$\ell^p$ version of singular values

Question

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

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 $s_k(M)=s_k(M^T)$. 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 though it is true for $p=2$ (take $G=F$) and for $k=1$ (Hahn-Banach).

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.