With and ordination, such as a PCA or CCA a large dataset with multiple species, sites or variables can be explored. The variables that explain most of the variation to PC or CA 1 and 2 can be explored in relation to the species or sites. This often explained as: species 1 or variable 1 has the highest eigenvalue (1.32) and explains most of the variation. However, this does not display how the eigenvalue relates to all the other variables in the ordination, the eigenvalue could just as well be 54.4. Is it possible to express these values in percentage as explained below?

For example, I have a dataset with 33 beetle genera (for simplicity only the first 5 genera are used) and six variables. I apply a CCA on the dataset and CA 1 explains 26% and CA 2 23% of the variation, thereof 49% of the variation is explained by CA1 and 2.

My first question is: Can I describe the variation between genera as a percentage, by expressing all eigenvalues (Table 1) as absolute values and dividing it by the sum of all absolute values under CCA 1 or 2 (Table 2): abs(Genus i)/sum(CCA i)*100%? If this this is possible, then could I say that Genus 3 is responsible of 4.90% of the 26% explained by CCA1 (0.049*26% = 1.3%). This, to me, seems more usable and clear than saying the eigenvalue is -0.24. It directly explains that this is higher than an equal distributed variation, since 1 out of the 33 genera explain 1/33*100% = 3%.

The second question: can the same principle – as described above – be applied on the eigenvalues of the variables (Table 3)? Then we can say that variable 4 creates the largest variation in the direction of CCA 1.

Third and last question: If both the first and second question are valid, can I then say that; the variables describing most of the variation in the direction of CCA 1 have the highest possibility of correlating with genera that have a relative high percentage of the eigenvalues with CCA 1. In this example, genus 3 would have the highest possibility of correlating with variable 4.

Thanks in advance for the effort of reading and answering this question.

Regards,

Wim Kaijser