# A matrix's element proof

Thanks again for copper.hat and Robert Israel's quick immediate reply. While I am modifying the questions, they've already given the answer. Now in this thread, I've changed it back to the original version, and I put the new version which is more challenging in the following thread: A hard proof of two matrix's elements. Hopefully, some brilliant experts could give me suggestions.

Given an constant $\alpha \in (0,1)$, and an $n \times n$ matrix $X$ whose all entries are between 0 and 1, and ${\|X\|}_{\infty} \le 1$. Suppose $$A=\sum_{i=1}^{\infty} {\alpha}^i X^i ,$$ $$B=\sum_{i=1}^{\infty} \frac {{\alpha}^i}{i!} X^i ,$$

I've done some experiments and found that :

• For each entry $(a,b)$, $[A]_{a,b} \ge [B]_{a,b} .$

(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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Sorry for the carelessness, I've modified the questions. – John Smith May 31 '12 at 7:14
I've put the modified version into a new thread math.stackexchange.com/questions/151921/… – John Smith May 31 '12 at 7:30

Note: This was an answer to the original question:

For each $i$, you have $[\frac {{\alpha}^i}{i!} X^i]_{j,k} \leq [{\alpha}^i X^i]_{j,k}$, since it is divided by $i!$.

Hence you have $[B]_{j,k} = [\sum \frac {{\alpha}^i}{i!} X^i]_{j,k} = \sum [\frac {{\alpha}^i}{i!} X^i]_{j,k} \leq \sum [{\alpha}^i X^i]_{j,k} = [\sum {\alpha}^i X^i]_{j,k} = [A]_{j,k}$.

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Sorry, I've modified the problem. Could you kindly read it again? Thanks! – John Smith May 31 '12 at 7:10

If $\alpha$ and all entries of $X$ are nonnegative, the same goes for all entries of $\alpha^i X^i$ for any nonnegative integer $i$. If $t \ge 0$, $t/i! \le t$. So each entry of each term of the $B$ series $\le$ the corresponding entry of the corresponding term of the $A$ series. And since the series converge, we get $B_{i,j} \le A_{i,j}$.

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Sorry, I've modified the problem. Could you kindly read it again? Thanks! – John Smith May 31 '12 at 7:10