So I am inspired by unitary matrices which preserve the $\ell^2$-norm of all vectors, so in particular the unit norm vectors. But then I saw that the $\ell^1$-norm of probability vectors is preserved by matrices whose columns are probability vectors. And this got me thinking: But what are the matrices preserving the $\ell^1$-norm of arbitrary real unit $\ell^1$-norm vectors? So basically we extend a probability vector to also allow a sign, but ignoring the signs, this should still be a probability vector; and then we ask for the corresponding structure-preserving matrices.


It is already clear that the columns of such a matrix should be this 'extended' kind of probability vector, because we can multiply the matrix with a standard basis vector which has $\ell^1$-norm 1. But not all of such matrices preserve this, take for example


$$ M = \frac{1}{2} \left(\begin{matrix} 1 & 1\\ 1 & -1 \end{matrix}\right) $$ 


and 


$$ x = \left( \begin{matrix} 0.3 \\ -0.7 \end{matrix} \right) $$ 


Then we have 


$$ Mx = \left(\begin{matrix} -0.2 \\ 0.5 \end{matrix}\right) $$ 


which fails the test.



What are the matrices preserving the $\ell^1$-norm?
As pointed out by YCor in the comments, the following theorem is true:


**Theorem 1** Let $p \in [1,\infty] \setminus \{2\}$. If a matrix $A \in \mathbb{R}^{n \times n}$ is an isometry on $\mathbb{R}^n$ with respect to the $p$-norm, then $A$ is a signed permutation matrix, i.e. a permutation matrix where some of the one's are replaced with $-1$.


For the proof, first note that the case $p = \infty$ follows from $p = 1$ by duality, so we only have to show the theorem for $\in [1,\infty) \setminus 2$.


Now we use the following lemma:


**Lemma 2** Let $p \in [1,\infty) \setminus \{2\}$ and let $(\Omega\_1,\mu\_1)$ and $(\Omega\_2,\mu\_2)$ be two measure spaces. If $T: L^p(\Omega\_1,\mu\_1) \to L^p(\Omega\_2,\mu\_2)$ is an isometric linear mapping, then $T$ is *disjointness preserving*, i.e. for all $f,g \in L^p(\Omega\_1,\mu\_1)$ which fulfil $fg = 0$, we also have $(Tf)(Tg) = 0$.


In a more general form, this lemma goes originally back to Lamperti ("On the isometries of certain function spaces", Pacific J. Math. 8 (1958), 459–466.). 


A very clear proof of the lemma in the above form can be found in Lemma 4.2.2 of S. Facklers PhD dissertation (DOI: 10.18725/OPARU-3268).


If we apply Lemma 2 to $L^p(\Omega\_1,\mu\_1) = L^p(\Omega\_2,\mu\_2) = \mathbb{R}^n$, it follows that every matrix $A \in \mathbb{R}^{n \times n}$ which is isometric with respect to the $p$-norm is automatically disjointness preserving. Hence, every row of $A$ contains exactly one non-zero entry. Since $A$ is invertible, this implies that every column of $A$ also contains exactly one non-zero entry. Thus, $A$ is of the form $A = DP$, where $P$ is a permutation matrix and $D$ is a diagonal matrix. Using again that $A$ is isometic, we can see that $D$ can only have the numbers $1$ and $-1$ on its diagonal.


**Remarks:** 


(a) Lemma 2 is of course quite general compared to the finite dimensional question. However, I don't think that a finite dimensional version of Lemma 2 is easier to prove.


(b) Using Lemma 2 above, one can also obtain a description of isometries on general $L^p$-spaces; see Theorem 3.1 in Lamperti's article quoted above.


