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Problem description:

(Informal). I simply want to know if there exists a necessary and (or at least) sufficient condition for an orthogonal matrix to map every point of the nonnegative orthant (say on $\mathbb{R}^m$) to itself.

(Formal). Let $\mathscr{D}_m = \{(x_1,\ldots,x_m)\in\mathbb{R}^m\,:\, x_i\geq 0,\ i=1,2,\ldots,m\}$. For a given $\mathbf{Q}$ an $m\times m$ real orthogonal matrix define the image $\mathscr{D}_m^{*} = \{\mathbf{y}\in\mathbb{R}^m\,:\, \mathbf{y}=\mathbf{Q}\mathbf{x},\, \mathbf{x}\in\mathscr{D}\}$. Is there a necessary and (or at least) sufficient condition on $\mathbf{Q}$ in order for $\mathscr{D}_m = \mathscr{D}^{*}_m$?

Motivation/Context of the problem: (Change-of-variable in Integration)

Let $\mathbf{S}$ be an $m\times m$ positive definite matrix. It is well-known that there exists an orthogonal matrix $\mathbf{Q}$ and a diagonal matrix $\boldsymbol{\Lambda}$ such that $\mathbf{Q}^{\top}\mathbf{S}\mathbf{Q} = \boldsymbol{\Lambda}$. With this, we have the following equality of $m$-variate integration via the classic (multivariate) change-of-variable, noting that the absolute value of the Jacobian is $|\text{det}(\mathbf{Q})| = 1$, $$ \int_{\mathscr{D}_m} \exp\{\mathbf{x}^{\top}\mathbf{S}\,\mathbf{x}\}\cdot\exp\{-\frac{1}{2}\mathbf{x}^{\top}\mathbf{x}\}\,\text{d}\mathbf{x} = \int_{\mathscr{D}^{*}_m} \exp\{\mathbf{y}^{\top}\boldsymbol{\Lambda}\,\mathbf{y}\}\cdot\exp\{-\frac{1}{2}\mathbf{y}^{\top}\mathbf{y}\}\,\text{d}\mathbf{y} $$ where $\mathscr{D}_m$ and $\mathscr{D}^{*}_m$ are given in the (formal) problem description. The left-hand-side can be calculated easily if $\mathscr{D}_m = \mathscr{D}^{*}_m$ which gives our motivation of the problem stated above.

Attempt:

Honestly speaking, I failed to find from the web for a simpler explanation. I came across with the article of Barker and Foran Self-Dual Cones in Euclidean Spaces with an abstract that contains the following line..

``...We begin by giving necessary and sufficient for a cone to be the orthogonal transform of the positive orthant...''

I humbly accept that I am new to this topic. It is even my first time to hear that the nonnegative orthant is a symmetric self-dual cone. I have basic Linear Algebra and Calculus as background. Hope anyone can help me with this.

EDIT: I found this link: Find a nonnegative basis of a matrix nullspace / kernel that relates to this problem.

venrey
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    There is something on this problem in the Perron-Frobenius theorem, do you know it? I am not too sure about this. – Giuseppe Negro Sep 11 '18 at 12:03
  • I am sorry. My answer is no. I'll try to search on that now. I am currently reading more background (introductory) material for the paper of Barker and Foran to understand the concepts which are really new to me. Maybe I could prove something about the necessary and sufficient conditions on the orthogonal matrix. – venrey Sep 11 '18 at 12:08
  • https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem But now that I see it, I don't think it can really help you, sorry. Yours looks like a not-so-easy problem unfortunately – Giuseppe Negro Sep 11 '18 at 12:09
  • Perron-Frobenius essentially says that, if all entries of your matrix are $>0$, then you have an eigenvector in the positive octant. – Giuseppe Negro Sep 11 '18 at 12:11
  • It says here https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem#No_other_non-negative_eigenvectors that some matrices contain at most one positive eigenvector.. This is quite a sad news. Though I really don't know how to connect it yet to my problem. – venrey Sep 11 '18 at 12:15
  • Somehow, the problem is quite trivial if all the rows of $\mathbf{Q}$ are nonnegative. In this case $\mathscr{D}^{*}_m \subseteq \mathscr{D}_m$. The true problem lies when we permit negative elements in the rows. – venrey Sep 11 '18 at 12:20
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    If the rows of an orthogonal matrix are non negative then the matrix is the identity. I suspect that no matrix except the identity fixes the positive orthant. – Giuseppe Negro Sep 11 '18 at 14:08
  • I have searched through the web regarding Perron-Frobenius theorem..and here's a paper which, of course as for now, I'm struggling to learn..https://www.math.wisc.edu/hans/paper_archive/other_papers/hs156.pdf

    the title of paper is "Matrices leaving a cone invariant" by Schneider and Tam

     – venrey Sep 11 '18 at 14:36
  • Well I was thinking the same thing..can you give some links about the diagonalization of orthogonal matrices? I can't find some decent sources. If $(\lambda_i,\mathbf{z}i)$, $i=1,2,\ldots,m$ are the eigenvalue-eigenvector pairs where the $\mathbf{z}_1,\ldots,\mathbf{z}_m$ form an othonormal basis, then it would suffice to show that each $\lambda_i\geq 0$ and the rows of each $\mathbf{E}_i = \mathbf{z}_i\mathbf{z}_i^{\top}$, $i=1,2,\ldots, m$ are nonnegative since by spectral decomposition: $$ \mathbf{y} = \mathbf{Q}\mathbf{x} = \sum{i=1}^{m} \lambda_i \mathbf{E}_i\mathbf{x} \geq 0 $$ – venrey Sep 11 '18 at 14:44
  • You say "map every point to itself", but then you have $\mathscr D^*=\mathscr D$, which only means the whole set maps to itself (individual points may move). In the latter case, diagonal reflections obviously work. For example, if $m=3$, there is $$Q=[[0,1,0];[1,0,0];[0,0,1]]$$ ...Am I misunderstanding the problem? – mr_e_man Sep 11 '18 at 15:39
  • It's not necessarily reflexive (i.e., identity) when I say "to itself". I said "nonnegative orthant to itself". Yes, diagonal reflections are possible. – venrey Sep 11 '18 at 15:55
  • I don't see how $\mathscr D^\subseteq\mathscr D$, unless $\mathscr D^=\mathscr D$; orthogonal matrices are invertible. And why do you require non-negative eigenvalues, which excludes reflection matrices? @GiuseppeNegro -- Here's a non-identity rotation matrix with non-negative rows, which also fixes $\mathscr D$ : $$Q=[[0,1,0];[0,0,1];[1,0,0]]$$ – mr_e_man Sep 11 '18 at 15:59
  • I am just considering matrices with non-negative eigenvalues for a possible sufficient condition on $\mathbf{Q}$. – venrey Sep 11 '18 at 16:01
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    @mr_e_man: Right, thank you. The corrected statement is: "every orthogonal matrix with non-negative entries is a permutation matrix", and the corrected conjecture is that only the only orthogonal matrices that fix $\mathscr D$ are permutations. – Giuseppe Negro Sep 11 '18 at 16:02
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    In 3D, it's obvious that the positive octant (a "corner of a cube") has exactly 6 symmetries: 3 reflections, 2 rotations (120deg), and the identity. I believe this generalizes to nD, with all symmetries being compositions of reflections that swap 2 axes. – mr_e_man Sep 11 '18 at 16:03
  • @Guiseppe Negro: Then, that solves the problem. Am I right? – venrey Sep 11 '18 at 16:04
  • Can anyone formalize some proof and put it as an answer so that we can finally close this problem? – venrey Sep 11 '18 at 16:07
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    I don't know how to formalize this, but the positive orthant is the convex hull of the positive coordinate axes, and the axes are the "edges" of the orthant, so they should have the same symmetries. A symmetry of the axes must send every axis to another axis, and must be invertible; these are indeed permutation matrices. – mr_e_man Sep 11 '18 at 16:17
  • Here's a proof on nonnegative orthogonal matrices being permutation matrices: https://math.stackexchange.com/questions/322514/nonnegative-orthogonal-matrices – venrey Sep 12 '18 at 03:08

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