When do two stochastic matrices multiply to the identity?

Question

In order to characterize equivalent experiments in the Blackwell order, I would be interested in what we can say about the problem in the question.

Similarly, this was already asked: When are the inverses of stochastic matrices also stochastic matrices?

I did not find any reference, so I proved the following, hopefully correctly (1,2 do not hinge on $C,G$ being stochastic)

Lemma Fix $n,m\in\mathbb{N}$. Let $C\in\mathbb{R}^{n\times m},G\in\mathbb{R}^{m\times n}$ be stochastic matrices (i.e. coloumns summing to 1, positive entries) such that $CG=Id_n$. Then:

1. $n\leq m$
2. $C$ and $G$ have full rank equal to $n$
3. $C$ has an entry equal to $1$ on each row
4. In particular, if $m=n$ $C,G$ are $n\times n$ permutation matrices

Proof Observe that, from linear algebra we know $$n=\mathsf{rank}(Id_n)=\mathsf{rank}(CG)\leq\min\{\mathsf{rank(C)},\mathsf{rank}(G)\}\leq min\{m,n\}$$ where the first inequality is a fact about rank of products and the second inequality we exploit the fact that the rank of a matrix is bounded by the minimum of its dimensions.\ If it was $m<n$, the inequality could not attain, and we can conclude $n\leq m$. In turn we can write: $$n=\mathsf{rank}(CG)\leq\min\{\mathsf{rank(C)},\mathsf{rank}(G)\}\leq\min\{m,n\}=n$$ and we can conclude that $\mathsf{rank(C)}=n=\mathsf{rank}(G)$. Then observe that, for any $j\in\{1,\dots,n\}$ $$1=(CG)_{jj}=\sum_{l=1}^m c_{jl}g_{lj}\leq\max_{l=1,\dots,m}\{c_{j,l}\}\sum_{l=1}^mg_{lj}=\max_{l=1,\dots,m}\{c_{j,l}\}\leq 1$$ where in the last equality we use the fact $G$ is stochastic, so its $j$-th coloumn sums to $1$, and in the last inequality that $C$ is stochastic, so that all its entries are in $[0,1]$. It follows that for each row $j$ of $C$ there is at least one entry which is $1$.
For the second fact, let $m=n$ and notice that by stochasticity if $c_{ij}=1$, then all other entries in the $j$-th coloumn of $C$ are $0$. It follows that the $n$ coloumns of $C$ are elements of the canonical basis of $\mathbb{R}^n$. Since $C$ has rank $n$, we can conclude that they are linear independent. In turn $G$ is the inverse of a permutation matrix hence a permutation matrix.

I would like to know if the proof is correct and I would be interested in if something more can be said on the structure of $C,G$ in the non square case (point 3).

Answer 1

The proof looks correct. It seems possible to derive some further interesting properties about $C$ and $G$ in the non $n\times n$-case:

From $$ 1=(CG)_{jj}=\sum_{l=1}^mc_{jl}\,g_{lj}\quad\quad\text{ and }\quad \sum_{l=1}^m\,g_{lj}=1 $$ we can conclude:

• If $c_{jl}<1$ for some $l$ then $g_{lj}=0\,.$

Otherwise we would get $$ 1=\sum_{l=1}^mc_{jl}\,g_{lj}<\sum_{l=1}^mg_{lj}=1\,. $$ Further:

• If $c_{jl}=1$ then $\sum_{i=1}^nc_{il}=1$ implies that for all rows $i\not=j$ the element $c_{il}$ must be zero. Therefore, $$ g_{li}=0\quad\quad\forall i\not=j\,. $$

Since we know that for each $j$ there must be at least one $l$ such that $c_{jl}=1$ we conclude:

• For every $j$ there must be at least one $l$ such that $g_{li}=0$ for all $i\not=j\,.$

Let's sort the columns of $C$ and the rows of $G$ such that the element $c_{ii}=1$ for all $i=1,...,n\,.$ We know that for each $i$ at least one such element must exists. Because every column of $C$ is stochastic the sorted matrices must be of the form $$ C=\left(\begin{matrix}I&C'\end{matrix}\right) $$ where $I$ is the $n\times n$-identity matrix and $C'$ an $n\times(m-n)$-matrix. The (row sorted) matrix $G$ must be of the form $$\tag{1} G=\left(\begin{matrix}D\\G'\end{matrix}\right) $$ where $D={\rm diag}(g_{11},...,g_{nn})$ and $G'$ is an $(m-n)\times n$-matrix.

Now:

• If $g_{ii}<1$ for some $i=1,...n$ then there must exist at least one $g_{li}>0$ (because $\sum_{l=1}^ng_{li}=1$). Because of (1) this $l$ must be greater than $n$. Therefore, $$ c_{li}=1\,, $$ and this is an element of $C'\,.$

So:

• If $D={\rm diag}(g_{11},...,g_{nn})$ is not the identity matrix then we can sort $C'$ as we sorted $C$ and find a (possibly smaller) identity matrix at the top left corner of $C'$ and a (possibly smaller) diagonal matrix at the top left corner of $G'$.

• Clearly, $C'G'$ must be a diagonal matrix that sums to the identity matrix with $D$: $$ I=CG=D+C'G'\,. $$

A typical example would be $$ C=\left(\begin{matrix}1&0&1&\frac{1}{2}\\0&1&0&\frac{1}{2}\end{matrix}\right)\,,\quad\quad G=\left(\begin{matrix}\frac{1}{2}&0\\0&1\\\frac{1}{2}&0\\0&0\end{matrix}\right) $$ where $$ C'=\left(\begin{matrix}1&\frac{1}{2}\\0&\frac{1}{2}\end{matrix}\right)\,,\quad D=\left(\begin{matrix}\frac{1}{2}&0\\0&1\end{matrix}\right)\,,\quad G'=\left(\begin{matrix}\frac{1}{2}&0\\0&0\end{matrix}\right)\,,\quad C'G'=\left(\begin{matrix}\frac{1}{2}&0\\0&0\end{matrix}\right)\,. $$ The question is if there are examples where $G'$ is not a diagonal matrix. The examples $$ C=\left(\begin{matrix}1&0&0&1\\0&1&1&0\end{matrix}\right)\,,\quad\quad G=\left(\begin{matrix}\frac{1}{2}&0\\0&1\\0&0\\\frac{1}{2}&0\end{matrix}\right) $$ or $$ C=\left(\begin{matrix}1&0&0&1\\0&1&1&0\end{matrix}\right)\,,\quad\quad G=\left(\begin{matrix}\frac{1}{2}&0\\0&\frac{1}{2}\\0&\frac{1}{2}\\\frac{1}{2}&0\end{matrix}\right) $$ are too trivial because they can be sorted so that $C'=I$ and $G'$ is diagonal.