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I am reading the paper "Understanding dimensional collapse in contrastive self-supervised learning." The authors identified a dimensional collapse phenomenon:

i.e. some dimension of embedding collapses to zero. They show this by collecting the embedding vectors on the validation set. Each embedding vector has a size of $d=128$, then compute the covariance matrix $C\in\mathbb{R}^{d\times d}$. Then the singular value decomposition is applied on the covariance matrix. They state that a number of singular values collapse to zero, thus representing collapsed dimensions.

Thus my questions are:

  1. What does singular value decomposition of covariance matrix represent?
  2. Why a number of singular value of covariance matrix collapse to zero can represent these dimension of embedding collapse?
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  • $\begingroup$ Does this answer your question? What is the intuitive relationship between SVD and PCA? $\endgroup$
    – Kurt G.
    Apr 20, 2022 at 17:51
  • $\begingroup$ Nope, The paper use the SVD of covariance. $\endgroup$
    – Noel
    Apr 22, 2022 at 0:52
  • $\begingroup$ "The paper use the SVD of covariance". The title of your question is "What does the singular value decomposition of covariance matrix represent?" . I was under the impression that SVD is the abbreviation of singular value decomposition. Please tell me now why exactly your question is not a duplicate? $\endgroup$
    – Kurt G.
    Apr 22, 2022 at 4:19
  • $\begingroup$ The answer said that the eigenvalue decomposition on covariance matrix is equivalent to SVD or original matrix. But I am confused by applying SVD on covariance matrix rather than original matrix. Thanks again. $\endgroup$
    – Noel
    Apr 22, 2022 at 8:10
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    $\begingroup$ Sure SVD is normally applied to the data matrix. Regarding your question: Isn't a PCA (diagonalization of the covariance matrix) not just a special case of an SVD of it ? It is probably time to post a link to the paper you are reading. The images tell me nothing. $\endgroup$
    – Kurt G.
    Apr 22, 2022 at 8:39

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I had the same question and found this to be a good answer that reflected my empirical observations of using the square root of singular values of the covariance matrix as the scale of major axes of variance, invariant to how the data is rotated: http://www.cs.utah.edu/~tch/CS4640F2019/resources/A%20geometric%20interpretation%20of%20the%20covariance%20matrix.pdf

Keep in mind that because all covariance matrix are symmetric and positive semi-definite, their singular values are the same as their eigenvalues. So you don't actually need to compute the SVD and can just directly compute the eigenvalues if you are interested in a rotation invariant measure of scale. As for $U$ and $V$, You can think of $V$ as rotating such that the axes are aligned to the major axes of variance, then we apply the eigenvalues / singular values to scale them. Because the covariance matrix is symmetric and positive semi-definite, $U = V$. I don't have a proof for it, but at least in this case, $U$ and $V$ appear to be involutory (their own inverse). So $U$ rotates back to the original space which isn't aligned on the major axes of variance.

Keep in mind my background is in engineering, so apologies for any lack of rigor here, especially in the second paragraph.

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