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Hello,

I'll state the problem first and than I'll a little bit of motivation.

Lets be given regular matrix $M \in \mathbb{R}^{n\times n}$ and norm $||.||$ in $\mathbb{R}^{n}$. Define $$ U =\{ L\in \mathbb{R}^{n\times n}: \forall x\in \mathbb{R}^{n} \; ||LMx||=||x|| \}$$ (all those "norm fix" matrices for $M$) (sorry I have problems with curly brackets). Now the point is to find the best "norm fixing" matrix. I decided that the best one, call it $\bar{L}$, should satisfy: $$\inf_{L\in U} \; \; \sup_{||x||=1} \; ||LMx-Mx|| \; \; = \; \; \sup_{||x||=1} \; ||\bar{L}Mx-Mx||$$

The problem is to find the matrix $\bar{L}$ explicitly, not sure if $\bar{L}$ is unique but it exists. I'm most interested for p-norm with p equal 1 or 2.

Motivation: I was simulating some physical phenomena on computer. And the final equation basically boiled down to $x_{n+1} = Ax_n$. Often $x$ represents some quantity which is conserved. So I came up with this idea how to fix existing numerical scheme to conservative one (with least damage possible)

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  • $\begingroup$ Edited your Latex, mostly on MO one needs double backslashes in \\{ and \\}, also put in spacing. Does the phrase "regular matrix" refer to a restriction of some kind? If your $M$ is invertible, and you are using $p=2,$ then all $L = O M^{-1}$ with $O \in O_n.$ $\endgroup$
    – Will Jagy
    Commented Jan 30, 2012 at 22:33
  • $\begingroup$ Meanwhile, the symmetry group of the "unit sphere" with $p=1$ or $p=\infty$ is finite. $\endgroup$
    – Will Jagy
    Commented Jan 30, 2012 at 22:36
  • $\begingroup$ Will: With regular matrix I meant invertible matrix(sorry i have still gaps in English terminology) and thanks for reply Survit: Why would I need ordering of $U$? $\inf_{L∈U}f(L)=\sup_{L∈U}−f(L)$ where $f$ is some function defined on $U$ $\endgroup$
    – Tomas Skrivan
    Commented Jan 31, 2012 at 11:25
  • $\begingroup$ sorry, i had previously misread your notation! $\endgroup$
    – Suvrit
    Commented Jan 31, 2012 at 13:13
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    $\begingroup$ You should step back and look at the literature on conservative numerical methods for PDE's. By asking your question in this narrow fashion, you're likely to miss out on broader answers. $\endgroup$
    – Brian Borchers
    Commented Mar 10, 2012 at 14:21

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