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 value13 dimension = 3;14 vectors = 3;15 16 %% In case of unpredicted problems, dont change the following code.17 if vectors > dimension18 disp(‘Please set vectors <= dimension.‘);19 end 20 X = randn(dimension,vectors); % every column is a vector21 X = X./repmat(sqrt(sum(X.^2,1)),dimension,1); % standardize22 if rank(X) ~= size(X,2)23 disp(‘These vectors are not independent. Run again.‘);24 end25 w = (X‘*X)\ones(vectors,1);26 u = X*w;27 w % the coefficient28 u % the equiangular vector29 X‘*u % the correlation value30 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