热转换不会给出预期的结果

Question

我想将hotelling转换应用于给予矢量并使自己练习自我，这就是为什么我在matlab中写下了遵循代码

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

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

肯定这不是对角线矩阵，那怎么了？预先感谢

Answer 1

当您处理行矢量而不是 column 向量时，您需要在eigenvalue/eigenvalue/eigenvector-decomposiiton中对其进行调整。而不是Y=A1*centered，您需要Y=centered*V。那你会得到

covariance_matrix =
    0.0000   -0.0000    0.0000
   -0.0000    1.5644   -0.0000
    0.0000   -0.0000  207.1022

因此，您将获得两个非零组件，这就是3D空间中仅三个点所能期望的。（它们只能形成平面，但不能形成音量。）