Let's assume I'm using principal component analysis to carry out clustering of a 2-d data set, using a non-normalized covariance matrix to carry out the operation. I then solve for the eigenvalues, and sort the list of eigenvalues in ascending order (i.e. highest absolute valued eigenvalues are first/top in my list).

Then, I calculate the eigenvectors associated with each eigenvalue. In a simple case with a single cluster of data from a 2-d data set, I expect the two vectors to give me the primary and secondary orientations of an ellipse. Using the centroid of the ellipse, and the orientation, I can generate an ellipse that can represent the data set in a general sense. However, I currently can only use the "mass" of the data set (i.e. sum of weighted data points) to estimate the size/area of the ellipse. I would like to be able to determine the dimensional variance in each dimension, so I could create "confidence ellipses" (i.e. an ellipse that encompasses 68%/1x-std.dev. of all data, another that encompasses 95%/2x-std.dev. of all data, etc.). I then plan to extend this to higher-dimensional data (i.e. 3-d, 4-d, 10-d, etc.).


  1. Does the magnitude of each eigenvector correlate with its respective eigenvalue (i.e. are the two proportional)?
  2. Does the magnitude of each eigenvector provide any additional information directly? (i.e. dimensional variance/standard-deviation)? If not, is there a normalization/calculation that can derive dimensional variance from the eigenvalue and/or eigenvector magnitude?

Thank you.


No, the magnitude of eigenvectors does not mean anything, only the direction matters. I think usually we denote the eigenvectors to be the one with length 1 by default.

If $\vec{v}$ is the eigenvector of eigenvalue $\lambda$, it is easy to see that $5\vec{v}$, $-\sqrt{2}\vec{v}$, $10^{28}\vec{v}$ etc. are all eigenvectors of eigenvalue $\lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.