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I ask this because from my understanding, they both are same, because they both denote variance along most important dynamics. So their corresponding values should be the same, but as clearly seen in matlab this is not the case.

why latent values returned by pca matlab are not singular values ?

user3086871
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    This http://stats.stackexchange.com/q/134282/3277 and other threads (search `PCA svd`) explain that topic. See also dense but important http://stats.stackexchange.com/q/141754/3277. Singular values are the sq. roots of the eigenvalues. In PCA, the data is typically column-centered; besides, it is typically normalized by sqrt(n) or sqrt(n-1) where n is the number of rows. For convenience+reasons: that corresponds to the eigendecomposition of the covariance matrix. – ttnphns Sep 22 '16 at 07:08
  • You could make sure that the two sets of singular values, - or, rather, eigenvalues, are linearly related, they are "same" in a sense. – ttnphns Sep 22 '16 at 07:11
  • @ttnphns +1 to your first comment, but the second one might be misleading to somebody because singular values of X and eigenvalues of cov(X) are not "linearly related"; they are quadratically related, as you know. – amoeba Sep 22 '16 at 09:53
  • @amoeba, by "two sets of eigenvalues" (or sing. values) I just meant one from X and the other from X/sqrt(n). They are proportionally same by factor n. Sorry if l wasn't clear. – ttnphns Sep 22 '16 at 10:43
  • **In the [duplicate post](http://stats.stackexchange.com/q/134282) point #3 answers your question. See also #5.** – amoeba Sep 22 '16 at 12:43

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