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.