我想把霍特林变换应用到给定向量中,让我自己练习,这就是为什么我在matlab中编写了下面的代码
function [Y covariance_matrix]=hotteling_trasform(X)
% this function take X1,X2,X3,,Xn as a matrix and apply hottleing
%transformation to get new set of vectors y1, y2,..ym so that covariance
%matrix of matrix consiist by yi vectors are almost diagonal
%% determine size of given matrix
[m n]=size(X);
%% compute mean of columns of given matrix
means=mean(X);
%% substract mean from given matrix
centered=X-repmat(means,m,1);
%% calculate covariance matrix
covariance=(centered'*centered)/(m-1);
%% Apply eigenvector decomposition
[V,D]=eig(covariance);
%% determine dimension of V
[m1 n1]=size(V);
%% arrange matrix so that eigenvectors are as rows,create matrix with size n1 m1
A1=zeros(n1,m1);
for ii=1:n1
A1(ii,:)=V(:,ii);
end
%% applying hoteling transformation
Y=A1*centered; %% because centered matrix is original -means
%% calculate covariance matrix
covariance_matrix=cov(Y);
然后我对给定的矩阵进行了测试
A
A =
4 6 10
3 10 13
-2 -6 -8
在运行代码之后
[Y covariance_matrix]=hotteling_trasform(A);
covariance_matrix
covariance_matrix =
8.9281 22.6780 31.6061
22.6780 66.5189 89.1969
31.6061 89.1969 120.8030
这绝对不是对角线矩阵,所以有什么问题吗?提前感谢
发布于 2016-12-25 19:59:47
在处理行向量而不是列向量时,需要在特征值/特征向量分解中对其进行调整。您需要的不是Y=A1*centered
,而是Y=centered*V
。然后你就会得到
covariance_matrix =
0.0000 -0.0000 0.0000
-0.0000 1.5644 -0.0000
0.0000 -0.0000 207.1022
因此,你会得到两个非零的成分,这是你可以期望的,从三维空间中的三个点。(它们只能形成一个平面,而不能形成一个体积。)
https://stackoverflow.com/questions/41323179
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