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Consider a finite point set $X$ in $\mathbb{R}^n$ and a $1$-Lipschitz map $\,f:\mathbb{R}^n \to \mathbb{R}^m$.

Is it true that the maximum eigenvalue of the centered covariance matrix of $f(X)$ is at most the maximum eigenvalue of the centered covariance matrix of $X$?

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  • $\begingroup$ Do I understand it right that $f$ is Lipschitz in the entire space, $f(0)=0$ and by the covariance matrix of the set $X_j$ you mean the matrix with entries $a_{i,j}=\langle X_i,X_j\rangle$? $\endgroup$
    – fedja
    Nov 18, 2017 at 1:16
  • $\begingroup$ $f$ is globally Lipschitz yes, but the covariance is the centered covariance (so no need to fix the image of the origin). edited the text. Also the matrix you describe is the Gram matrix, not the covariance, but it has the same non zero eigenvalues as the non centered covariance matrix $\endgroup$
    – alesia
    Nov 18, 2017 at 1:24
  • $\begingroup$ Just meaning you subtract the center of mass before taking scalar products? (sorry for stupid questions; I just think that to ask is easier than to guess or to google :-) ) $\endgroup$
    – fedja
    Nov 18, 2017 at 1:28
  • $\begingroup$ yes, that's what I mean by centering. Then you can take scalar products instead of covariance if you prefer $\endgroup$
    – alesia
    Nov 18, 2017 at 1:30
  • $\begingroup$ Then the answer is "no". Let $e_j$ be some orthonormal set ($j=1,\dots,n$).Take all pairs $e_j,-e_j$ and $2n$ zero vectors. Define $f(x)=\|x\|$ (mapping to $\mathbb R^1$). If $n$ is not too small, you are in trouble. $\endgroup$
    – fedja
    Nov 18, 2017 at 1:39

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