This is not duplicate of A matrix's element proof, but it is harder than that one.
Given an constant $\alpha \in (0,1)$, and an $n \times n$ matrix $X$ whose all entries are between 0 and 1, and each row sum of $X$ is 1, and ${\|X\|}_{\infty} \le 1$. Suppose $A=\sum_{i=0}^{\infty} {\alpha}^i X^i ,$ $B=\sum_{i=0}^{\infty} \frac {{\alpha}^i}{i!} X^i ,$
I've done some experiments and found that :
For every two entries $(a,b)$ and $(c,d)$ ,
- if $[A]_{a,b} \ge [A]_{c,d}$, then $[B]_{a,b} \ge [B]_{c,d}.$
(Note that I use $[A]_{i,j}$ to denote the $(i,j)$-entry of the matrix $A$)
How can I prove this result mathmetrically? Any suggestions are warmly welcome.
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** I leak out one condition that each row sum of $X$ is 1.
** The subscript of sum should be starting from 0 (rather than 1) that is, $A=\sum_{i=0}^{\infty} {\alpha}^i X^i $