I am sorry that the following question is elementary. I have not received an answer from my post at Math Stack Exchange.

In the following question, all cones are convex and contain the origin. Let $C \subset \mathbb{R}^{m}$ be a cone consisting of $m$-dimensional column vectors, and $D \subset \mathbb{R}^{n}$ another cone of $n$-dimensional column vectors.

Given any subsets $A \subset \mathbb{R}^{m}$ and $B\subset \mathbb{R}^{n}$, define $A \otimes B \subset \mathbb{R}^{m \times n}$ as the set of $m \times n$ matrices of the form $x_1 {y_1}^T + \dotsb + x_q {y_q}^T$, where $x_1, \dotsc, x_q \in A$ and $y_1, \dotsc, y_q \in B$, $q\geq 1$ is an arbitrary positive integer, and $^T$ denotes the transpose.

Is it true that $$(C \otimes D)^\circ = C^\circ \otimes D^\circ,$$ where $^\circ$ denotes the interior relative to the respective euclidean topologies?

Here's an example where the said equality holds. Take $C = {\mathbb{R}_{\ge 0}}^m$ to be the cone of all $m$-dimensional column vectors whose every entry is a nonnegative real number. Similarly, take $D = {\mathbb{R}_{\ge 0}}^n$. Then $C \otimes D = {\mathbb{R}_{\ge 0}}^{m \times n}$ is the cone of $m \times n$ matrices whose every entry is nonnegative. Indeed, it suffices to show that $\mathbf{E}_{ij}$, the matrix whose $(i, j)$th entry is $1$ and has $0$s everywhere else, lies in $C \otimes D$, because $C \otimes D$ is closed under taking nonnegative linear combinations. But $\mathbf{E}_{ij} = \vec{e_i}\vec{e_j}^T$, where $\vec{e_i} \in C$ and $\vec{e_j} \in D$. Therefore $(C \otimes D)^\circ = {\mathbb{R}_{> 0}}^{m \times n}$ is the set of all $m \times n$ matrices whose every entry is strictly positive.

Noting that $C^\circ = {\mathbb{R}_{> 0}}^m$ consists of all $m$-dimensional vectors whose entries are all strictly positive real numbers, and similarly $D^\circ = {\mathbb{R}_{> 0}}^n$, we get $C^\circ \otimes D^\circ = {\mathbb{R}_{> 0}}^{m \times n}$ also.

  • $\begingroup$ If $x_1\in C$ and $y_1\in D$, then $x_1$ is $m\times1$ and $y_1^T$ is $n\times1$. What can then $x_1y_1^T$ mean? $\endgroup$ Apr 17, 2023 at 20:20
  • $\begingroup$ Iosif, thanks for pointing out my error. The transpose should not be there. $\endgroup$
    – Colin Tan
    Apr 18, 2023 at 4:48

1 Answer 1


Yes, $(C \otimes D)^\circ = C^\circ \otimes D^\circ$ is correct.

Let us start by proving the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$. To this end, it is enough to show that $C^\circ \otimes D^\circ$ is open. For a given point $z = x_1 {y_1}^T + \dotsb + x_q {y_q}^T \in C^\circ \otimes D^\circ$, we can use convexity and openness to replace $x_1$ by a convex combination of $n$ linearly independent points in $C$, and similarly for $y_1$. In this way, we can assume without loss of generality that the $x_i$'s span $\mathbb{R}^m$, and similarly the $y_i$'s span $\mathbb{R}^n$. Therefore by nudging each one of these points a little (without leaving $C^\circ$ or $D^\circ$), we can move in any direction in $C^\circ \otimes D^\circ$, and it follows that the latter set is open.

Furthermore, the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$ is dense, because $C^\circ \otimes D^\circ$ is dense even in $C \otimes D$ by the obvious termwise approximation argument.

To now arrive at the claimed equality, recall that every convex open set is the interior of its closure. This applies in particular to both $C^\circ \otimes D^\circ$ and $(C \otimes D)^\circ$. This gives $$ C^\circ \otimes D^\circ = \overline{C^\circ \otimes D^\circ}^\circ = \overline{(C \otimes D)^\circ}^\circ = (C \otimes D)^\circ, $$ where the second step is the density from the previous paragraph.

  • $\begingroup$ For the first paragraph, to prove that $C^\circ \otimes D^\circ$ is open, do we actually need the stronger assumption that $x_i \otimes y_j = x_i {y_j}^T$ span $\mathbb{R}^n \otimes \mathbb{R}^m \cong \mathbb{R}^{m \times n}$? $\endgroup$
    – Colin Tan
    Apr 27, 2023 at 6:51
  • $\begingroup$ I was trying to write out the argument explicitly. Say the nudging is in the direction $x \otimes y$, where $x = \sum_i \lambda_i x_i$ and $y = \sum_j \mu_j y_j$ for scalars $\lambda_i, \mu_j$. Then $x \otimes y = \sum_{i, j} \lambda_i \mu_j x_i \otimes y_j$. So what do I do with the cross-terms $x_i \otimes y_j$ where $i \neq j$? $\endgroup$
    – Colin Tan
    Apr 27, 2023 at 6:53
  • $\begingroup$ Ah sorry, I think my formulation of that didn't fully capture what I actually meant. What I meant is that we can assume without loss of generality that $z = w + \sum_{i,j} \beta_{i,j} x_i \otimes y_j$, where the $x_i$ and the $y_j$ are bases, $\beta_{i,j} > 0$ for all $i$ and $j$, and $w$ contains all the remaining terms. Is this clearer? Then you can move in all direction simply by perturbing the coefficients $\beta_{i,j}$. $\endgroup$ Apr 27, 2023 at 12:21

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