RE: [R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-08 Thread Liaw, Andy
There *is* a Matlab newsgroup for you to ask Matlab questions. From the latest Matlab digest: MATLAB Usenet Group Celebrates Its 10th Anniversary The MATLAB Usenet group, comp.soft-sys.matlab (CSSM), celebrated its 10th anniversary this month. CSSM is a collaboration space where thousands of

Re: [R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-07 Thread antonio rodriguez
Hi Feng, AFIK SVD analysis provides a one-step method for computing all the components of the eigen value problem, without the need to compute and store big covariance matrices. And also the resulting decomposition is computationally more stable and robust. Cheers, Antonio Rodriguez -

Re: [R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-07 Thread Feng Zhang
Thanks for those replies. But I tested several cases, and found the two percentage from SVD and EVD are not the same. So how to explain the difference and which one should be the right one for use in PCA? - Original Message - From: antonio rodriguez [EMAIL PROTECTED] To: Feng Zhang

Re: [R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-07 Thread Feng Zhang
I used Matlab to do this case study. x = randn(200,3); %%generating a 200x3 Gaussian matrix [a,b,c]=svd(x); %%SVD composition S=diag(b) S =[15.6765 14.8674 13.4016]' S(1)^2/sum(S.^2); 0.3802 ZeroedX = X - repmat(mean(X),200,1); %%ZeroedX is now zero centered data C = cov(ZeroedX);

[R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-06 Thread Feng Zhang
Hey, All In principal component analysis (PCA), we want to know how many percentage the first principal component explain the total variances among the data. Assume the data matrix X is zero-meaned, and I used the following procedures: C = covriance(X) %% calculate the covariance matrix;

RE: [R] Confused by SVD and Eigenvector Decomposition in PCA

2003-02-06 Thread Liaw, Andy
If I'm not mistaken, for positive semi-definite matrices, the eigenvalues are equal to squared singular values, so you should get the same answer either way. The code you shown is definitely not R (looks like Matlab), so why are you posting to R-help? Andy -Original Message- From: