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I have the following implication. How can I show that it holds true?

Let $K$ be a commutative ring. For any integer $p$, let $M_p := \{R \in K^{n \times n}: r_{ij} = 0,\ \mathrm{if}\ i > j - p\}$, where $r_{ij}$ denotes the $(i,j)$-th entry of the matrix $R$.

Show that $R \in M_p \wedge S \in M_q \Longrightarrow RS \in M_{p+q}$.

This is a generalization of the known fact (obtained by setting $p = 0$ and $q = 0$) that the product of two upper-triangular matrices is upper-triangular.

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Let $R \in M_p$ and $S \in M_q$. If $i > j-(p+q)$, then the $i,j$ entry of $RS$ is given by $(RS)_{ij} = \sum_{k=1}^n r_{ik} s_{kj}$; we would like to show that this sum is equal to zero. Because $R \in M_p$, we will have $r_{ik} = 0$ whenever $i > k - p \iff k < i+p$. Similarly, $s_{kj} = 0$ whenever $k > j - q$.

So, we can rewrite $$ (RS)_{ij} = \sum_{k=1}^n r_{ik} s_{kj} = \sum_{(j-q)\geq k \geq (i+p)} r_{ik}s_{kj}. $$ However, if $i > j - (p+q)$, then there are no integers $k$ such that $(j-q)\geq k \geq (i+p)$.

darij grinberg
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Ben Grossmann
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  • Thank you for your answer! However, could you please elaborate why we would like to show that the sum is equal to zero? The implication doesn't hold true, right? –  Dec 27 '19 at 22:04
  • Also why there aren't such integers k if $i > j − (p + q)$? –  Dec 27 '19 at 22:29
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    @philk982 to your first comment: to show that $RS \in R_{p+q}$, we need to show that $(RS)_{ij} = 0$ (when $i > j - (p+q)$). The entry which must be equal to zero is equal to that sum – Ben Grossmann Dec 27 '19 at 23:00
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    @philk982 to your second comment: if $i > j − (p + q)$, then $(j-q) > (i+p)$. – Ben Grossmann Dec 27 '19 at 23:03
  • I have taken the liberty to rename $R_{p+q}$ by $M_{p+q}$ in order to disentangle notations for matrices and matrix spaces. The comments above should be translated accordingly. – darij grinberg Dec 28 '19 at 02:07