Let's say I have a matrix $X$ with $4$ variables and $n$ observations and I run PCA, so I compute the eigenvalues/eigenvectors of the sample covariance matrix, and then I test $H_0: \lambda_1=\lambda_2=\lambda_3=\lambda_4$, and I find that I don't reject $H_0$. So I don't have evidence to say that there's a pair of different eigenvalues. What can I say about the estimation of the principal components $\hat{v}_i = \text{eigenvectors}\left(\widehat{\text{cov}}(X)^T\right)(X_i-\bar{X})$?

  • $\begingroup$ A lot can be said--but do you want an answer that utilizes only the stated information? In that case, the components are completely indeterminate. $\endgroup$
    – whuber
    Oct 22, 2021 at 15:16
  • $\begingroup$ @whuber Yes, I will like to draw a conclusion only based on the fact that I don't reject the test. What do you mean by "completely indeterminate"? $\endgroup$
    – Ejrionm
    Oct 22, 2021 at 20:47
  • $\begingroup$ Any orthogonal frame could be the principal components. $\endgroup$
    – whuber
    Oct 22, 2021 at 21:27
  • 2
    $\begingroup$ When you cannot reject the hypothesis that all eigenvalues are degenerate, that is tantamount to saying any orthogonal basis qualifies as a series of principal components. To put it another way, you are interpreting the specific PCs that were found as possibly resulting purely from noise rather than reflecting any underlying structure. $\endgroup$
    – whuber
    Oct 22, 2021 at 23:29
  • 1
    $\begingroup$ I don't want it to seem like there's some obscure insight here: I'm just applying basic linear algebra. Good search words are "degenerate eigenvalues." $\endgroup$
    – whuber
    Oct 25, 2021 at 13:33


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