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Let's say I have a SVD of a matrix $A = U \Sigma V^T$, $A \in \mathbb{R}^{n \times m}$, and I'm using top-k components corresponding to $\sigma_1, ..\sigma_k$, the k largest values on the diagonal of $\Sigma$. They are also square roots of k largest eigenvalues of $A^T A$.

I encountered the sum of $$\frac{\sum_1^k \sigma_i^2}{\sum_1^n \sigma_i^2}$$ being interpreted as "preserved variability". This makes sense if $A^T A$ were the variance - but it's not, not unless the mean along the rows of A is zero. What am I missing here? Is there an inherent assumption that SVD is only applied to "normalized" $A$?

Sycorax
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Hicjo
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  • Does this answer your question? [Relationship between SVD and PCA. How to use SVD to perform PCA?](https://stats.stackexchange.com/questions/134282/relationship-between-svd-and-pca-how-to-use-svd-to-perform-pca) see amoeba's answer – Aksakal May 28 '20 at 21:37
  • sort of! it opens with "Let us assume that it is centered" :) If that is indeed the default assumption always, I guess the answer to my question is "$A$ is assumed to be centered" – Hicjo May 28 '20 at 21:43
  • see #5 in amoeba's summary – Aksakal May 28 '20 at 21:44
  • do you know if any useful interpretation exists if A is not centered? – Hicjo May 28 '20 at 21:50
  • interpretation of what? – Aksakal May 28 '20 at 21:53
  • $\frac{\sum_1^k \sigma_i^2}{\sum_1^n \sigma_i^2}$ – Hicjo May 28 '20 at 21:56
  • I'm going to ask that as a separate question - thanks for this :) – Hicjo May 28 '20 at 23:11

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