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Conceptually, what would be the meaning of performing PCA on the $M'$ of initial matrix $M=[n \times m]$ where $n$ rows refer to observations and $m$ columns refer to variables?

The problem

Perform PCA for $N=10$ observations of a network/graph $G_n=[p \times p]$. PCA has been performed on $M'$ where $M= [N \times p^2]$, where each $G_n$ has been put into $M$ by vectorisation.

[U S V] = svd(cov(M'))
PC = U'*M

By convention the columns indicate different variables, and the rows the different observations but I performed PCA on the transposed matrix and interesting results came up. Are these by any chance interpretable, or just meaningless?

kjetil b halvorsen
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hH1sG0n3
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  • Did you center your data matrix $M$? If so, how? – amoeba Mar 30 '16 at 11:46
  • Thanks for your comment. Yes, by subtracting the mean of each $M_n$, where $M_n = [v_1, v_2, ... , v_p]$ the vectorised $G_n$ by $M_n=G_n(:)$; – hH1sG0n3 Mar 30 '16 at 11:53
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    You might want to add this to the question because the issue of centering is crucial here. You are centering rows (whereas usually in PCA you would be centering columns). – amoeba Mar 30 '16 at 15:41
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    You may want to check https://stats.stackexchange.com/a/7144/77888. Also, Google "PCA transpose trick". – bers Feb 06 '19 at 14:52

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