A stochastic matrix, with elements $\in[0,1]$ and rows summing to 1 are known to have one eigenvalue 1 (stationary distribution) and the rest of lower magnitude. However I don't know about many results regarding their inverses.
- In which cases is it possible to find another stochastic matrix being the inverse to the first?
- If we can not find an inverse fulfilling the requirements, can we find some "pseudo" inverse?
- Can we "expand" the probability space to find a larger space where it is possible to find a matrix ?
EDIT: The closest I have come so far is to expand the space to double number of states and allowing elements > 1. Let us call the $N\times N$ probability transition matrix $\bf P$, then assuming $\bf P$ nonsingular we can calculate ${\bf P}^{-1}$, build the new matrices :
$${\bf P_e = P \otimes I}_2\hspace{1cm}{\bf P_{ie}} = |{\bf P}^{-1}|\otimes ({{\bf 1}_2{\bf 1}_2}^T-{\bf I}_2)^{(1-\text{sgn}({\bf P}))\otimes {\bf I}_2}$$
Where the $|\cdot|$ and power are scalar wise operations.
If we do this then of course $\bf P_{ie}P_e \neq I\hspace{1cm}\bf P_{e}P_{ie} \neq I$, but that is because we are not finished yet!
If we now turn each $2\times2$ block $\bf A$ into a scalar by the following calculation: $[1,0] {\bf A} [1,-1]^T$
And systematically apply to all blocks:
$$\cases{({\bf I}_N \otimes [1,0])({\bf P_{ie}P_e})({\bf I}_N \otimes [1,-1]^T)\\
({\bf I}_N \otimes [1,0])({\bf P_{e}P_{ie}})({\bf I}_N \otimes [1,-1]^T)}$$
- Now we do indeed get identities!
- Row sums of $\bf P_{ie}$ equals 1 if we treat every second bin as $-1$ (as the construction above does).
- But how to interpret the elements of $\bf {P_{ie}}$ which can be >1 ?
Furthermore $\sum_j |({\bf P_{ie}})_{ij}|$ seems to be a measure of the confusion, it goes to $\infty$ as ${\bf P} \to \frac{1}{N^2}{{\bf 1}_N} {{\bf 1}_N}^T$ which is indeed the maximally confusing distribution as we lose all information except the mean values ( all other eigenvalues except stationary distribution will be 0 ).
Best Answer
The only stochastic matrices with a stochastic inverse are the permutation matrices. If $A$ and $B$ are stochastic, then all entries of $AB$ are non-negative. The only way $(AB)_{ij}$ can be zero is if the "support" of row $i$ of $A$ and column $j$ of $B$ are disjoint. So if $AB=I$ we need row $1$ of $A$ to be orthogonal to columns $2,\ldots,n$ of $B$. As these columns must be linearly independent, then row $1$ of $A$ can only have one non-zero entry. Continuing, all rows of $A$ have only one nonzero entry etc.