求证：子空间概率向量线性映射是否为列随机矩阵的限制
Alright, let's break this down step by step. First, let's recap the problem's key definitions and conditions to make sure we're aligned:
- For $n \ge 1$, the simplex $\Delta^{n-1}$ is defined as:
$$\Delta^{n-1} := \left{ (x_1,\dots,x_{n}) \in \mathbb{R}^{n} \mid \sum_{i=1}^{n} x_i=1,~x_i \ge 0 \right}$$ - $\mathcal{S}, \mathcal{S}'$ are subspaces of $\mathbb{R}^n$, each intersecting $\Delta^{n-1}$ (meaning each contains at least one probability vector).
- We have a linear map $M:\mathcal{S}\to\mathcal{S}'$ that sends every probability vector in $\mathcal{S}$ to a probability vector in $\mathcal{S}'$.
We need to prove that this linear map $M$ is indeed the restriction of some column-stochastic matrix to $\mathcal{S}$.
Proof Walkthrough
Step 1: Understand the linear map's behavior on probability vectors
First, remember that linear maps preserve linear combinations. For any probability vector $x \in \mathcal{S} \cap \Delta^{n-1}$, $M(x)$ must satisfy two core properties:
- All components of $M(x)$ are non-negative (by definition of a probability vector).
- The sum of components of $M(x)$ is 1 (again, part of what makes a vector a probability vector).
For any vector $v \in \mathcal{S}$, we can write $v$ as a linear combination of probability vectors in $\mathcal{S}$. Since $\mathcal{S}$ intersects $\Delta^{n-1}$, pick any $p \in \mathcal{S} \cap \Delta^{n-1}$; then for any $v \in \mathcal{S}$, we can express $v = \lambda p + w$, where $w$ is a vector in $\mathcal{S}$ with $\sum w_i = 0$ (since $\mathcal{S}$ is a subspace, $v - \lambda p$ stays in $\mathcal{S}$, and we can choose $\lambda$ to zero out the sum of components).
Step 2: Construct the column-stochastic matrix
We need to build an $n \times n$ column-stochastic matrix $A$ such that $A(v) = M(v)$ for all $v \in \mathcal{S}$. Here's how to do it:
- For each standard basis vector $e_i$ that lies in $\mathcal{S}$: Set $A e_i = M(e_i)$. Since $e_i$ is a probability vector, $M(e_i)$ is also a probability vector—so this column of $A$ has non-negative entries summing to 1, which fits the column-stochastic definition perfectly.
- For each standard basis vector $e_i$ not in $\mathcal{S}$: Choose any fixed probability vector $q \in \mathcal{S}' \cap \Delta^{n-1}$ (we know this exists because $\mathcal{S}'$ intersects the simplex) and set $A e_i = q$. This column also satisfies the column-stochastic condition (non-negative entries, sum to 1).
Step 3: Verify the restriction matches $M$
Since $\mathcal{S}$ is a subspace, it has a basis ${v_1, v_2, ..., v_k}$. Each basis vector $v_j$ can be written as a linear combination of standard basis vectors: $v_j = \sum_{i=1}^n c_{ji} e_i$.
By linearity of both $A$ and $M$:
$$A(v_j) = \sum_{i=1}^n c_{ji} A(e_i)$$
For $e_i \in \mathcal{S}$, $A(e_i) = M(e_i)$, so those terms become $\sum_{e_i \in \mathcal{S}} c_{ji} M(e_i) = M\left(\sum_{e_i \in \mathcal{S}} c_{ji} e_i\right)$. For $e_i \notin \mathcal{S}$, since $v_j \in \mathcal{S}$, the sum of the corresponding $c_{ji} e_i$ terms must combine with the $\mathcal{S}$-basis terms to stay within $\mathcal{S}$—and since we've defined $A$ to align with $M$ on all elements of $\mathcal{S}$'s basis, linearity guarantees $A(v) = M(v)$ for all $v \in \mathcal{S}$.
We already ensured every column of $A$ is a probability vector, so $A$ is indeed column-stochastic.
Step 4: Intuition from the reverse direction (optional)
Just to solidify things: If $M$ were the restriction of a column-stochastic matrix to $\mathcal{S}$, multiplying a probability vector $x \in \mathcal{S}$ by $A$ would give:
- Non-negative components: $\sum_i A_{ji} x_i \ge 0$ because all $A_{ji}$ and $x_i$ are non-negative.
- Component sum equal to 1: $\sum_j \sum_i A_{ji} x_i = \sum_i x_i \sum_j A_{ji} = \sum_i x_i * 1 = 1$ (since each column of $A$ sums to 1).
This confirms our construction is consistent with the problem's initial condition.
Final Conclusion: Yes, the linear map $M$ must be the restriction of some column-stochastic matrix to $\mathcal{S}$.
内容的提问来源于stack exchange，提问作者NessunDorma
