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In this post, it is stated that

In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $XX^\top$ can cause loss of precision

Why is this the case? Isn't $XX^\top$ just a matrix multiplication, what makes this operation so disastrous?

Rodrigo de Azevedo
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  • This has been asked before [here](https://math.stackexchange.com/questions/359397/why-is-svd-on-x-preferred-to-eigendecomposition-of-xx-top-in-pca); see also [this stats.SE question](https://stats.stackexchange.com/questions/79043/why-pca-of-data-by-means-of-svd-of-the-data). – angryavian Feb 19 '22 at 03:44
  • Because the singular bases are orthogonal while the eigenbasis does not have to be, and so can be arbitrarily close to a non-basis. – Conifold Feb 19 '22 at 04:55
  • Not exactly an answer, but the singular values are Lipschitz continuous but the eigenvalues need not be. See https://mathoverflow.net/q/88132/31729 – copper.hat Feb 19 '22 at 06:19
  • This specific question (not the title) is just about the instability created by multiplying out $A A^T$ or $A^T A$, and in these cases the eigenbasis you want to find is orthogonal anyway. – Ian Feb 19 '22 at 10:16

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