If $M$ maps all probability vectors on a subspace to some probability vectors, is $M$ the restriction of a column stochastic matrix?

Question

For $n \ge 1$, let $$\Delta^{n-1} := \left\{ (x_1,\dots,x_{n}) \in \mathbb{R}^{n} \mid \sum_{i=1}^{n} x_i=1,~x_i \ge 0 \right\}$$ and let $\mathcal{S},~\mathcal{S}'\subset\mathbb{R}^n$ be subspaces such that $\mathcal{S} \cap \Delta^{n-1} \ne \emptyset$ and $\mathcal{S}' \cap \Delta^{n-1} \ne \emptyset$. Consider a linear map $M:\mathcal{S}\to\mathcal{S}'$ such that $M(\mathcal{S} \cap \Delta^{n-1}) \subseteq \mathcal{S}' \cap \Delta^{n-1}$. The problem is to prove whether or not there exists a linear map $T_Q:\mathbb{R}^n\to\mathbb{R}^n$ given by some column stochastic matrix ($Q_{ij}\geq0$, $\sum_iQ_{ij}=1$, in the standard basis), such that $Q v=M(v)$ for every $v\in \mathcal{S} \cap \Delta^{n-1}$. Has this extension problem been already studied?

Answer 1

Here is a minimal counterexample for $n = 3$.

Let $S$ and $S'$ be the subspaces defined by $x_1 = x_2 + x_3$ and $x_1 = 0$ respectively. It is easy to see that $S \cap \Delta$ and $S' \cap \Delta$ are two segments with endpoints $a_2 := (1/2, 1/2, 0), a_3 := (1/2, 0, 1/2)$ and respectively $b_2 := (0,1,0), b_3 := (0,0,1)$.

Clearly the linear map $M\colon S \to S'$ defined by $M(x_1, x_2, x_3) := (0, 2x_2, 2x_3)$ satisfies the condition $M(S \cap \Delta) \subset S' \cap \Delta$ as $M$ sends $a_2$ to $b_2$ and $a_3$ to $b_3$.

It is then easy to see that if a $3 \times 3$ non-negative matrix $Q$ satisfies $$\begin{pmatrix}Q_{11} & Q_{12} & Q_{13} \\ Q_{21} & Q_{22} & Q_{23}\\ Q_{31} & Q_{32} & Q_{33}\end{pmatrix}\begin{pmatrix}1/2 & 1/2 \\ 1/2 & 0\\ 0 & 1/2\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 1 & 0\\ 0 & 1\end{pmatrix},$$ then $Q_{ij} = 0$ for all $(i,j) \neq (2,2), (3,3)$. Hence there is no column stochastic matrix $Q$ that sends $a_2$ to $b_2$ and $a_3$ to $b_3$.