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I am embarassed to pose this question. It is a generalization of a question asked less than 24 hours ago by an unknonw (Google), which has been deleted since then, presumably by its author (her/)himself.

Let $M\in M_n(\mathbb R)$ be given. A version of the question can be written for non-square matrices. Complex entries may be considered as well. Its singular values $s_1\ge s_2\ge \cdots\ge s_n(\ge0)$ are given by the Rayleigh--Weyl formula $$s_k=\inf_{\dim F=n+1-k}\sup_{\quad x\in F|x\ne0} \frac{\|Mx\|_2}{\|x\|_2}=\sup_{\dim F=k}\inf_{\quad x\in F|x\ne0} \frac{\|Mx\|_2}{\|x\|_2},$$

where $\|\cdot\|_p$ stands for the $\ell^p$-norm over $\mathbb R$.

When $p\in[1,+\infty]$, $\ell^p$ version of the singular values, denoted $s_{k,p}(M)$, could be defined the same way, but replacing the $\ell^2$-norm by the $\ell^p$ one. When $p=2$, we know $s_kM)=s_k(M^T)$. Hence the question:

Is there a relation between $s_{k,p}(M)$ and $s_{k,p'}(M^T)$, when $p$ and $p'$ are conjugate exponent ?

A first attempt, unsuccessful, is to pretend that given $F$ or dimension $k$, there exists a $G$ of the same dimension such that for every $x\in F$, one has $$\|x\|_{p}=\sup_{y\in G|y\neq 0}\frac{y^Tx}{\|y\|_{p'}}.$$ Unfortunately, this is false in most cases, even thoug it is true for $p=2$ (take $G=F$) and for $k=1$ (Hahn-Banach).

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Edit: LaTeX. – Andrey Rekalo Dec 22 '10 at 15:26
Look at books on Banach space theory that have information on s-numbers (Tomczak-Jaegermann, Pietsch, Diestel-Jarchow-Tonge, and HERMANN KONIG Eigenvalue distribution of compact operators). Sorry I can't give more information from where I am now. – Bill Johnson Dec 22 '10 at 18:41
@Bill. Thank you, I'll have a look. Your last sentence looks that one used by Jean Leray in his papers written during the time he was a war prisonner. – Denis Serre Dec 22 '10 at 18:44
Leray was imprisoned on Maui, Denis?!?! :) – Bill Johnson Dec 22 '10 at 19:33
No, he was in Austria. An interesting tale, because he switched to pure maths (algebraic topology), to be sure that his science could not be used by the german army. Sure, he would have prefered to be in Maui. But if he had been, this would have meant that Japan and Germany had won the war. – Denis Serre Dec 22 '10 at 20:03

Here is a reference in the spirit of Bill Johnson's comment. The book

A. Pietsch, Operator Ideals, North-Holland, Amsterdam, 1980

contains a chapter entitled "s-Numbers of Operators on Banach Spaces". The constructions are too complicated to review here.

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Sorry, this does not answer the original question. But the remarks below might still be relevant. So here goes.

  1. The cases $p=1$ and $p=\infty$ might be the only "useful" (numerically) cases beyond $p=2$, because afaik, computing these "generalized" singular values for $p \neq 1, 2, \infty$ will be NP-Hard (e.g., see this preprint, which is also avail in a journal now)

  2. This preprint on arXiv (which actually considers a $\|Mx\|_p / \|x\|_q$) might also be interesting (it also mentions very nice connections to related work.

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It might be more useful to pose the problem as follows. Let $X = (\mathbb{R}^n, \| \cdot \| _p )$ and $X^\ast = (\mathbb{R}^n, \| \cdot \| _{p^\ast})$, where $p^\ast$ is the conjugate exponent to $p$. Rather than considering $M$ as a map from $ X \to X $, it may be more useful to treat it as $M \colon X \to X^\ast $. (Of course, when $p=p^{*}=2$, these are the same.) In that case, one can make sense of the compositions $M^\ast M \colon X \to X$ and $M M^\ast \colon X ^\ast \to X^\ast $, and take the singular values as the square root of the eigenvalues of these maps.

EDIT: This is equivalent to looking at $\| Mx \| _{p^\ast} / \| x \|_p $ instead of $\| Mx \| _p / \| x \|_p $, so it ties into the work that Suvrit mentioned in his response.

EDIT 2: Sorry, I made a stupid mistake in the struck-out sentence above. Of course, if $ M \colon X \to X^\ast $, then we again have $ M^\ast \colon X \to X^\ast $ -- not $X^\ast \to X$ as I had written above. Ultimately, you may have to resort to the fact that $\ell^p$ is isomorphic to $\ell^2$ (since $n$ is finite), so one can map between $X$ and $X^\ast$ -- but this has gotten sufficiently far from my original answer that I'll just stop at that.

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Singular values of $A: X \to X$ are the eigenvalues of $A^{*} A$, which implies $$ A v = \sum_{n} s_n e_n(v) f_n $$ for some bases $e_n \in X^{*} $, $f_n \in X$. Independence on $p$ is then obvious. At least for reflexive Banach spaces.

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$A^*A$ makes sense only on a Hilbert space, Helge. – Bill Johnson Dec 22 '10 at 19:32

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