Which similarity transformations preserve stochasticity of a matrix.

Question

An $n\times n$ real matrix is called stochastic (aka Markov) if it is entry-wise non-negative and it's rows sum to 1.

Let $\mathcal{M}_n^+$ be the set of stochastic matrices. I am interested in the following question:

What are the conditions on $S\in GL(n,\mathbb{C})$ such that $S^{-1}\mathcal{M}_{n}^+S\subseteq\mathcal{M}_n^+.$

The first thing I note is that conjugation is invertible, so this condition automatically implies that $S^{-1}\mathcal{M}_{n}^+S=\mathcal{M}_n^+.$

There are also a few further simplifications we can make: The set of stochastic matrices can be expressed as the convex hull of stochastic matrices whose entries are 0 or 1. This is a result analogous to the Birkhoff–von Neumann theorem, and is proved for example in Lippold (2024). Thus a necessary and sufficient condition is that $S$ conjugates any 0/1 stochastic matrix to a stochastic matrix.

My initial guess was that any such $S$ must be a permutation matrix (or the negative of a permutation matrix), however the following $2\times 2$ example shows that this is not necessarily the case: Consider $S= \begin{bmatrix} \sqrt{3} & 2\\2& \sqrt{3}\end{bmatrix}$, whose inverse is $S^{-1}=\begin{bmatrix}-\sqrt{3} & 2\\ 2 & -\sqrt{3}\end{bmatrix}$. To see that conjugation by $S$ preserves stochasticity it suffices to show that conjugation by $S$ sends the following 0/1 stochastic matrices to stochastic matrices: $\begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix},\begin{bmatrix}1&0\\1&0\end{bmatrix}, \begin{bmatrix}0&1\\0&1\end{bmatrix}.$ This can be done by routine calculation.

If anyone knows of a characterization of which similarity transformations preserve stochasticity, or can provide a reference to relevant literature I would be much obliged.

Answer 1

Let $S\in GL_n(\mathbb R)$ and $f:X\mapsto S^{-1}XS$. Here are some partial results:

1. If $f(\mathcal M_n^+)\subseteq\mathcal M_n^+$, then $S$ is a scalar multiple of a row-stochastic matrix.
2. Furthermore, the $S$ above is, up to scaling, either a permutation matrix or a positive row-stochastic matrix.
3. If $f(\mathcal M_n^+)=\mathcal M_n^+$, then $S$ is a scalar multiple of permutation matrix.
4. When $n=2$, $f(\mathcal M_2^+)\subseteq\mathcal M_2^+$) if and only if $S$ is a nonzero scalar multiple of an invertible doubly stochastic matrix.

Proofs.

1. Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$ and $e=\sum_ie_i$ be the vector of ones. Since $ee_i^T$ is row-stochastic, so is $S^{-1}ee_i^TS$. Therefore $S^{-1}ee_i^TSe=e$ for every $i$. Hence $e_1^TSe=\cdots=e_n^TSe=k$ and $S^{-1}e=\frac{1}{k}e$ for some constant $k$. It follows that the row-stochastic matrix $S^{-1}ee_i^TS$ is identical to $\frac{1}{k}ee_i^TS$. Hence $\frac{1}{k}e_i^TS$ is a probability vector for each $i$. Thus $\frac{1}{k}S$ is row-stochastic.

2. By absorbing $\frac{1}{k}$ into $S$, we may assume that $S$ itself is row-stochastic. If $S^{-1}$ is nonnegative, then $S$ must be a permutation matrix. If $S^{-1}$ has some negative entries instead, let $(S^{-1})_{kl}=-m<0$ be its smallest element. For any $i\ne j$, let $B = (e-e_l)e_i^T + e_le_j^T$. Then $B$ is row-stochastic and so is $S^{-1}BS$. In particular, the $k$-th row of the latter must be nonnegative, i.e., \begin{align*} 0\le e_k^TS^{-1}BS &=e_k^TS^{-1}[(e-e_l)e_i^T + e_le_j^T]S\\ &=[(1+m)e_i^T - me_j^T]S\\ &=(1+m)S_{i\ast} - mS_{j\ast}.\\ \end{align*} Therefore $0\le S_{j\ast}\le\frac{1+m}{m}S_{i\ast}$. It follows that if $S$ has a zero entry, then all entries on the same column are zero. Yet this is impossible, because $S$ is invertible. Hence $S$ must be entrywise positive.

3. If $f(\mathcal M_n^+)=\mathcal M_n^+$, then by result 1, we may assume that $S$ is row-stochastic. Also, we have $f^{-1}(\mathcal M_n^+)\subseteq\mathcal M_n^+$. Therefore $S^{-1}$ is also row-stochastic. Hence $S$ must be a permutation matrix.

4. Proof omitted.

Remarks.

• The converse of result 1 is not true. E.g. when $n\ge3$, some doubly stochastic matrices may fail to conjugate row-stochastic matrices to row-stochastic matrices: consider $$ S=\frac{1}{32}\pmatrix{14&9&9\\ 9&17&6\\ 9&6&17}, \quad S^{-1}=\frac{1}{55}\pmatrix{253&-99&-99\\ -99&157&-3\\ -99&-3&157}. $$ Then $$ S^{-1}\pmatrix{1&0&0\\ 1&0&0\\ 0&1&0}S =\frac{1}{1760}\pmatrix{1265&-297&792\\ 785&471&504\\ -15&1751&24} \not\ge0. $$
• In view of the previous counterexample and result 4, it is interesting to ask whether $S$ must be doubly stochastic when $f(\mathcal M_n^+)\subseteq\mathcal M_n^+$.
• $f$ is an example of a linear preserver of $\mathcal M_n^+$. In Li et al., Linear maps preserving permutation and stochastic matrices, Linear Algebra and its Applications, 341 (2002) 5-22, the authors have shown (cf. prop. 4.3, pp. 16-17) that a linear operator $T:M_n(\mathbb R)\to M_n(\mathbb R)$ preserves all row-stochastic matrices if $T(\mathcal M_n^+)=\mathcal M_n^+$, or $T(X)=x_{i1}R_1+\cdots+x_{in}R_n$ for some fixed index $i$ and for some extreme points $R_1,\ldots,R_n$ of $\mathcal M_n^+$. Unfortunately, this result is an ‘if’ but not an ‘only if’. It is not a complete characterisation of linear preservers of $\mathcal M_n^+$ and I cannot use it to filter out those $S$ that conjugate stochastic matrices to stochastic matrices.