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It's easy to show that orthogonal/unitary matrices preserve the $L_2$ norm of a vector, but if I want a transformation that preserves the $L_1$ norm, what can I deduce about the matrices that do this? I feel like it should be something like the columns sum to 1, but I can't manage to prove it.

EDIT:

To be more explicit, I'm looking at stochastic transition matrices that act on vectors that represent probability distributions, i.e. vectors whose elements are positive and sum to 1. For instance, the matrix

$$ M = \left(\begin{array}{ccc}1 & 1/4 & 0 \\0 & 1/2 & 0 \\0 & 1/4 & 1\end{array}\right) $$

acting on

$$ x=\left(\begin{array}{c}0 \\1 \\0\end{array}\right) $$

gives $$ M \cdot x = \left(\begin{array}{c}1/4 \\1/2 \\1/4\end{array}\right)\:, $$ a vector whose elements also sum to 1.

So I suppose the set of vectors whose isometries I care about is more restricted than the completely general case, which is why I was confused about people saying that permutation matrices were what I was after.

Sooo... given the vectors are positive and have entries that sum to 1, can we say anything more exact about the matrices that preserve this property?

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Look at the extremal points of the unit ball. –  t.b. Apr 6 '12 at 14:51
    
My first guess is that such matrices need to correspond to signed permutations of the coordinates, i.e. a finite group. –  Jyrki Lahtonen Apr 6 '12 at 14:54
    
@Jyrki: your first guess is perfectly right and in fact it holds for all $\ell^p$-spaces (and in an appropriate reformulation also for $L^p[0,1]$) with $p \neq 2$, of course -- this is a theorem due to Banach and Lamperti, see also here. –  t.b. Apr 6 '12 at 14:57
    
Thanks @t.b., the unit ball does it in this case. I will study your link for the general $\ell^p, p\neq 2$. –  Jyrki Lahtonen Apr 6 '12 at 14:59
    
Thanks for your comments -- I saw that thread and another one talking about isometries, but wasn't sure if this was exactly the same question, because the answers for those seemed to talk about permutation matrices, and I'm not sure if that's what I'm after. Sorry if I'm misusing technical terms; I'll edit my question to provide an example of what I'm talking about. –  Rory Apr 6 '12 at 16:03
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up vote 3 down vote accepted

The matrices that preserve the set $P$ of probability vectors are those whose columns are members of $P$. This is obvious since if $x \in P$, $M x$ is a convex combination of the columns of $M$ with coefficients given by the entries of $x$. Each column of $M$ must be in $P$ (take $x$ to be a vector with a single $1$ and all else $0$), and $P$ is a convex set.

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Thanks for your answer. I don't think I quite see why it's obvious though, as I'm not handy with convexity arguments. Can you explain a bit more why each column of $M$ must be in $P$? –  Rory Apr 6 '12 at 16:39
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For example, the first column of $M$ is $M e$ where $e_1 = 1$ and $e_2 = \ldots = e_n = 0$. –  Robert Israel Apr 6 '12 at 16:44
    
Hi Robert. Thanks. I guess I see why that example works, but not why it's necessarily the most general case. Sorry if I'm being obtuse. –  Rory Apr 6 '12 at 16:48
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Another way to put it is that the columns of $P$ are discrete distributions. Multiplying such a matrix by another distribution will yield a mixture of the columns, where the mixture weights sum to 1, hence that mixture must also be a discrete distribution. –  EMS Apr 6 '12 at 17:37
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Once you know each column of $M$ is in $P$, for any $x \in P$ you have $Mx \in P$ because a convex set contains all convex combinations of its members. –  Robert Israel Apr 7 '12 at 0:56
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