Chen-Kou-Lyu theory: I. Convex cones in finite dimensions and their cross-dual matrix cones

In the July 2026 post, I explained how Yuxu Chen, Hui Kou and Zhenchao Lyu managed to show that the category of RB-domains is a proper subcategory of that of FS-domains, solving a long-standing open problem [5]. At the end of that same post, I mentioned that they had also solved the question whether the probabilistic powerdomain of an RB-domain is an RB-domain. The answer is no [4], and they even showed more: the finite posets P whose probabilistic powerdomain (precisely, whose dcpos of subprobability valuations) is an RB-domain are exactly the finite trees.

I am planning to explain how this works, but I will not explain everything in one post: today I will focus on the theory of (cross-dual) matrix cones that they use. This is a well-known theory: see Barker [1] for a survey and a list of references, and see Barker and Loewy [2] for slightly more specific material. I will also refer to one result due to Berman and Gaiha [3]. So yes, calling this post “Chen-Kou-Lyu theory: I.” is a bit of a misnomer: for now, I will merely explain a lot of basic results that they use, but which had been known since the (early) 1970s.

I must apologize if you see all this on a topology blog: today the material will be algebra, and topology will play only a very minor role. In fact, the main bit of topology we will use can be summed up in the following remark.

Remark A. Let n ∈ N. We write ||_|| for the usual Euclidean norm on Rⁿ, namely, if u is the vector with components u₁, …, u_n, then ||u|| ≝ $\sqrt {u_1^2 + \cdots + u_n^2}$. We let d be the associated metric: d(u,v) ≝ ||u–v||. The open ball topology of (Rⁿ, d) is the same as the product topology, where R is given its standard topology.
A subset C of Rⁿ is bounded if and only if there is an a ∈ R₊ such that ||u|| ≤ a for every u ∈ C. Then the compact subsets of Rⁿ are its closed bounded subsets. There are many possible proofs of this. One goes as follows. A closed bounded subset C of Rⁿ, with a as above, is included in the compact set [0,a]ⁿ, hence is compact. Conversely, a compact subset C of Rⁿ is closed since compact subsets of Hausdorff spaces are closed. It is bounded because ||_||: Rⁿ → R is continuous, hence reaches its supremum on C.

I will assume some basic knowledge in linear algebra: finite-dimensional vector spaces, linear bases, independent sets of vectors, matrices, transposes and product of matrices mostly.

Cones in Rⁿ, and matrix cones

Let n ∈ N, which will remain fixed for the rest of this post. A cone in Rⁿ is a non-empty subset C of Rⁿ such that for every a ∈ R₊ and every x ∈ C, ax ∈ C. Here ax is scalar multiplication of the vector x by a. Writing x as the column vector $\left(\begin{matrix}x_1 \\ \vdots \\ x_n \end{matrix}\right)$, ax is the column vector $\left(\begin{matrix}ax_1 \\ \vdots \\ ax_n \end{matrix}\right)$. Instead of $\left(\begin{matrix}x_1 \\ \vdots \\ x_n \end{matrix}\right)$, we will most often use the notation (x₁, …, x_n)^T. The T superscript denotes the transpose of a matrix.

Note that we assumed C to be non-empty, and this is equivalent to saying that C contains the zero vector 0.

A ray R_x is a set of the form {ax | a ∈ R₊} for some x ∈ Rⁿ that is non-zero. Hence a cone is either the trivial cone {0}, or a non-empty union of rays.

A subset C of Rⁿ is convex if and only if for all x, y ∈ C and α ∈ [0,1], the convex combination αx + (1−α)y is in C. In other words, given any two points x and y in C, C must also contain the whole line segment from x to y. All our cones will be convex, but I will keep saying “convex cone” so that you won’t confuse them with the non-convex cones that occur in the literature.

A convex cone is the same thing as a cone C that is closed under addition. Indeed, if C is closed under addition, then for all x, y ∈ C and α ∈ [0,1], αx and (1−α)y are in C, and their sum is in C, too. Conversely, if C is convex, then for all x, y ∈ C, the sum x + y is equal to 1/2(2x) + 1/2(2y), which is in C.

The set M_n of n × n matrices of real numbers is a real vector space, which we can equate with Rⁿ². We form the following convex cone in M_n.

Definition B (Crossed Matrix cone). Given two convex cones C and C′ in a finite dimensional real vector space Rⁿ, the crossed matrix cone R_C,C′ is the collection of finite sums ∑_i=1^m u_i v_i^T, where m ∈ N and u_i ∈ C, v_i ∈ C′. We topologize it with the subspace topology induced by the inclusion in Rⁿ², with its usual metric topology.

We take the convention that u_i and v_i are column vectors, and this has a somewhat confusing consequence. Writing u_i as the column vector $\left(\begin{matrix}a_1 \\ \vdots \\ a_n \end{matrix}\right)$, namely as (a₁, …, a_n)^T, and v_i as the column vector $\left(\begin{matrix}b_1 \\ \vdots \\ b_n \end{matrix}\right)$, namely as (b₁, …, b_n)^T, v_i^T is the row vector (b₁, …, b_n), without any superscript. Then u_i v_i^T is the matrix $\left(\begin{matrix}a_1b_1 & \ldots & a_1 b_n \\ \vdots & \cdots & \vdots \\ a_n b_1 & \cdots & a_n b_n \end{matrix}\right)$.

Such matrices u_i v_i^T are very special. Notably, they have rank at most 1, meaning that no two of its rows (equivalently, no two of its columns) are linearly independent. Our main objective in this post is to show that, under some conditions on C and C′, R_C,C′ is a closed convex cone, which contains the identity matrix only in very specific cases. This is exactly what we studied in the July 2026 post: the cone there was R_C,C, where C was the Lorentz cone in dimension 3, and one of the final aspects of the proof was to show that R_C,C did not contain the identity matrix.

We start with the easy part.

Lemma C. For all convex cones C and C′ in Rⁿ, R_C,C′ is a convex cone in M_n. Topologically, R_C,C′ is Hausdorff, and the algebraic operations: addition, subtraction, scalar multiplication, are all (jointly) continuous.

Proof. For every α ∈ R₊, for every element w ≝ ∑_i=1^m u_i v_i^T of R_C,C′, where u_i ∈ C and v_i ∈ C′, αw = ∑_i=1^m (αu_i) v_i^T: since αu_i ∈ C, αw is in R_C,C′. Therefore R_C,C′ is a cone. R_C,C′ is clearly closed under addition, hence under convex combinations. R_C,C′ is Hausdorff as a subspace of M_n ≅ Rⁿ², which is itself Hausdorff. The continuity properties follow from similar continuity properties on Rⁿ², which are obvious. ☐

Showing that R_C,C′ is closed, under some conditions, is harder. One of these conditions will be that C′ be the dual cone C* of C, and that C be a proper cone, all notions that we will introduce below.

The fact that R_C,C* is closed—under these conditions—is a bit surprising (unless you have already read about a similar example of a closed matrix cone in the July 2026 post, for example). For example, one may form a sequence of elements w_n of R_C,C* by letting w_n ≝ ∑_i=1ⁿ u_i v_i^T, where the number of summands n goes to infinity. By taking u_n and v_n decreasing fast enough to 0, (w_n)_n∈N can be made to tend to some matrix w ∈ M_n, which one would be tempted to write as ∑_i=1^∞ u_i v_i^T. The fact that R_C,C* is closed implies that this infinite sum can be rewritten as a finite sum. The key to this is a compactness argument hidden in the fact that all our cones live in some finite-dimensional space. (And, as in the July 2026 post, we will use Carathéodory’s theorem. I will take the opportunity to prove it, too.)

Full-dimensional cones

One of the notions we will need is that of a proper cone in Rⁿ, and a proper cone is full-dimensional in the following sense.

Definition D (Full-dimensional cone). A cone C in Rⁿ is full-dimensional if and only if its interior in Rⁿ is non-empty.

That definition may seem either obvious or completely obscure. Hopefully the following lemma will show that this is equivalent to something that you find more intuitive.

Lemma E. For every convex cone C in Rⁿ, the following are equivalent:

1. C is full-dimensional;
2. C−C = Rⁿ, where C−C ≝ {u−v | u, v ∈ C};
3. C spans the whole of Rⁿ, namely every element of Rⁿ is a linear combination of elements of C;
4. there is a basis of Rⁿ consisting of vectors of C.

Proof. 1 ⇒ 2. Let u be in the interior int(C) of C. There is an ϵ > 0 such that the open ball B^d_u,<ϵ with center u and radius ϵ is included in C. For every non-zero vector v ∈ Rⁿ, w ≝ u + $\frac \epsilon {2 ||v||}$ v is in B^d_u,<ϵ, hence in int(C) ⊆ C. Therefore we can write v as $\frac {2 ||v||} \epsilon w – \frac {2 ||v||} \epsilon u$, which is in C−C.

2 ⇒3 is trivial.

3 ⇒ 4. We build linearly independent vectors u_i (1 ≤ i ≤ n) in C, by induction on i. Explicitly, assuming that we have already built linearly independent vectors u₁, . . . , u_i–1 in C, and if i ≤ n, we claim that we can find u_i ∈ C such that u₁, . . . , u_i–1, u_i are linearly independent. Since i ≤ n, u₁, . . . , u_i–1 cannot generate the whole of Rⁿ, so there is a vector v ∈ Rⁿ that cannot be expressed as a linear combination of u₁, . . . , u_i–1. By assumption, v is a linear combination ∑_j=1^m α_j v_j of elements v_j of C. If every v_j could be expressed as a linear combination of u₁, . . . , u_i–1, then v could, too, and that is impossible. Hence some v_j is not a linear combination of u₁, . . . , u_i–1, and it suffices to take u_i ≝ v_j.

We now have a linearly independent family of vectors u₁, . . . , u_n in C. Since n is the dimension of Rⁿ, they must form a basis of Rⁿ.

4 ⇒ 1. Let u₁, . . . , u_n be a basis of Rⁿ consisting of vectors of C. Let u ≝ ∑_i=1ⁿ u_j. Since C is a convex cone, u is in C. Let ϵ ≝ 1 /2. The set U of all sums ∑_i=1ⁿ β_i u_i where 1−ϵ < β_i < 1 + ϵ is open in Rⁿ, because it is equal to ∩_i=1ⁿ p_i^–1 (]1−ϵ,1 + ϵ[), where p_i is the linear (hence continuous, in finite dimensions) map that sends every vector to its coordinate number i in the basis u₁, . . . , u_n. U contains u, and is included in C, since C is closed under scalar multiplication by non-negative reals and under sums. Therefore C has non-empty interior. ☐

For example, in R², the ray R_(0,1) ≝ {(x,y) | x=0, y≥0} is not full-dimensional. In general, given n≥1, the rays in Rⁿ that are full-dimensional are those such that n=1, and there are only two.

Remark F. One might think that the basis of Lemma E, item 4 can be taken to generate the cone C as well, in the sense that every element of C would be a linear combination of elements of the basis with non-negative coefficients, but that is wrong. For example, R itself is full-dimensional in R, but whatever basis you take, it will consist of one non-zero real number a, and you cannot write –a as a non-negative real number times a.

Example G. Generalizing a construction we have seen in the July 2026 post, the Lorentz cone Lⁿ⁺¹ in Rⁿ⁺¹, where n ≥ 1 is the collection of vectors (t, s₁, ···, s_n)^T such that $t \geq \sqrt {s_1^2 + \cdots + s_n^2}$. See the picture above for the case n=2. It is practical to equate such points with pairs (t, s) where t ∈ R and s ≝ (s₁, ···, s_n) ∈ Rⁿ, and the latter condition is t ≥ ||s||. The interior of Lⁿ⁺¹ is the collection of pairs (t, s) such that t > ||s||, and is non-empty, so Lⁿ⁺¹ is full-dimensional. We can also obtain the same results by using Lemma E, item 4: writing e_i for the standard basis vector of Rⁿ, with a 1 at position i and zeros at all other positions, the vectors (1, e₁), …, (1, e_n) and (1, –e₁) form a linear basis of Rⁿ⁺¹ that consists of vectors of Lⁿ⁺¹.

Pointed cones

A proper cone will not only need to be full-dimensional, but also pointed in the following sense.

Definition H (Pointed cone). A cone C in Rⁿ is pointed if and only if the only point x such that x and −x are both in C is the zero vector 0.

More succinctly, C is pointed if and only if C ∩ (–C) = {0}.

If n≥ 1, then Rⁿ itself is a cone in Rⁿ that is not pointed. In general, a cone C in Rⁿ is pointed if and only if it does not contain any line {αx | α ∈ R}, for any non-zero vector x in Rⁿ. The Lorentz cone Lⁿ⁺¹ is pointed: if (t, s) ∈ Lⁿ⁺¹ and (–t, –s) ∈ Lⁿ⁺¹, then t ≥ ||s|| and –t ≥ ||–s|| = ||s||; in particular t ≥ 0 and t ≤ 0, so t=0, and then t ≥ ||s|| implies that s=0, too.

The main value of the notion lies in the following lemma.

Lemma I. Every convex cone C in Rⁿ induces a preordering ≤_C on Rⁿ, defined by x ≤_C y if and only if y−x ∈ C, which is a partial ordering if and only if C is pointed. In that case, the zero vector 0 is the least element of C.

Proof. We have x–x=0 ∈ C, so ≤_C is reflexive. If x ≤_C y and y ≤_C z, then y−x and z–y are in C, hence their sum is, to, so that x ≤_C z. Hence ≤_C is a preordering. If x ≤_C y and y ≤_C y, then y−x and x–y are in C, so if C is pointed then they are both equal to 0, namely x=y: ≤_C is a partial ordering. Conversely, if ≤_C is a partial ordering, then for every point x such that both x and –x are in C, we have 0 ≤_C x and x ≤_C 0, hence x=0, so C is pointed. Finally, since 0 ≤_C x for every x ∈ C, 0 is the least element of C. ☐

Dual cones

For any two vectors z ≝ (z₁, …, z_n)^T and x ≝ (x₁, …, x_n)^T in Rⁿ, we will often have to compute z^T x: this is equal to ∑_i=1ⁿ z_i x_i, and this is the scalar product of z and x. This scalar product is usually written as z · x or as ⟨z, x⟩, but we will simply use the notation z^T x. We also note that z^T x = x^T z, and that, being just a real number, z^T x is its own transpose.

Definition J (Dual cone). The dual cone C* of a cone C in Rⁿ is {z ∈ Rⁿ |∀x ∈ C, z^T x ≥ 0}.

Lemma K. For every convex cone C in Rⁿ, its dual cone C* is a closed convex cone.

Proof. The fact that C* is a convex cone is clear. It is closed because it is the intersection of the sets f_x⁻¹ ([0, ∞[) over all x ∈ C, where f_x is the map z ↦ z^T x; f_x is linear on a finite-dimensional space, hence is continuous, so f_x⁻¹ ([0, ∞[) is closed for every x ∈ C. ☐

Example L. The Lorentz cone (see Example G) is its own dual: (Lⁿ⁺¹)* = Lⁿ⁺¹. Indeed, let (t, s), (t’, s’) ∈ Lⁿ⁺¹. Then (t, s)^T (t’, s’) = tt′ + s^T s′ ≥ ||s|| ||s′||+ s^T s′. The Cauchy-Schwarz inequality, applied to –s and s’, states that (−s)^T s′ ≤ ||−s|| ||s′|| = ||s|| ||s′||, so ||s|| ||s′||+ s^T s′ ≥ 0. Therefore (t, s)^T (t’, s’) ≥ 0. This shows that Lⁿ⁺¹ ⊆ (Lⁿ⁺¹)*.
Conversely, let (t, s) ∈ (Lⁿ⁺¹)* . The vector (1,0) is in Lⁿ⁺¹, so (t, s)^T (1,0) ≥ 0, whence t ≥ 0. In particular, (t, s) is in Lⁿ⁺¹ in the special case where s = 0. We now assume s ≠ 0, so that ||s|| ≠ 0. The vector (||s||, –s) is in Lⁿ⁺¹, so (t, s)^T (||s||, –s) ≥0. In other words, t ||s|| ≥ s^T s′ = ||s||². Dividing out by ||s||, we obtain that t ≥ ||s||, so (t, s) ∈ Lⁿ⁺¹.

Not every convex cone C in Rⁿ is its own dual: by Lemma K, C* is always closed, so a non-closed cone C in Rⁿ cannot be its own dual. An example of such a cone is the set {(x, y) ∈ R₊² | x = y = 0 or x, y > 0}, which is a convex cone, which is not closed: the points (1, 1/2ⁿ) form a sequence of points in the cone that converges to (1, 0), which is not in the cone.

We will need the following result, which is a baby version of classical separation results, and which usually follows from the Hahn-Banach theorem.

Lemma M (Separation). Let C be a non-empty closed, convex subset of Rⁿ and z ∈ Rⁿ – C. Then there is a vector x ∈ Rⁿ and a real number b such that z^T x > b and y^T x ≤ b for every y ∈ C.
If C is a closed convex cone in Rⁿ, then we can even require b = 0.

The vector x and the number b define a hyperplane {y ∈ Rⁿ | y^T x = b}. The lemma states that one can choose them in such a way that z be on one side of the hyperplane (and outside of it) and the whole of C lies on the other side (possibly touching it), see the picture below.

Proof. Let us remark that for every point y ∈ Rⁿ, ||y|| = $\sqrt {y^\top y}$. We will use this below.

Let f : Rⁿ → R map every y ∈ Rⁿ to ||z−y||. Since C is non-empty, let us pick an element x₀ ∈ C. We let a ≝ inf {f (x) | x ∈ C} ≤ f (x₀). Then a is also equal to inf {f (x) | x ∈ C₀}, where C₀ ≝ {x ∈ C | ||x|| ≤ ||x₀||}. Since ||_|| is continuous, C₀ is closed. It is bounded, hence compact by Remark A. Since ||_|| is continuous, so is f, so f reaches its infimum at some point y₀ of C₀ (hence in C): ||z−y₀||= a. In other words, y₀ is the closest point to z in C, or one of the closest. See the picture below.

Since z ∉ C, we must have ||z−y₀|| > 0, so a >0.

We claim that (z−y₀)^T (y−y₀) ≤ 0 for every y ∈ C. Geometrically, this means that the angle between the segments [z, y₀] and [y, y₀] is larger than 90 degrees. In order to see this, we realize that, since C is convex, αy + (1−α)y₀ is in C for every α ∈]0,1]. Since a realizes the minimum of f on C, a ≤ f (αy + (1−α)y₀), in other words, a ≤ ||z − (αy + (1−α)y₀)||= ||(z−y₀) − α(y−y₀)||. Squaring both sides, we obtain that ||z−y₀||² ≤ ((z−y₀)−α(y−y₀))^T ((z−y₀)−α(y−y₀)) = ||z−y₀||² + α²||y−y₀||² − 2α(z−y₀)^T (y−y₀). Therefore 2α(z−y₀)^T (y−y₀) ≤ α²||y−y₀|². This holds for α>0 (and ≤1), so (z−y₀)^T (y−y₀) ≤ α/2 ||y−y₀||². Letting α tend to 0, we obtain that (z−y₀)^T (y−y₀) ≤0.

Let x ≝ z−y₀. The inequality we have just proved rewrites as x^T (y−y₀) ≤ 0, for every y ∈ C. Let b ≝ x^T y₀. For every y ∈ C, y^T x = x^T y is then less than or equal to x^T y₀ = b. Finally, z^T x − b = x^T z − x^T y₀ = x^T x = ||x||² is non-negative. It is non-zero, otherwise x would be equal to 0, so z would be equal to y₀, which is impossible since z ∉ C and y₀ ∈ C. Therefore z^T x − b > 0, namely z^T x > b.

Finally, let us assume that C is a closed convex cone. Since 0 ∈ C, the inequality y^T x ≤ b for every y ∈ C implies that b ≥ 0. If there was any y ∈ C such that y^T x > 0, then, letting α ∈ R₊ be so large that α y^T x > b, we would obtain a contradiction from the fact that αy ∈ C, hence (αy)^T x ≤ b. Therefore y^T x ≤0 for every y ∈ C. Since z^T x > b ≥0, in particular z^T x > 0. Hence we can replace b by 0 in the inequalities z^T x > b and y^T x ≤ b (y ∈ C). ☐

Proposition N. Let C be a cone in Rⁿ. The bidual cone C** contains C, and is equal to it if C is closed and convex.

Proof. For every x ∈ C, for every z ∈ C*, x^T z = z^T x ≥ 0, so x ∈ C**. Conversely, let us assume that C is a closed convex cone. If C is a proper subset of C**, then there is a point z ∈ C** that is not in C. By the last part of Lemma M, there is a vector x ∈ Rⁿ such that z^T x > 0 and y^T x ≤ 0 for every y ∈ C. In particular, y^T (−x) ≥ 0 for every y ∈ C, so −x ∈ C*, by definition of C*. Since z ∈ C**, we must have z^T (−x) ≥ 0, but this is impossible since z^T x >0. ☐

Proposition O. Let C be a cone in Rⁿ. Then C* is a closed convex cone in Rⁿ, which is pointed if C is full-dimensional, and full-dimensional if C is pointed, closed and convex.

Proof. If C is full-dimensional, let x₁, . . . , x_n be a basis of Rⁿ consisting of elements of C, using Lemma E, item 4. Let z ∈ C* be such that −z is also in C*. For every i ∈ {1, ···, n}, z^T x_i ≥0 and (−z)^T x_i ≥0, so z^T x_i = 0. Since x₁, . . . , x_n is a basis of Rⁿ, z^T x = 0 for every x ∈ Rⁿ. Taking x equal to the standard basis vector e_i, this entails that the ith component of z is equal to 0. This holds for every i ∈{1, ···, n}, so z = 0. Hence C* is pointed.

If C is pointed, closed and convex, we claim that C* is full-dimensional. Let us assume that C* is not full-dimensional. C*−C* is a linear subspace of Rⁿ, as one sees easily, and one that is different from Rⁿ by Lemma E, item 2. Hence a basis z₁, . . . , z_m of C*−C* must consist of strictly less than n vectors. Because of this, there must be a non-zero vector x ∈ Rⁿ such that z_i^T x = 0 for every i ∈ {1, ···, m}: in order to find one, we need to solve a system of m < n equations in n variables, and such a system has at least n−m independent solutions. From z_i^T x = 0 for every i ∈ {1, ···, m}, we deduce that z^T x = 0 for every z ∈ C*−C*. Hence x^T z = z^T x ≥ 0 (indeed, is equal to 0) for every z ∈ C*, and also (–x)^T z = z^T (–x) ≥ 0 (indeed, again equal to 0) for every z ∈ C*. This shows that both x and −x are in C**. Since C is closed and convex, we have C** = C by Proposition N, so x and −x are both in C. This is impossible since C is pointed. ☐

Proper cones

Have you forgotten what we were at? I guess so. We were trying to define proper cones, and we are there at last.

Definition P (Proper cone). A proper cone in Rⁿ is a pointed full-dimensional closed convex cone in Rⁿ.

By Proposition O and Proposition N, for every proper cone C in Rⁿ, C* is also a proper cone in Rⁿ, and C** = C. We sum this up in the following proposition, together with some information we will need later on. We will also reuse the notations z₀, x₀, C₁ and C*₁ in at least two occasions, so try to keep them in mind.

Proposition Q. For every proper cone C in Rⁿ, C* is also a proper cone in Rⁿ, and C** = C.
For every z₀ ∈ int(C*) (and such a point exists), the set C₁ ≝ {x ∈ C | z₀^T x = 1} is compact, and if n≥1, then C₁ is non-empty and every element of C can be written as αx for some x ∈ C₁ and α ∈ R₊.
For every x₀ ∈ int(C) (and such a point exists), the set C*₁ ≝ {z ∈ C* | z^T x₀ = 1} is compact, and if n≥1, then C*₁ is non-empty and every element of C* can be written as αz for some z ∈ C*₁ and α ∈ R₊.

Proof. We have already argued that the first part is by Proposition O and Proposition N. Let us turn to the second part.

Since C* is proper, int(C*) is non-empty. We fix an arbitrary point z₀ ∈ int(C*), and we let C₁ ≝ {x ∈ C | z₀^T x = 1}. This is the intersection of the closed set C with the inverse image of the closed set {1} under the (linear hence) continuous map x ↦ z₀^T x, hence is closed.

If n=0, then C, C* and Rⁿ are reduced to one point, and so are their interiors. Then C₁ and C₁* are empty, and therefore compact. We now assume n≥1.

Since C is closed, the set C’₁ ≝ {x ∈ C | ||x||= 1} is closed and bounded, hence compact by Remark A. C cannot be reduced to {0}, since its interior would be empty, considering that n≥1. Hence there is a non-zero element u in C, and then u₁ ≝ u/||u|| is in C’₁. showing that C’₁ is non-empty. Since x ↦ z₀^T x is continuous, it maps C’₁ to a compact set of real numbers, which is non-empty since C’₁ is non-empty. In particular, there is a point x₀ ∈C’₁ such that z₀^T x₀ = inf_x∈C’1 z₀^T x, namely, the infimum is attained. Let a be this infimum. Since z₀ ∈int(C*), z₀−ϵx₀ is still in int(C*), hence in C*, for ϵ > 0 small enough. By definition of C*, and since x₀ ∈ C’₁ ⊆ C, this entails that (z₀−ϵx₀)^T x₀ ≥0, hence that a = z₀^T x₀ ≥ ϵ ||x₀||² = ϵ. Therefore a > 0.

For every element x ∈ C₁, x is not equal to zero since z₀^T x = 1. Therefore x/||x|| is well-defined, and in C’₁. By definition of a, z₀^T (x/||x||) ≥ a, so ||x|| ≤ 1/a. This implies that the closed set C₁ is bounded, hence compact by Remark A.

For every element y of C, z₀^T y ≥ 0 since z₀ ∈ C*. If z₀^T y > 0, then y can be written as αx where α ≝ z₀^T y and x ≝ (1/α)y ∈ C₁. If z₀^T y = 0, then we can write y as αx where α ≝ 0 and x is any point of C₁. Such a point exists: take (1/a)x₀, for example. In particular, we have shown that C₁ is non-empty.

We reason similarly in order to establish the final part of the Proposition, swapping C and C* (thanks to the fact that C** = C) and x₀ and z₀. ☐

The cross-dual matrix cone R_C,C* and the cone Γ(C)

The special case of a crossed matrix cone R_C,C′ where C′ = C* is called a cross-dual matrix cone. Every matrix M ∈ R_C,C* is in the cone Γ(C) of matrices that preserve C, in the sense that the associated linear operator maps vectors of C to vectors of C, as item 2 of the following lemma shows, among other things.

One may legitimately ask whether R_C,C* = Γ(C), and item 3 gives a partial answer; we will give a fuller answer in Theorem Zc, the final theorem of this post.

Lemma R. Let C be a convex cone in Rⁿ, and Γ(C) denote the set of matrices M ∈ M_n that preserve C, in the sense that Mu ∈ C for every u ∈ C. Then:

1. Γ(C) is a convex cone, which is closed if C is closed;
2. R_C,C* ⊆ Γ(C);
3. If C = R₊ⁿ, the cone of vectors with non-negative entries, then Γ(C) = R_C,C*, and is the cone of matrices with non-negative entires, which we may equate with R₊ⁿ².

Proof. 1. That Γ(C) is a convex cone is easy. If C is closed, then Γ(C) is closed because it is equal to ∩_u∈C f_u⁻¹ (C), where f_u : M ↦ Mu is linear hence continuous for every u ∈ C.

2. Let M ≝ ∑_i=1^m u_i v_i^T be an arbitrary element of R_C,C*, where m ∈ N and u_i ∈ C, v_i ∈ C*. For every u ∈ C, Mu = ∑_i=1^m u_i v_i^T u = ∑_i=1^m (v_i^T u) u_i is in C. Therefore M ∈ Γ(C).

3. Let C ≝ R₊ⁿ. Then C* is the collection of vectors u ∈ Rⁿ such that u^T v ≥ 0 for every v ∈ Rⁿ. This holds in particular for v equal to the standard basis vector e_i, so every element u ≝ (u₁, …, u_n)^T of C* must satisfy u_i ≥ 0 for every i ∈{1, ···, n}. That is, C* ⊆ C. Conversely, for every u ∈ C, we have u^T v ≥ 0 for every v ∈ Rⁿ, since u^T v is a sum of products of non-negative numbers. Hence C ⊆ C*. We conclude that C* = C = R₊ⁿ.

In particular, R_C,C* = R_C,C, which consists of matrices of the form ∑_i=1^m u_i v_i^T where u_i, v_i ∈ R₊ⁿ. Any such matrix has non-negative entries. Conversely, for every matrix A ∈ M_n with non-negative entries a_ij, 1 ≤ i, j ≤ n, we can write A as ∑_i,j=1ⁿ a_ij e_i e_j^T, namely as ∑_i,j=1ⁿ (a_ij e_i) e_j^T, showing that A is in R_C,C*. ☐

The following is a special case of Theorem 3.1 (i) of [3], and gives another relation between Γ(C) and cross-dual matrix cones. Equating matrices A, B in M_n with vectors u_A, u_B in Rⁿ², there is a funny way of computing the scalar product u_A^T u_B, as Tr (AB^T). (The operator Tr is the trace operator, which computes the sum of the diagonal entries.) Indeed, if A = (a_ij)_{1 ≤ i, j ≤ n} and B = (b_ij)_{1 ≤ i, j ≤ n}, then AB^T is the matrix whose i,j-entry is ∑_k=1ⁿ a_ik b_jk, so its trace is ∑_i,k=1ⁿ a_ik b_ik, and that is exactly u_A^T u_B. Using this formula is sometimes helpful, especially once we are aware of the following three facts: for every real number a, seen as a 1 × 1 matrix, Tr (a)=a; for all matrices A and B of respective dimensions m × n and n × m, Tr (AB) = Tr (BA); and Tr is linear.

Lemma S. For every closed convex cone C in Rⁿ, Γ(C)= R_C*,C*.

Proof. For every A ∈ R_C*,C*, by definition Tr (AB^T) ≥ 0 for every B ∈ R_C*,C, in particular when B = u v^T where u ∈ C* and v ∈ C. Then Tr (AB^T)= Tr (Avu^T)= Tr (u^TAv) = u^TAv (a 1 × 1 matrix) = (Av)^T u. This holds for every u ∈ C*, so Av ∈ C**. Since C is closed, C** = C by Proposition N, so Av ∈ C. This holds for every v ∈ C, so A ∈ Γ(C).

Conversely, let A ∈ Γ(C). For all u ∈ C* and v ∈ C, and letting B ≝ u v^T, Tr (AB^T)= Tr (Avu^T) = u^TAv = u^T(Av) ≥0, since u ∈ C* and Av ∈ C. Since Tr is linear, the same inequality holds for every sum of terms u v^T, namely for every element B of R_C*,C, so A ∈ R_C*,C*. ☐

Carathéodory’s theorem

It will be practical to introduce the notion of standard simplices. For every n≥ 1, let ∆_n be the set of vectors of n non-negative real numbers α ≝ (α₁, α₂, ···, α_n)^T such that that ∑_i=1ⁿ α_i = 1. See the following picture for the cases n=1, 2, 3.

A word of warning, though: ∆_n is usually denoted as ∆_n–1, and called the standard (n−1)-simplex, in the algebraic topology literature.

We will need the following important observation, due to Constantin Carathéodory. We already mentioned this result in the July 2026 post, and we give a proof of it here.

Proposition T (Carathéodory’s theorem). For every n ∈ N, for every subset A of Rⁿ, every point of the form ∑_i=1^m α_i x_i with m≥1, (α₁, α₂, ···, α_m)^T ∈ ∆_m and x_i ∈ A for every i ∈{1, ···, m} can be written in the same way, with the additional constraint that m ≤ n + 1.

The collection of elements of the form ∑_i=1^m α_i x_i as above is the convex hull of A, namely the smallest convex subset of Rⁿ that contains A. Carathéodory’s theorem states that every point in the convex hull of a subset A of Rⁿ, however large, is already in the convex hull of at most n+1 points of A.

Proof. Let x be any point in the convex hull of A, and let us write x as ∑_i=1^m α_i x_i, where m≥1, (α₁, α₂, ···, α_m)^T ∈ ∆_m and x_i ∈ A for every i ∈{1, ···, m}, in such a way that m is minimal. We claim that m≤n+1, and to see this, we assume that m>n+1 and we reach for a contradiction. Since m>n+1, the vectors x₂−x₁, x₃−x₂, . . . , x_m−x_m–1 form a list of strictly more than n vectors in Rⁿ, so they must be linearly dependent. In other words, there are real coefficients β₁, ···, β_m–1, not all equal to 0, such that ∑_i=1^m–1 β_i (x_i+1−x_i) = 0. We rewrite this as ∑_i=1^m γ_i x_i = 0, where γ₁ ≝ –β₁, γ_i ≝ β_i–1 – β_i for every i ∈{2, ···, m−1} and γ_m ≝ β_m–1. We note that ∑_i=1^m γ_i = 0. The latter implies that at least one number γ_i is strictly positive (in particular, non-zero): otherwise, β₁ = −γ₁ ≥0, β₂ = β₁−γ₂ ≥ β₁, . . . , β_m−1 = β_m−2 −γ_m−1 ≥ β_m−2, and finally β_m−1 = γ_m−1 ≤0, so we would have 0 ≤ β₁ ≤ β₂ ≤ ··· ≤ β_m−1 ≤0, which would entail that every β_i is equal to 0.

For every t ∈ R, we have x = ∑_i=1^m α_i x_i – t ∑_i=1^m γ_i x_i (since ∑_i=1^m γ_i x_i = 0) = ∑_i=1^m (α_i – t γ_i) x_i. We will fix t shortly. The set I of indices i ∈ {1, ···, m} such that γ_i > 0 is finite and, as we have seen, non-empty. Therefore it makes sense to let t ≝ min_i∈I (α_i/γ_i). Let i₀ ∈ I be such that t = α_i0/γ_i0. Since every α_i is in R₊ and since i₀ ∈ I, hence γ_i0 > 0, we note that t ≥ 0. In the sum ∑_i=1^m (α_i – t γ_i) x_i, the term with i=i₀ is equal to 0; the terms with i ∈ I–{i₀} are such that t ≤ α_i/γ_i by definition of t as a minimum, so α_i – t γ_i ≥0; the terms with i ∉ I are such that α_i – t γ_i ≥ α_i (since t ≥ 0 and γ_i ≤ 0) ≥ 0 (since α_i ∈ R₊). It follows that x is equal to the sum of (α_i – t γ_i) x_i over all i ∈ {1, ···, m}–{i₀}, a sum of at most m−1 vectors x_i from A multiplied by non-negative coefficients. Additionally, the sum of these coefficients is equal to ∑_i=1^m (α_i – t γ_i) (since the coefficient obtained with i=i₀ is equal to 0) = ∑_i=1^m α_i – t ∑_i=1^m γ_i = ∑_i=1^m α_i (since ∑_i=1^m γ_i = 0) = 1, so the vector of these coefficients is in ∆_m–1. This contradicts the minimality of m, proving the theorem. ☐

The cross-dual matrix cone of a proper cone is closed

We arrive at the promised result that R_C,C′ is a closed convex cone “under some conditions”, as we said earlier. The precise result is that this holds if C’ = C* and C is a proper cone. We had proved this in the special case where C is the 3-dimensional Lorentz cone L³ in the July 2026 post (see Proposition 4.1 there), considering that the cone R_C studied there is R_C,C, and that the Lorentz cone is its own dual (see Example L).

We equip the standard simplex ∆_n with the subspace topology from Rⁿ. As such, this is a compact Hausdorff space.

Theorem U. For every proper cone C in Rⁿ, the cross-dual matrix cone R_C,C* is closed in M_n ≅ Rⁿ².

Proof. If n=0, this is clear, since R_C,C* contains just the zero matrix. We therefore assume n≥1.

Let z₀, x₀, C₁ and C*₁ be defined as in Proposition Q. Let us say that an element of R_C,C* is normalized if and only if it is of the form ∑_i=1^m α_i u_i v_i^T where m=n²+1, (α₁, α₂, ···, α_m)^T is in ∆_m, each u_i is in C₁ and each v_i is in C*₁. Let us write R¹_C,C* for the set of normalized elements of R_C,C*.

We observe that R¹_C,C* is compact in M_n. Indeed, let f : ∆_m × C^m ×(C*)^m → M_n map ((α₁, ···, α_m)^T,(u₁, ···, u_m), (v₁, ···, v_m)) to ∑_i=1^m α_i u_i v_i^T. The function f is continuous, since composing it with each projection onto coordinates j, k yields the continuous map ((α₁, ···, α_m)^T,(u₁, ···, u_m), (v₁, ···, v_m)) ↦ ∑_i=1^m α_i u_ij v_ik^T (writing the jth component of u_i as u_ij and the kth component of v_i as v_ik). ∆_m is compact, and so are C₁ and C*₁ by Proposition Q, so ∆_m × C₁^m ×(C*₁)^m is compact. Therefore R¹_C,C* = f [∆_m × C₁^m ×(C*₁)^m] is compact.

We also note that every element w ≝ ∑_i=1^m u_i v_i^T of R_C,C* (with m≥1 arbitrary, u_i ∈ C and v_i ∈ C*) is equal to αw₁ for some α ∈ R₊ and some w₁ ∈ R¹_C,C*; we will call αw₁ the standard form of w. Indeed, first, by Carathéodory’s theorem (Proposition T) we can require m≤n²+1. We can in fact require that m=n²+1, by adding dummy zero terms. By Proposition Q, we can write each u_i as β_i u¹_i and each v_i as γ_i v¹_i, where β_i, γ_i ≥ 0, u¹_i ∈ C₁ and v¹_i ∈ C*₁. Therefore w =∑_i=1^m β_i γ_i u¹_i (v¹_i)^T. Let α ≝ ∑_i=1^m β_i γ_i. If α = 0, then we can take any element of R¹_C,C* for w₁. Such an elements exists: by Proposition Q, C₁ and C*₁ are non-empty, then take w₁ ≝ u v^T where u ∈ C₁ and v ∈ C*₁. If α > 0, then w = αw₁ where w₁ ≝ ∑_i=1^m (β_i γ_i / α) u¹_i (v¹_i)^T. Now (β₁ γ₁ / α, …, β_m γ_m / α) is in ∆_m, so w₁ ∈ R¹_C,C*.

We can now prove that R_C,C* is closed. Since Rⁿ² is a metric space, it suffices to show that the limit of every convergent sequence of elements of R_C,C* is in R_C,C*. Let therefore (α_nw_n)_n∈N be a sequence of elements of R_C,C*, all written in standard form, and let us assume that it converges to some element w_∞ of M_n ≅ Rⁿ². If w_∞= 0, then w_∞ is in R_C,C*, and we conclude. Hence we assume that w_∞≠0 from now on.

Since R¹_C,C* is compact, there is a subsequence (w_nk)_k∈N of (w_n)_n∈N (with n₁ < … < n_k < …) that converges to some element w of R¹_C,C*. Additionally, (α_nkw_nk)_k∈N still converges to w_∞. Since w_∞≠0, we have w_nk≠0 for infinitely many values of k; otherwise w_nk would be equal to 0 for k large enough, and this would imply w_∞= 0. Hence, extracting a further subsequence if necessary, we may assume that w_nk≠0 for every k ∈ N. Since Euclidean norm ||_|| is continuous, the terms ||α_nkw_nk||= α_nk ||w_nk|| converge to ||w_∞||. Since division is continuous on R–{0} with its usual metric topology, the scalars α_nk = ||α_nkw_nk||/||w_nk|| converge to α ≝ ||w_∞||/||w||. This makes sense since w is in R¹_C,C*, hence is non-zero, so that ||w||≠0. Now the scalars α_nk converge to α, the matrices w_nk converge to w, so the matrices α_nkw_nk converge to αw. In a Hausdorff space such as Rⁿ², limits are unique, so w_∞= αw. We have α ∈ R₊ and w ∈ R¹_C,C*, so w_∞ is in R_C,C*. ☐

Simplicial cones, extreme rays

We will now embark on characterizing when the identity matrix is in the cross-dual matrix cone R_C,C*. The answer rests on the notion of simplicial cones, defined as follows.

Definition V (Polyhedral, simplicial cones). A cone C in Rⁿ is polyhedral if and only if it is generated by finitely many vectors x₁, . . . , x_m ∈ Rⁿ, in the sense that C = {∑_i=1^m α_i x_i | α₁, . . . , α_m ∈ R₊}. It is simplicial if and only if, in addition, the vectors x₁, . . . , x_m can be taken to form a linear basis of Rⁿ (in particular, m=n).

Every polyhedral cone is closed and convex, every simplicial cone is polyhedral. A line {αx |α ∈ R}, where x≠0, is polyhedral since it is generated by x and −x, but is not simplicial.

Definition W. Given a convex cone C in Rⁿ, an extreme ray of C is a ray R_x, for some x ∈ C–{0}, such that if we can write x as a sum u + v with u, v ∈ C, then both u and v must be in R_x.

A typical example of extreme rays is given by the rays R_x1, . . . , R_xn of a simplicial cone in Rⁿ, as we will illustrate in Lemma X below. Before that, let us notice that the notion of extreme ray only depends on the ray R_x itself, not on the particular element x. Indeed, let us assume that R_x = R_y for two non-zero vectors x and y. Then y = αx for some α>0. Let us assume that x is such that the only way to write x as u + v with u, v ∈ C is to take u, v ∈ R_x. Then, if we can write y as u′ + v′ with u’, v’ ∈ C, we will have x = u + v where u ≝ (1/α)u′ and v ≝ (1/α)v′, both of them in C; so u, v ∈ R_x = R_y, and therefore u′, v′ ∈ R_y as well.

We will see that a polyhedral cone can only have finitely many extreme rays. Some convex cones have infinitely many. For example, the extreme rays of the Lorentz cone Lⁿ⁺¹ (with n≥1) are the rays R_(1,s) where s ranges over the unit sphere of Rⁿ, namely the set of vectors s of Rⁿ such that ||s||= 1, and there infinitely many, in fact uncountably many. I will leave this as an exercise. As hints, you should first show that every ray of Lⁿ⁺¹ can be written as R_(1,s) where 0 < ||s|| ≤ 1. In order to show that this is not an extreme ray if ||s||<1, write s as the sum of (1/2, (1+ϵ)/2 s) and of (1/2, (1−ϵ)/2 s) for ϵ > 0 small enough. When ||s||=1, imagine that (1, s) is the sum of two elements (t₁, s₁) and (t₂, s₂) of Lⁿ⁺¹, show that t₁ = ||s₁|| and t₂ = ||s₂||, then use the (full) Cauchy-Schwarz inequality: s₁^T s₂ ≤ ||s₁|| ||s₂||, with equality if and only if s₁ and s₂ are linearly dependent, with positive coefficients (i.e., s₁=0 or s₂=0 or s₂ = λ s₁ for some λ>0).

At the other end of the spectrum, some convex cones, even some non-trivial convex cones (i.e., some convex cones that are different from {0}) have no extreme ray at all. For example, taking C ≝ R² as a convex cone in R², every non-zero vector x ≝ (x₁, x₂) ∈C can be written as the sum of 1/2 (x₁−x₂, x₁+x₂) and of 1/2 (x₁+x₂, x₂−x₁) , which are both in C, but neither of them is in R_x; in fact neither of them is even a scalar multiple of x. Therefore no ray of C is extreme.

Lemma X. Let C be a polyhedral cone in Rⁿ, generated by a family of vectors x₁, . . . , x_m. The extreme rays of C are among R_x1, . . . , R_xm. If the vectors x₁, . . . , x_m are linearly independent, then the extreme rays are exactly R_x1, . . . , R_xm.

Proof. Let R_x be an extreme ray of C, where x ∈ C–{0}. We write x as ∑_i=1^m α_i x_i, where each α_i is in R₊. We pick i so that α_i > 0; this must exist, otherwise x would be equal to 0. Since x = α_i x_i + ∑_j≠i α_j x_j and R_x is extreme, α_i x_i is in R_x, so x_i ∈ R_x, and hence R_x = R_xi.

We now assume that x₁, . . . , x_m are linearly independent. Let i ∈{1, ···, m}, and let us assume that we can write x_i as u + v with u, v ∈ C . Since x₁, . . . , x_m generate C, we can write u as ∑_i=1^m α_i x_i and v as ∑_i=1^m β_i x_i where α_i, β_i ≥ 0. From x_i = u + v and the fact that the vectors x₁, . . . , x_m are linearly independent, we obtain that 1 = α_i + β_i and α_j + β_j = 0 for every j≠i. Since α_j, β_j ≥0, we must have α_j = β_j = 0 for every j≠i. Therefore u = α_i x_i and v = β_i x_i are in R_xi. This shows that R_xi is an extreme ray of C. ☐

Lemma X implies that a polyhedral cone can only have finitely many rays. We have seen that the Lorentz cone Lⁿ⁺¹ (with n≥1) has uncountably many, hence Lⁿ⁺¹ is not polyhedral.

Lemma Y. Given a pointed polyhedral cone C in Rⁿ, generated by vectors x₁, . . . , x_m, C is already generated by those vectors x_i such that R_xi is an extreme ray of C.

Proof. Let C be generated by vectors x₁, . . . , x_m, with m ∈ N. By Lemma X, all the extreme rays of C are among R_x1, . . . , R_xm. Up to permutation, we will assume that they are those numbered from 1 to p, where 0 ≤ p ≤ m.

For all i, j ∈{1, ···, m}, let us write i ◁ j if and only if x_j is in the cone generated by x₁, . . . , x_i, namely if x_j can be written as a linear combination of x₁, . . . , x_i with non-negative coefficients. Clearly, i ◁ i for every i ∈{1, ···, m}. We claim that: (∗) if i ◁ i+1 ◁ ··· ◁ j, then i ◁ j. This is by induction on j−i. The case j−i=0 is trivial. Otherwise, x_j can be written as ∑_k=1^j–1 α_k x_k where each α_k is in R₊, since j−1 ◁ j. By induction hypothesis, for every k ∈ {i + 1, ···, j−1} we have i ◁ k, so we can write x_k as ∑_l=1ⁱ β_kl x_l where each β_kl is in R₊. Therefore x_j = ∑_l=1ⁱ (∑_k=1^j–1 α_k β_kl) x_l, so i ◁ j.

We claim that k ◁ k+1 ◁ ··· ◁ m for every k ∈ {p, ···, m}. We prove this by induction on m−k. If m−k=0, this is clear. Otherwise, we show that k ◁ k+1 under the induction hypothesis k+1 ◁ k + 2 ◁ ··· ◁ m, where p ≤ k < m. Using the induction hypothesis and (∗), we have k+1 ◁ k+2, k+1 ◁ k+3, . . . , k+1 ◁ m, so x_k+2, . . . , x_m can all be written as linear combinations of x₁, . . . , x_k+1 with non-negative coefficients. It follows that every element of C, which is a linear combination of x₁, . . . , x_m with non-negative coefficients, can be rewritten as a linear combination of x₁, . . . , x_k+1 instead, still with non-negative coefficients. Since p ≤ k < m, R_xk+1 is not an extreme ray of C. Therefore we can write x_k+1 as u + v where u, v ∈ C, and u or v is not in R_xk+1. By symmetry, let us assume that v ∉ R_xk+1. We can write u and v as linear combinations of x₁, . . . , x_k+1, with non-negative coefficients, as we have seen above. Hence let us write u as ∑_i=1^k+1 α_i x_i and v as ∑_i=1^k+1 β_i x_i, where α_i, β_i ∈ R₊. We then have x_k+1 =∑_i=1^k+1 (α_i+β_i) x_i. If α_k+1+β_k+1 ≥ 1, then (1–α_k+1–β_k+1) x_k+1 is in –C, and is equal to ∑_i=1^k (α_i+β_i) x_i, which is in C. Since C is pointed, both are equal to 0. In particular, ∑_i=1^k (α_i+β_i) x_i = 0. The latter implies that ∑_i=1^k α_i x_i, which is in C, is equal to –∑_i=1^k β_i x_i, which is in –C. Since C is pointed, both are equal to 0. Therefore v = ∑_i=1^k+1 β_i x_i = β_k+1 x_k+1 is in R_xk+1, and that is impossible. Hence we must conclude that α_k+1+β_k+1 <1, and therefore x_k+1 =∑_i=1^k $\frac {\alpha_i+\beta_i} {1-\alpha_{k+1}-\beta_{k+1}}$ x_i, so k ◁ k+1.

Now that we know that k ◁ k+1 ◁ ··· ◁ m for every k ∈ {p, ···, m}, we have p ◁ p+1 ◁ ··· ◁ m, so p ◁ l for every l ∈ {p+1, ···, m} by (*). In other words, every x_l with l ∈ {p+1, ···, m} can be expressed as a linear combination of x₁, . . . , x_p with non-negative coefficients. Since every element x of C can be expressed as a linear combination of x₁, . . . , x_p, x_p+1, . . . , x_m with non-negative coefficients, we can rewrite that linear combination as a linear combination of x₁, . . . , x_p alone, still with non-negative coefficients. ☐

The only if direction of the following theorem appears as Lemma 5.1 of [4], and is the only direction that we will be interested in future posts. Both directions were proved by Barker and Loewy [2, Proposition 3.1], see also the survey [1, Theorem 2.B.5]. Barker and Loewy actually consider the case where C is not necessarily proper, in which case R_C,C* must be replaced by its closure.

Theorem Z. Let C be a proper cone in Rⁿ. The identity matrix I_n is in the cross-dual matrix cone R_C,C* if and only if C is simplicial.

Before we prove this, let us take the example of the Lorentz cone Lⁿ⁺¹, with n≥1. We have seen that Lⁿ⁺¹ is not polyhedral, hence certainly not simplicial. This implies that, when C ≝ Lⁿ⁺¹, R_C,C* cannot contain the identity matrix. We have seen in Example L that C ≝ Lⁿ⁺¹ is its own dual, so R_C,C = R_C,C* cannot contain the identity matrix. This is what we proved near the end of the July 2026 post, in the special case n=2.

Proof. If C is simplicial, then it is generated by a linear basis x₁, . . . , x_n of Rⁿ. Let A be the matrix obtained by putting x₁, . . . , x_n side by side, as its columns. Since x₁, . . . , x_n form a linear basis, A is invertible. We can write I_n as AI_nA⁻¹, and rewrite the middle I_n as ∑_i=1ⁿ e_i e_i^T, so I_n = ∑_i=1ⁿ A e_i e_i^T A⁻¹. But A e_i = x_i is in C, and (e_i^T A⁻¹)^T is in C*, since for every j ∈{1, ···, n}, (e_i^T A⁻¹)^TT x_j = e_i^T A⁻¹ x_j = e_i^T e_j, which is equal to 1 if i=j and to 0 otherwise, and is therefore non-negative in any case. In other words, I_n = ∑_i=1ⁿ u_i v_i^T where u_i ≝ A e_i ∈ C and v_i ≝ (e_i^T A⁻¹)^T ∈ C*, showing that I_n ∈ R_C,C*.

We now assume that C is a proper cone such that I_n ∈ R_C,C*, and we aim to show that C is simplicial. We reuse z₀, x₀, C₁ and C*₁ from Proposition Q. Since I_n ∈ R_C,C*, we can write I_n as ∑_i=1^m u_i v_i^T where m≥1, u_i ∈ C and v_i ∈ C*. Without loss of generality, we will assume that no u_i and no v_i is equal to 0. We can also assume that R_ui≠R_uj for all i≠j: otherwise u_j=βu_i for some β>0, and we can rewrite u_i v_i^T + u_j v_j^T as u_i (v_i + βv_j)^T.

We can write each u_i as α_i u’_i for some α_i ∈ R₊ and u’_i ∈ C₁. We have an explicit formula for α_i: since u’_i ∈ C₁, by definition of C₁, z₀^T u’_i = 1, so z₀^T u_i = α_i. No α_i is equal to 0, and we can write u_i v_i^T as u’_i v’_i^T, where v’_i ≝ α_i v_i. Hence I_n = ∑_i=1^m u’_i v’_i^T, where each u’_i is in C₁. Additionally, the rays R_u’i are pairwise distinct, because R_u’i=R_ui.

For every x₁ ∈ C₁, x₁ = I_n x₁ = ∑_i=1^m u’_i v’_i^T x₁ = ∑_i=1^m (v’_i^T x₁) u’_i. We note that the coefficients v’_i^T x₁ are non-negative, since v’_i ∈ C* and x₁ ∈ C. Their sum ∑_i=1^m (v’_i^T x₁) is equal to (∑_i=1^m v’_i^T) x₁. But v’_i^T = α_i v_i^T = z₀^T u_i v_i^T, so ∑_i=1^m v’_i^T = z₀^T ∑_i=1^m u_i v_i^T = z₀^T I_n = z₀^T. Therefore ∑_i=1^m (v’_i^T x₁) = z₀^T x₁, which is equal to 1 since x₁ ∈ C₁. Hence (v’₁^T x₁, ···, v’_m^T x₁) is in ∆_m. We have shown that every element x₁ ∈ C₁ can be written as ∑_i=1^m β_i u’_i, for some (β₁, ···, β_m) ∈ ∆_m, namely β_i ≝ v’_i^T x₁.

Every element x of C can be written as αx₁ for some α ∈ R₊ and x₁ ∈ C₁. Then x = ∑_i=1^m αβ_i u’_i. This shows that C is polyhedral. It is pointed since proper, so by Lemma Y, C is generated by those vectors u’_i such that R_u’i is an extreme ray of C. Up to permutation, we will assume that these vectors are numbered u’₁, . . . , u’_p.

For every j ∈ {1, ··· , p}, u’_j is in C₁, hence can be written as ∑_i=1^m β_i u’_i for some (β₁, ···, β_m) ∈ ∆_m. We have in fact seen that we can take β_k ≝ v’_k^T u’_j for every k ∈ {1, ···, m}. For every k ∈ {1, ···, m}–{j}, we can split the sum ∑_i=1^m β_i u’_i into the sum of β_k u’_k and of ∑_i≠k β_i u’_i. Since R_u’i is an extreme ray of C, both terms must be in R_u’i, in particular the first one. If β_k≠0, this would imply that R_u’i=R_u’k, but we have made sure that the rays R_u’i are pairwise distinct, so this is impossible. It follows that β_k=0 for every k ∈ {1, ···, m}–{j}, and then that β_j=1, since (β₁, ···, β_m) ∈ ∆_m. In other words, v’_k^T u’_j = 0 for every k ∈ {1, ···, m}–{j}, and v’_j^T u’_j = 1.

Now let us imagine that ∑_i=1^p α_i u’_i = 0,  for arbitrary real numbers α₁, . . . , α_p. For every k ∈ {1, ···, p}, we have 0 = v’_k^T ∑_i=1^p α_i u’_i = ∑_i=1^p α_i v’_k^T u’_i = α_k, so all the coefficients α_k are equal to zero. This shows that the vectors u’_i, 1 ≤ i ≤ p, are linearly independent. We remember that C is generated (as a cone) by these vectors. C is proper, hence full-dimensional, hence by Lemma E, every vector of Rⁿ is a linear combination of vectors of C, which are all linear combinations of u’₁, . . . , u’_p. This shows that these vectors form a linear basis of Rⁿ, hence that C is simplicial. ☐

When does R_C,C* = Γ(C) hold?

We finally answer the question in which cases R_C,C* = Γ(C). We had given a partial answer to that question in Lemma R, item 3.

For every A ∈ M_n, for every cone K in Rⁿ, we write AK for {AB | B ∈ K}. Similarly, KB = {AB | A ∈ K}. We will apply this notably to the case K ≝ R₊ⁿ.

Lemma Za. Let A ∈ M_n and K be a cone in Rⁿ. If A is invertible, then (AK)* = (A⁻¹)^T K*.

Proof. Since A is invertible, A^T is invertible, too, and (A^T)⁻¹ = (A⁻¹)^T. Indeed, A^T (A⁻¹)^T = (A⁻¹A)^T = I_n^T = I_n and (A⁻¹)^T A^T = (AA⁻¹)^T = I_n^T = I_n.

The elements w of (AK)* are those vectors such that w^T Au ≥0 for every u ∈ K. Writing w as (A^T)⁻¹ v = (A⁻¹)^T v, w^T Au = v^T A⁻¹ Au = v^T u, so the elements of (AK)* are the vectors (A^T)⁻¹ v such that v^T u ≥0 for every u ∈ K. In other words, (AK)* = (A⁻¹)^T K*. ☐

Lemma Zb. Let A ∈ M_n and K be a cone in Rⁿ. If A is invertible, then Γ(AK)= A Γ(K) A⁻¹ and R_AK,(AK)* = A R_K,K* A⁻¹.

Proof. For every matrix M ∈ M_n, M ∈ Γ(AK) if and only if for every u ∈ AK, Mu ∈ AK, if and only if for every v ∈ K, MAv ∈ AK, if and only if for every v ∈ K, A⁻¹ MAv ∈ K, if and only if A⁻¹MA ∈ Γ(K), if and only if M ∈ A Γ(K) A⁻¹.

An element of R_AK,(AK)* is a finite sum ∑_i=1^m A u_i w_i^T where u_i ∈ K and w_i ∈ (AK)*. By Lemma Za, (AK)* = (A⁻¹)^T K*, so the elements of R_AK,(AK)* are the finite sums ∑_i=1^m A u_i ((A⁻¹)^T v_i)^T, where u_i ∈ K and v_i ∈ K*. Such a finite sum is equal to ∑_i=1^m A u_i v_i^T A⁻¹ = A (∑_i=1^m u_i v_i^T) A⁻¹. Therefore R_AK,(AK)* = A R_K,K* A⁻¹. ☐

Theorem Zc. For a proper cone C in Rⁿ, the following are equivalent:

1. R_C,C* = Γ(C);
2. C is simplicial;
3. there is an invertible matrix A ∈ M_n such that C = A R₊ⁿ.

Proof. 1⇒2. The identity matrix I_n is in Γ(C), so if R_C,C* = Γ(C), then C is simplicial by Theorem Z.

2⇒3. Let us assume that C is simplicial. As in the beginning of the proof of Theorem Z, the matrix A whose columns are any given linear basis x₁, . . . , x_n that generate C is invertible. The elements of C are the linear combinations ∑_i=1ⁿ α_i x_i where u ≝ (α₁, ···, α_n) ∈ R₊ⁿ, namely the vectors Au with u ∈ R₊ⁿ. Hence C = A R₊ⁿ.

3⇒1. Let us write C as AK, where A is an invertible matrix and K ≝ R₊ⁿ. Then Γ(C) = Γ(AK)= A Γ(K) A⁻¹ by Lemma Zb, Γ(K) = R_K,K* by Lemma R, item 3, so Γ(C) = A R_K,K* A⁻¹. The latter is equal to R_AK,(AK)*, namely to R_C,C* , by Lemma Zb. ☐

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— Jean Goubault-Larrecq (September 20th, 2026)