調和矩陣

在圖論中，調和矩陣（harmonic matrix），也稱拉普拉斯矩陣或拉氏矩陣（Laplacian matrix）、離散拉普拉斯（discrete Laplacian），是圖的矩陣表示。^[1]

調和矩陣也是拉普拉斯算子的離散化。換句話說，調和矩陣的縮放極限是拉普拉斯算子。它在機器學習和物理學中有很多應用。

定義

若G是簡單圖，G有n個頂點，A是鄰接矩陣，D是度數矩陣，則調和矩陣是^[1]

${\displaystyle L=D-A}$

${\displaystyle L_{i,j}:={\begin{cases}\deg(v_{i})&{\mbox{if}}\ i=j\\-1&{\mbox{if}}\ i\neq j\ {\mbox{and}}\ v_{i}{\mbox{ is adjacent to }}v_{j}\\0&{\mbox{otherwise}}\end{cases}}}$

動機

這跟拉普拉斯算子有什麼關係？若f 是加權圖G的頂點函數，則^[2]

${\displaystyle E(f)=\sum w(uv)(f(u)-f(v))^{2}/2=f^{t}Lf}$

w是邊的權重函數。u、v是頂點。f = (f(1), ..., f(n)) 是n維的矢量。上面泛函也稱為Dirichlet泛函。^[3]

接續矩陣

而且若K是接續矩陣（incidence matrix），則^[2]

${\displaystyle K_{ev}=\left\{{\begin{array}{rl}1,&{\text{if }}v=i\\-1,&{\text{if }}v=j\\0,&{\text{otherwise}}.\end{array}}\right.}$

${\displaystyle L=K^{t}K}$

Kf 是f 的圖梯度。另外，特徵值滿足

${\displaystyle {\begin{aligned}\lambda _{i}&=\mathbf {v} _{i}^{\textsf {T}}L\mathbf {v} _{i}\\&=\mathbf {v} _{i}^{\textsf {T}}M^{\textsf {T}}M\mathbf {v} _{i}\\&=\left(M\mathbf {v} _{i}\right)^{\textsf {T}}\left(M\mathbf {v} _{i}\right)\\\end{aligned}}}$

舉例

圖	度矩陣	鄰接矩陣	調和矩陣
	${\textstyle \left({\begin{array}{rrrrrr}2&0&0&0&0&0\\0&3&0&0&0&0\\0&0&2&0&0&0\\0&0&0&3&0&0\\0&0&0&0&3&0\\0&0&0&0&0&1\\\end{array}}\right)}$	${\textstyle \left({\begin{array}{rrrrrr}0&1&0&0&1&0\\1&0&1&0&1&0\\0&1&0&1&0&0\\0&0&1&0&1&1\\1&1&0&1&0&0\\0&0&0&1&0&0\\\end{array}}\right)}$	${\textstyle \left({\begin{array}{rrrrrr}2&-1&0&0&-1&0\\-1&3&-1&0&-1&0\\0&-1&2&-1&0&0\\0&0&-1&3&-1&-1\\-1&-1&0&-1&3&0\\0&0&0&-1&0&1\\\end{array}}\right)}$

其他形式

對稱正規化調和矩陣

${\displaystyle L^{\text{sym}}:=D^{-{\frac {1}{2}}}LD^{-{\frac {1}{2}}}=I-D^{-{\frac {1}{2}}}AD^{-{\frac {1}{2}}}}$

${\displaystyle L_{i,j}^{\text{sym}}:={\begin{cases}1&{\mbox{if }}i=j{\mbox{ and }}\deg(v_{i})\neq 0\\-{\frac {1}{\sqrt {\deg(v_{i})\deg(v_{j})}}}&{\mbox{if }}i\neq j{\mbox{ and }}v_{i}{\mbox{ is adjacent to }}v_{j}\\0&{\mbox{otherwise}}.\end{cases}}}$

注意^[4]

${\displaystyle \lambda \ =\ {\frac {\langle g,L^{\text{sym}}g\rangle }{\langle g,g\rangle }}\ =\ {\frac {\left\langle g,D^{-{\frac {1}{2}}}LD^{-{\frac {1}{2}}}g\right\rangle }{\langle g,g\rangle }}\ =\ {\frac {\langle f,Lf\rangle }{\left\langle D^{\frac {1}{2}}f,D^{\frac {1}{2}}f\right\rangle }}\ =\ {\frac {\sum _{u\sim v}(f(u)-f(v))^{2}}{\sum _{v}f(v)^{2}d_{v}}}\ \geq \ 0,}$

隨機漫步

${\displaystyle L^{\text{rw}}:=D^{-1}L=I-D^{-1}A}$

${\displaystyle L_{i,j}^{\text{rw}}:={\begin{cases}1&{\mbox{if }}i=j{\mbox{ and }}\deg(v_{i})\neq 0\\-{\frac {1}{\deg(v_{i})}}&{\mbox{if }}i\neq j{\mbox{ and }}v_{i}{\mbox{ is adjacent to }}v_{j}\\0&{\mbox{otherwise}}.\end{cases}}}$

動力學和微分方程

例如，離散的冷卻定律使用調和矩陣^[5]

${\displaystyle {\begin{aligned}{\frac {d\phi _{i}}{dt}}&=-k\sum _{j}A_{ij}\left(\phi _{i}-\phi _{j}\right)\\&=-k\left(\phi _{i}\sum _{j}A_{ij}-\sum _{j}A_{ij}\phi _{j}\right)\\&=-k\left(\phi _{i}\ \deg(v_{i})-\sum _{j}A_{ij}\phi _{j}\right)\\&=-k\sum _{j}\left(\delta _{ij}\ \deg(v_{i})-A_{ij}\right)\phi _{j}\\&=-k\sum _{j}\left(\ell _{ij}\right)\phi _{j}.\end{aligned}}}$

使用矩陣矢量

${\displaystyle {\begin{aligned}{\frac {d\phi }{dt}}&=-k(D-A)\phi \\&=-kL\phi \end{aligned}}}$

${\displaystyle {\frac {d\phi }{dt}}+kL\phi =0}$

解是

${\displaystyle {\begin{aligned}{\frac {d\left(\sum _{i}c_{i}\mathbf {v} _{i}\right)}{dt}}+kL\left(\sum _{i}c_{i}\mathbf {v} _{i}\right)&=0\\\sum _{i}\left[{\frac {dc_{i}}{dt}}\mathbf {v} _{i}+kc_{i}L\mathbf {v} _{i}\right]&=\\\sum _{i}\left[{\frac {dc_{i}}{dt}}\mathbf {v} _{i}+kc_{i}\lambda _{i}\mathbf {v} _{i}\right]&=\\{\frac {dc_{i}}{dt}}+k\lambda _{i}c_{i}&=0\\\end{aligned}}}$

${\displaystyle c_{i}(t)=c_{i}(0)e^{-k\lambda _{i}t}}$

${\displaystyle c_{i}(0)=\left\langle \phi (0),\mathbf {v} _{i}\right\rangle }$

平衡舉動

當${\textstyle \lim _{t\to \infty }\phi (t)}$的時候，

${\displaystyle \lim _{t\to \infty }e^{-k\lambda _{i}t}=\left\{{\begin{array}{rlr}0&{\text{if}}&\lambda _{i}>0\\1&{\text{if}}&\lambda _{i}=0\end{array}}\right\}}$

${\displaystyle \lim _{t\to \infty }\phi (t)=\left\langle c(0),\mathbf {v^{1}} \right\rangle \mathbf {v^{1}} }$

${\displaystyle \mathbf {v^{1}} ={\frac {1}{\sqrt {N}}}[1,1,...,1]}$

${\displaystyle \lim _{t\to \infty }\phi _{j}(t)={\frac {1}{N}}\sum _{i=1}^{N}c_{i}(0)}$

MATLAB代碼

N = 20;%The number of pixels along a dimension of the image
A = zeros(N, N);%The image
Adj = zeros(N*N, N*N);%The adjacency matrix

%Use 8 neighbors, and fill in the adjacency matrix
dx = [-1, 0, 1, -1, 1, -1, 0, 1];
dy = [-1, -1, -1, 0, 0, 1, 1, 1];
for x = 1:N
   for y = 1:N
       index = (x-1)*N + y;
       for ne = 1:length(dx)
           newx = x + dx(ne);
           newy = y + dy(ne);
           if newx > 0 && newx <= N && newy > 0 && newy <= N
               index2 = (newx-1)*N + newy;
               Adj(index, index2) = 1;
           end
       end
   end
end

%%%BELOW IS THE KEY CODE THAT COMPUTES THE SOLUTION TO THE DIFFERENTIAL
%%%EQUATION
Deg = diag(sum(Adj, 2));%Compute the degree matrix
L = Deg - Adj;%Compute the laplacian matrix in terms of the degree and adjacency matrices
[V, D] = eig(L);%Compute the eigenvalues/vectors of the laplacian matrix
D = diag(D);

%Initial condition (place a few large positive values around and
%make everything else zero)
C0 = zeros(N, N);
C0(2:5, 2:5) = 5;
C0(10:15, 10:15) = 10;
C0(2:5, 8:13) = 7;
C0 = C0(:);

C0V = V'*C0;%Transform the initial condition into the coordinate system 
%of the eigenvectors
for t = 0:0.05:5
   %Loop through times and decay each initial component
   Phi = C0V.*exp(-D*t);%Exponential decay for each component
   Phi = V*Phi;%Transform from eigenvector coordinate system to original coordinate system
   Phi = reshape(Phi, N, N);
   %Display the results and write to GIF file
   imagesc(Phi);
   caxis([0, 10]);
   title(sprintf('Diffusion t = %3f', t));
   frame = getframe(1);
   im = frame2im(frame);
   [imind, cm] = rgb2ind(im, 256);
   if t == 0
      imwrite(imind, cm, 'out.gif', 'gif', 'Loopcount', inf, 'DelayTime', 0.1); 
   else
      imwrite(imind, cm, 'out.gif', 'gif', 'WriteMode', 'append', 'DelayTime', 0.1);
   end
end

GIF：離散拉普拉斯過程，使用拉普拉斯矩陣

應用

• Kirchoff定理：調和矩陣的行列式
• 人工智慧
• 非線性降維
• 譜聚類
• Cheeger不等式：調和矩陣的第二大的特徵值跟最大割問題有關
• 圖Signal processing^[3]