# Is PCA (by eigendecomposition) or SVD better in decorrelating the predictors of a machine learning model?

Is there any reason to think that SVD is better than PCA (by eigendecomposition) in decorrelating the predictors of a machine learning model?

• I'm leaving this as an answer because I don't have enough reputation here to comment: 1. As elucidated in the accepted answer to this same question over at stats, the question doesn't quite make sense. 2. I recommend checking out this very useful answer that should clear out some further confusions you may have about PCA, SVD and their connection. Oct 25 '18 at 17:05
• @BrianK, thanks for your comment. 1. I find this answer pretty witty but not very serious. To start with, when I am talking about PCA then I am talking about PCA with eigendecomposition. When we are talking in a short way about PCA then we usually mean this. 2. That's a very nice answer it does not directly answer my question. So let's please stay on topic: any specific answer to my question? Oct 26 '18 at 9:10

As a final remark, let’s discuss the numerical advantages of using SVD. A basic approach to actually calculating PCA on a computer would be to perform the eigenvalue decomposition of $$X^TX$$ directly. It turns out that doing so would introduce some potentially serious numerical issues that could be avoided by using SVD.