# A hard proof of two matrix's elements

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, $$A=\sum_{i=0}^{\infty} {\alpha}^i X^i$$

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I edited your other question and corrected the typos, perhaps you would do the same for this version of your question! – Gigili May 31 '12 at 7:58
The assumption seems to be the same, but the conclusion is much harder for me to prove. – John Smith May 31 '12 at 8:03

It's not true. Note that $A = (I-\alpha X)^{-1} - I$ and $B = \exp(\alpha X) - I$. A consequence of your claim would be that if $A_{ab} = A_{cd}$ then $B_{ab} = B_{cd}$. Try $X = \pmatrix{1/10 & 29/40\cr 1/5 & 1/5\cr}$ and $\alpha = 4/5$. I get $A = \pmatrix{4/17 & 29/34\cr 4/17 & 6/17\cr}$ so $A_{11} = A_{21}$. But $B \approx \pmatrix{.135321868 & .6642856308 \cr .1832512085 & .226947472\cr}$ so $B_{11} \ne B_{21}$.

EDIT: If the sum starts at $0$, then $A = (I-\alpha X)^{-1}$ and $B = \exp(\alpha X)$. Let's try for a $3 \times 3$ matrix where $A_{12} = A_{23}$ with $\alpha = 1/2$. $$X = \left( \begin {array}{ccc} \frac15&\frac15&\frac35\\ {\frac {23}{60}}&\frac25&{\frac {13}{60}}\\ \frac{1}{10}&\frac{1}{2}&\frac25 \end {array} \right),\ A = \left( \begin {array}{ccc} {\frac {2942}{2445}}&{\frac {248}{815}}&{\frac {1204}{2445}}\\ {\frac {254}{815}}&{\frac { 1128}{815}}&{\frac {248}{815}}\\ {\frac {422}{2445}} &{\frac {368}{815}}&{\frac {3364}{2445}}\end {array} \right)$$ $$\ B \approx \left( \begin {array}{ccc} 1.127701142& 0.1620598118& 0.3589603173\\ 0.2284419794& 1.252387244& 0.1678920478 \\ 0.0872176063& 0.3115915010& 1.249912163 \end {array} \right)$$

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Sorry, I leak out that each row sum of $X$ is 1. – John Smith May 31 '12 at 8:33
I also corrected one mistake that the index of summations in $A$ and $B$ should go from 0 (rather than 1). This would ensure that $A$ and $B$ are diagonally dominant. I think in this case, your counterexample won't apply. – John Smith May 31 '12 at 9:07
@Robert: I copied your TeX for your matrices to write mine, and it's different from how I usually make matrices. Does \pmatrix just know how big to make the matrix, and then put parentheses around it? And what is \cr? – mixedmath May 31 '12 at 10:08
\pmatrix is plain TeX rather than LaTeX. It just happens to be what I'm used to. It makes a matrix delimited by parentheses. \cr goes to the next row. – Robert Israel May 31 '12 at 16:21

The result is still not true.

Suppose $X = \pmatrix{1/10 & 9/10 \cr 1/2 & 1/2 \cr}$ and $\alpha = 4/5$.

Then I get

$$A \approx \pmatrix{2.272 & 2.727 \cr 1.515 & 3.484 \cr} \qquad B \approx \pmatrix{1.261 & 0.963 \cr .535 & 1.690 \cr}$$

And $[A]_{(1,2)} > [A]_{(1,1)}$ but $[B]_{(1,2)} <[B]_{(1,1)}$

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I corrected the index $i$ goes from 0, not start from 1. So $A$ and $B$ should be diagonally dominant. – John Smith May 31 '12 at 9:53
@John: These are the sums from $0$. I'll note that in this case, even $\sum_0^8 \alpha^i X^i$ has that $[A]_{(1,2)} > [A]_{(1,1)}$ – mixedmath May 31 '12 at 10:00