Question

我想将HOTELLING TRANSFORMATION应用于给定向量并使我的自我练习，这就是为什么我在matlab中编写了以下代码

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);

然后我测试了它给定的矩阵

A

A =

     4     6    10
     3    10    13
    -2    -6    -8

并在运行代码

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

绝对不是对角矩阵，所以有什么不对？提前谢谢

Answer 1

当您处理行向量而不是列向量时，您需要在特征值/特征向量分解中对其进行调整。而不是Y=A1*centered您需要Y=centered*V。然后你会得到

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，这是您在3D空间中仅有三个点所期望的。（它们只能形成一个平面，但不能形成一个体积。）

hotteling转换不会产生预期的结果

1 个答案: