# What are the linear isometries on $R^n$, equipped with the $l_1$ norm?

Which conditions must the matrix entries satisfy, and what would be an interpretation of the row and column sums of the matrix?

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## 1 Answer

Hint: Such an isometry would have to map the $l^1$-unit-sphere onto itself. How does this "sphere" look like?

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Hmm, trying some unit vectors of $R^n$ I figure it is necessary that the absolute values of a column must sum up to 1. But is this sufficient? –  Karsten W. Jan 20 '11 at 12:45
The matrix could consist of equal columns and would have not full rank. So would it suffice if I request linear independence of the columns (additionally to the absolute column sum being 1 for each column)? –  Karsten W. Jan 20 '11 at 13:03
@Karsten: to take Christian's hint further. Draw a picture of the "unit sphere" of $\mathbb{R}^2$ with $\ell_1$ norm. You'll notice certain points sticking out at you. Prove that those points are also "special" in the $\mathbb{R}^n$ case. What would happen if a linear transformation carries one of those special points to a non-special point? –  Willie Wong Jan 20 '11 at 14:12
Ok, am I correct if I argue that a linear isometry is characterized by that it takes extreme points to extreme points, and, since a) the extreme points in this case are characterized by $\|x\|_\infty=1$, and b) $\|Ax\|_\infty$ is just the absolute column maximum for one of the columns of A for any extreme point, it follows that A must consists of zeros and ones, with exactly one 1 in each column and have full rank, i.e. A must be a permutation matrix? –  Karsten W. Jan 20 '11 at 14:59