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假设这里有一组向量$\left\{ x_i \right\}_{i=1}^n$和一个待投影的向量$u$。假设$u$和每个$x_i$的内积都为正数，也就是说$u$和每个$x_i$的夹角都小于90度。那么当我们把$u$投影到$\left\{ x_i \right\}_{i=1}^n$上时，理所应当地每个$x_i$的系数$\beta_i$也都应该大于零：$$u = x_1\beta_1+\cdots+x_n\beta_n,\beta_i\geq0$$不知道读者们的空间直觉怎么样，反正我最开始就是这么天真的认为的。最近看了Efron的“Least Angle Regression”后，才明白原来不是这样的，自己以前too young了。有些时候系数会变成负的。下面贴一个小MATLAB代码，自己去体会吧！
1 %{ 2 % This small matlab program show you a unexpected result in 3 % high-dimensional geometry: for any set of n-dimensional vectors 4 % {x_1,...,x_n}, if these vectors are indepentent, then you can always find 5 % an equiangular vector u, so that the inner product (x_i,u)=1 for all i. 6 % BUT, if we project u into {x_i}, some coefficients may be ‘negative‘! 7 %} 8 9 %% 10 clear; 11 close all; 12 %% You can change the following two variable‘s value 13 dimension = 3; 14 vectors = 3; 15 16 %% In case of unpredicted problems, dont change the following code. 17 if vectors > dimension 18 disp(‘Please set vectors <= dimension.‘); 19 end 20 X = randn(dimension,vectors); % every column is a vector 21 X = X./repmat(sqrt(sum(X.^2,1)),dimension,1); % standardize 22 if rank(X) ~= size(X,2) 23 disp(‘These vectors are not independent. Run again.‘); 24 end 25 w = (X‘*X)\ones(vectors,1); 26 u = X*w; 27 w % the coefficient 28 u % the equiangular vector 29 X‘*u % the correlation value 30 if dimension == 3 % ‘3‘ is the upper bound of dimensions of humans where we can draw. 31 quiver3(0,0,0,X(1,1),X(2,1),X(3,1),‘r‘),hold on; 32 quiver3(0,0,0,X(1,2),X(2,2),X(3,2),‘r‘),hold on; 33 quiver3(0,0,0,X(1,3),X(2,3),X(3,3),‘r‘),hold on; 34 quiver3(0,0,0,u(1),u(2),u(3),‘b‘),hold off; 35 end
总结一句话：高维空间有危险，忽久留= =||
PS：ta今天一共和我说了30个字。
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原文地址：http://www.cnblogs.com/mashiqi/p/4226692.html
