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Recently I've been interested in applying PCA to a dataset I have and I wanted to develop a deep understanding of what I would actually be doing when I implement it.

Today I encountered two confronting answers to the question of what is the maximum number of principal components. The two answers are these ones:

Do any of you know what is the meaning of that extra component that sklearn's PCA is offering?

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    $\begingroup$ If the number of samples $n$ is less than or equal to the number of features, the $n$-th PC will be constant zero (eigenvalue = 0). This is what sklearn will presumably return. The number of non-trivial PCs is $n-1$ as per the linked answer. $\endgroup$
    – amoeba
    Mar 16, 2018 at 16:27
  • $\begingroup$ Are you doing PCA with or without centering? $\endgroup$
    – whuber ♦
    Mar 16, 2018 at 16:31
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    $\begingroup$ @whuber I don't think sklearn.decomposition.PCA can do PCA without centering. I don't see such an option in the documentation. $\endgroup$
    – amoeba
    Mar 16, 2018 at 16:41
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    $\begingroup$ @amoeba Completely on point! I just checked what is the n-th principal component of my data and it is always 0! Problem solved :) $\endgroup$
    – Marcos Galletero Romero
    Mar 16, 2018 at 17:37
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    $\begingroup$ If you use scikit's PCA then it does centering for you. $\endgroup$
    – amoeba
    Mar 16, 2018 at 22:12

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Per @amoeba's comments:

If the number of samples $n$ is less than or equal to the number of features, the $n$-th PC will be constant zero (eigenvalue = 0). This is what sklearn will presumably return. The number of non-trivial PCs is $n−1$ as per the linked answer.

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  • $\begingroup$ Could you please explain why this is so? $\endgroup$
    – Sos
    Feb 29, 2020 at 10:01
  • $\begingroup$ Because the rank of PC's matrix should be less than or equal to n. Recall the linear independence. $\endgroup$
    – kurtkim
    Jun 23, 2020 at 17:54

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