# PCA, test for eigen values

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 $$\hat{v}_i = \text{eigenvectors}\left(\widehat{\text{cov}}(X)^T\right)(X_i-\bar{X})$$?

• 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.
– whuber
Oct 22, 2021 at 15:16
• @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"? Oct 22, 2021 at 20:47
• Any orthogonal frame could be the principal components.
– whuber
Oct 22, 2021 at 21:27
• 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.
– whuber
Oct 22, 2021 at 23:29
• 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."
– whuber
Oct 25, 2021 at 13:33