For my univerity studies I had to prove this:

$ Let \,\, n \,\ \in N \,\, and \,\, L1,\, L2 \, \in R(n \times n) \,\, be \,\, both \,\, lower \,\, triangular \,\, matrices.$ $ Show \,\, that \,\, L := L1L2 \,\, is \,\, also \,\, a \,\, lower \,\, triangular \,\, matrix.$

I proved it like this (and I need some verification for the proof):

$L_{ij} := (L1L2)_{ij}$

$Now \,\, just \,\, look \,\, at \,\, (L1L2)_{ij} \,\, where \,\, j > i.$

$(L1L2)_{ij} = \sum_{r=1}^n l1_{ir}l2_{rj} = \sum_{r=j}^i l1_{ir}l2_{rj} = \sum_{r=1}^n l1_{ir}l2_{rj} = \begin{cases} 0, & \text{if $j>i$ is even} \\ a \in R, & \text{else} \end{cases} $

$\Rightarrow \text{L is lower triangular}$


  • 2
    $\begingroup$ This is incomprehensible. You have $\sum_{r=1}^nL1_{ir}L2_{rj}=\sum_{r=j}^iL1_{ir}L2_{rj}$ which obviously is unjustified and then you turn it back to the original indexing with utter no reason for doing this at all. The is you claim for no reason whatsever if $0$ if $j> i$ is even (which is imparsible; I assume you mean if $j$ is even) although there is no reason why that would be-- and is some identified constant $a$ if not. What is $a$ and why is it equal to $a$? And then you claimed this means it is lower triangular but those $a$ terms mean it is not. $\endgroup$
    – fleablood
    Dec 5, 2018 at 18:10

2 Answers 2


Let me call the matrices $A$ and $B$ to avoid too many subscripts. Fix $i<j$. Then $$ (AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj}. $$ But either $i<k$ or $k<j$ holds (because the negation $k\leq i<j\leq k$ is impossible), so $A_{ik}=0$ or $B_{kj}=0$. Therefore all the addends are zero and $(AB)_{ij}=0$.

Analysis of the OP's proof

When you write $$ \sum_{r=1}^n L1_{ir} L2_{rj} = \sum_{r=j}^i L1_{ir} L2_{rj}, $$ what are you trying to say? Notice also that the starting index of the summation is larger than the ending one. What is it that you want to express?

The next step in your computation then is just reverting back to the previous formula. Why is that?

Finally, what does "$j>i$ is even" mean? And what about $a\in R$?

In summary, I would not consider your proof a proof at all.

  • $\begingroup$ I mean thank you for your time and showing me another way of proving this but my question was if my prove is rigerous like I formulated it? $\endgroup$ Dec 5, 2018 at 17:49
  • $\begingroup$ Well in that case, not so much. I'll expand my answer $\endgroup$
    – Federico
    Dec 5, 2018 at 17:51
  • $\begingroup$ @Mathmeeeeen What I showed you is not "another" way of proving it. It's the way to express rigorously this simple fact. What you wrote cannot be considered a proof $\endgroup$
    – Federico
    Dec 5, 2018 at 17:59
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    $\begingroup$ It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly. $\endgroup$
    – fleablood
    Dec 5, 2018 at 18:14
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    $\begingroup$ alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof". $\endgroup$ Dec 5, 2018 at 20:27

Let $A=[a_{ij}]$ and $B=[b_{ij}]$ be two $n \times n$ lower triangular matrices.

By definition

\begin{equation} a_{ij}=b_{ij}=0~~\text{if}~~~i<j. \end{equation}

Let $AB=[c_{ij}]$. It is sufficient to show that $c_{ij}=0$ whenever $i<j$ where

\begin{equation} c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}. \end{equation} Case I: $k \leq i <j$.

In this case, $b_{kj}=0$ for $k,j\in \{1,2,\ldots,n\}$ with $k<j$.

Case II: $i < k$

In this case, $a_{ik}=0$ for $i,k\in \{1,2,\ldots,n\}$ with $i<k$.

Hence, for $i<j$, we have \begin{eqnarray} c_{ij}&=&\sum_{k=1}^i a_{ik}b_{kj}+\sum_{k=i+1}^n a_{ik}b_{kj}\\ &=& 0+0\\ &=&0 \end{eqnarray}


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