A question on matrix multiplication Let say I have set of real valued square matrices $\left(A_0, A_1, A_2, ..., A_n \right)$. All these matrices have constraint such that all row-sum of individual matrices equal to 1.
My question the row-sum of a matrix which is computed by multiplication of any subset of above set of matrices will remain to be 1? If yes, is there any matrix rule that govern this?
 A: The matrices you are referring to are row-stochastic. You can prove that the sum of each row will remain equal to one in many different ways. I will show it based on convex optimization arguments. Notice that you just need to show that the product of two of such matrices is one, since if it holds for 2 matrices, it also holds for $n$ matrices.
Consider then $A=X_1$ and $X_2$. Pick one row of $A$, let's name it $\alpha$. Then:
$$ \alpha^\top X= \alpha_1 x_1^\top + \ldots + \alpha_n x_n^\top$$
where $x_i^\top$ represents the $i$th row of matrix $X$.  Since $\sum_{i=1}^n \alpha_i =1 $  what you are actually doing is a convex combination of vectors $x_i^\top$ which have sum-entries equal to one. As a result, the vector $\alpha^\top X$ has also a row-sum equal to one.
This is easy to show for instance if the number of row vectors in $X$ is $n=2$, because you would have:
$$\alpha_1 x_1^\top + (1-\alpha_1) x_2^\top$$ and if you sum the entries of this vector you obtain (assume each $x_n$ has $m$ entries):
$$\sum_{l=1}^m \alpha_1 x_1(m) + (1-\alpha_1) x_2(m)=  \alpha_1 \sum_{l=1}^m x_1(m) + (1-\alpha_1) \sum_{l=1}^mx_2(m)= \alpha+ 1 - \alpha=1$$
