx_kkx_1=2,x_2=1,x_3=3,x_4=23142kk141334214jj1x_1=x_3+x_4x_3x_4x_1x_3=x_1+x_2+x_4jn_jkkx_j/n_jx_jnL_k\subset{1,2,\cdots,n}kkx_k=\sum{j\in Lk}\frac{x_j}{n_j}\tag{2.1}n_jjj\in L_kn_jjk1343142x_1=x_3/1+x_4/2x_2=x_1/3, x_3=x_1/3+x_2/2+x_4/2, x_4=x_1/3+x_2/2\begin{split}x_1 &= x_3/1+x_4/2\x_2 &= x_1/3 \ x_3&=x_1/3+x_2/2+x_4/2\ x_4&=x_1/3+x_2/2\end{split}\pmb{x}=\begin{bmatrix}x_1\x_2\x_3\x_4\end{bmatrix}\pmb{Ax}=\pmb{x}\pmb{A}\pmb{A}=\begin{bmatrix}0&0&1&\frac{1}{2}\\frac{1}{3}&0&0&0\\frac{1}{3}&\frac{1}{2}&0&\frac{1}{2}\\frac{1}{3}&\frac{1}{2}&0&0\end{bmatrix}\tag{2.2}\pmb{A}\pmb{x}1\pmb{A}\pmb{A}1nji\pmb{A}A{ij}=1/njA{ij}=0\pmb{A}j1/n_j011\pmb{A}1\pmb{A}n\times n\pmb{e}n1\pmb{A}^T\pmb{e}=\pmb{e}1\pmb{A}^T\pmb{A}\pmb{A}^T\pmb{A}1V_1(\pmb{A})\pmb{A}1V_1(\pmb{A})1\pmb{x}\sum_ix_i=1\pmb{A}W_112W_23、4、5\pmb{A}=\begin{bmatrix}0&1&0&0&0\1&0&0&0&0\0&0&0&1&\frac{1}{2}\0&0&1&0&\frac{1}{2}\0&0&0&0&0\end{bmatrix}V_1(\pmb{A})2\pmb{x}=\begin{bmatrix}\frac{1}{2}\\frac{1}{2}\0\0\0\end{bmatrix},\pmb{y}=\begin{bmatrix}0\0\\frac{1}{2}\\frac{1}{2}\0\end{bmatrix}V_1(\pmb{A})\frac{3}{4}\pmb{x}+\frac{1}{4}\pmb{y}=\begin{bmatrix}\frac{3}{8}\\frac{3}{8}\\frac{1}{8}\\frac{1}{8}\0\end{bmatrix}\dim(V_1(\pmb{A}))\gt1WrW_1,W_2,\cdots,W_r\dim(V_1(\pmb{A}))\ge r\pmb{x}\in V_1(\pmb{A})\sum_ix_i=1WrW_1,W_2,\cdots,W_r$$ ,那么将无法找到一个可以比较一个子网和另外一个子网中页面的页面分数。

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作者: 老齐
链接: http://math.itdiffer.com/google.html
来源: 老齐教室-机器学习数学基础
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