我们继续进行简化,时刻记住要将邻接矩 A AA阵和度矩阵 D DD凑进去:
( A 1 X A 2 X ⋮ A N X ) = A N × N X N × 1 = A X \begin{pmatrix} A_1 X \\ A_2 X \\ \vdots \\ A_N X \end{pmatrix} = A^{N\times N} X^{N\times 1} = AX⎝⎜⎜⎜⎛A1XA2X⋮ANX⎠⎟⎟⎟⎞=AN×NXN×1=AX
( x 1 D 1 x 2 D 2 ⋮ x i D N ) = ( D 1 0 0 ⋯ 0 0 D 2 0 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ D N ) N × N X N × 1 = D N × N X N × 1 = D X \begin{pmatrix} x_1 D_{1} \\ x_2 D_{2} \\ \vdots \\ x_i D_{N} \end{pmatrix} = \begin{pmatrix} D_1 & 0 & 0 & \cdots & 0 \\ 0 & D_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & D_N\\ \end{pmatrix} ^{N\times N} X^{N\times 1} = D^{N\times N} X^{N \times 1} = DX⎝⎜⎜⎜⎛x1D1x2D2⋮xiDN⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛D10⋮00D2⋮000⋮0⋯⋯⋱⋯00⋮DN⎠⎟⎟⎟⎞N×NXN×1=DN×NXN×1=DX
综上,得到了拉普拉斯算子在图上的等式, L LL被称作为拉普拉斯矩阵
Δ f = D X − A X = ( D − A ) X L = D − A \begin{aligned} \Delta f &= DX - AX = (D - A) X \\ L &= D - A \end{aligned}ΔfL=DX−AX=(D−A)X=D−A

推导-方法2:直接干脆

基于上述推导过程,以及单个节点的定义,我们可以再反过来直接从定义入手,快速推演一遍;
先看原本的定义公式,可以展开,假如节点 i ii的邻域内有 N i N_{i}Ni个节点,即节点的度 D i D_iDi,那么:
Δ f x i = ∑ j ∈ N i ( x i − x j ) = N i ∗ x i − x j 1 − x j 2 ⋯ − x j N = D i x i − x j 1 − x j 2 ⋯ − x j N \begin{aligned} \Delta f_{x_i} &= \sum_{j \in N_i} (x_i - x_j) \\ &= N_{i} * x_i - x_{j1} - x_{j2} \cdots -x_{jN} \\ &= D_i x_i - x_{j1} - x_{j2} \cdots -x_{jN} \end{aligned}Δfxi=j∈Ni∑(xi−xj)=Ni∗xi−xj1−xj2⋯−xjN=Dixi−xj1−xj2⋯−xjN
上式中的 x j 1 , x j 2 , ⋯ , x j N x_{j1}, x_{j2} , \cdots , x_{jN}xj1,xj2,⋯,xjN来自于 X = [ x 1 , x 2 , ⋯ , x N ] T X=[x_1,x_2, \cdots, x_N]^TX=[x1,x2,⋯,xN]T,为了理解方便下列等式方便,只假设节点 1 11的有两个邻居: 2 , N − 1 2,N-12,N−1, 节点 2 22的也有2邻居为 1 , 3 1,31,3,:
Δ f = ( Δ f x 1 Δ f x 2 ⋮ Δ f x N ) = ( ∑ j ∈ N 1 ( x 1 − x j ) ∑ j ∈ N 2 ( x 2 − x j ) ⋮ ∑ j ∈ N N ( x N − x j ) ) = ( D 1 ∗ x 1 − x 2 − x N − 1 D 2 ∗ x 2 − x 1 − x 3 ⋮ D N ∗ x N − x j 1 − x j 2 ⋯ − x j N ) = ( D 1 − 1 0 ⋯ − 1 − 1 D 2 − 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ D N ) ( x 1 x 2 ⋮ x N ) = ( D − A ) X = L X \begin{aligned} \Delta f &= \begin{pmatrix} \Delta f_{x_1} \\ \Delta f_{x_2} \\ \vdots \\ \Delta f_{x_N} \end{pmatrix} = \begin{pmatrix} \sum_{j \in N_{1}} (x_1 - x_j) \\ \sum_{j \in N_{2}} (x_2 - x_j) \\ \vdots \\ \sum_{j \in N_{N}} (x_N - x_j) \end{pmatrix} \\& = \begin{pmatrix} D_1 * x_1 - x_2 - x_{N-1} \\ D_2 * x_2 - x_1 - x_3 \\ \vdots \\ D_N * x_N - x_{j1} - x_{j2} \cdots - x_{jN} \end{pmatrix} \\ &= \begin{pmatrix} D_1 & -1 & 0 & \cdots & -1 \\ -1 & D_2 & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & D_N\\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_{N} \end{pmatrix} \\ &= (D - A) X \\ &= LX \end{aligned}Δf=⎝⎜⎜⎜⎛Δfx1Δfx2⋮ΔfxN⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛∑j∈N1(x1−xj)∑j∈N2(x2−xj)⋮∑j∈NN(xN−xj)⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛D1∗x1−x2−xN−1D2∗x2−x1−x3⋮DN∗xN−xj1−xj2⋯−xjN⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛D1−1⋮0−1D2⋮00−1⋮0⋯⋯⋱⋯−10⋮DN⎠⎟⎟⎟⎞⎝⎜⎜⎜⎛x1x2⋮xN⎠⎟⎟⎟⎞=(D−A)X=LX