Product of two lower triangular matrices is a lower triangular matrix












0














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}
$



$Rightarrow text{L is lower triangular}$



$square$






linear-algebra matrices proof-verification







share|cite|improve this question


















  • 1




    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.
    – fleablood
    Dec 5 '18 at 18:10


















0














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}
$



$Rightarrow text{L is lower triangular}$



$square$






linear-algebra matrices proof-verification







share|cite|improve this question


















  • 1




    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.
    – fleablood
    Dec 5 '18 at 18:10
















0












0








0







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}
$



$Rightarrow text{L is lower triangular}$



$square$






linear-algebra matrices proof-verification







share|cite|improve this question













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}
$



$Rightarrow text{L is lower triangular}$



$square$






linear-algebra matrices proof-verification




linear-algebra matrices proof-verification






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asked Dec 5 '18 at 17:36









MathmeeeeenMathmeeeeen

133




133








  • 1




    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.
    – fleablood
    Dec 5 '18 at 18:10
















  • 1




    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.
    – fleablood
    Dec 5 '18 at 18:10










1




1




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.
– fleablood
Dec 5 '18 at 18:10






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.
– fleablood
Dec 5 '18 at 18:10












1 Answer
1






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3














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 $kleq i<jleq k$ 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 $ain R$?



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






share|cite|improve this answer























  • 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?
    – Mathmeeeeen
    Dec 5 '18 at 17:49










  • Well in that case, not so much. I'll expand my answer
    – Federico
    Dec 5 '18 at 17:51










  • @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
    – Federico
    Dec 5 '18 at 17:59






  • 1




    It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
    – fleablood
    Dec 5 '18 at 18:14










  • alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
    – Mathmeeeeen
    Dec 5 '18 at 20:27











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1 Answer
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3














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 $kleq i<jleq k$ 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 $ain R$?



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






share|cite|improve this answer























  • 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?
    – Mathmeeeeen
    Dec 5 '18 at 17:49










  • Well in that case, not so much. I'll expand my answer
    – Federico
    Dec 5 '18 at 17:51










  • @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
    – Federico
    Dec 5 '18 at 17:59






  • 1




    It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
    – fleablood
    Dec 5 '18 at 18:14










  • alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
    – Mathmeeeeen
    Dec 5 '18 at 20:27
















3














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 $kleq i<jleq k$ 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 $ain R$?



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






share|cite|improve this answer























  • 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?
    – Mathmeeeeen
    Dec 5 '18 at 17:49










  • Well in that case, not so much. I'll expand my answer
    – Federico
    Dec 5 '18 at 17:51










  • @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
    – Federico
    Dec 5 '18 at 17:59






  • 1




    It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
    – fleablood
    Dec 5 '18 at 18:14










  • alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
    – Mathmeeeeen
    Dec 5 '18 at 20:27














3












3








3






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 $kleq i<jleq k$ 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 $ain R$?



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






share|cite|improve this answer














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 $kleq i<jleq k$ 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 $ain R$?



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 17:57

























answered Dec 5 '18 at 17:45









FedericoFederico

4,859514




4,859514












  • 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?
    – Mathmeeeeen
    Dec 5 '18 at 17:49










  • Well in that case, not so much. I'll expand my answer
    – Federico
    Dec 5 '18 at 17:51










  • @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
    – Federico
    Dec 5 '18 at 17:59






  • 1




    It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
    – fleablood
    Dec 5 '18 at 18:14










  • alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
    – Mathmeeeeen
    Dec 5 '18 at 20:27


















  • 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?
    – Mathmeeeeen
    Dec 5 '18 at 17:49










  • Well in that case, not so much. I'll expand my answer
    – Federico
    Dec 5 '18 at 17:51










  • @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
    – Federico
    Dec 5 '18 at 17:59






  • 1




    It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
    – fleablood
    Dec 5 '18 at 18:14










  • alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
    – Mathmeeeeen
    Dec 5 '18 at 20:27
















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?
– Mathmeeeeen
Dec 5 '18 at 17:49




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?
– Mathmeeeeen
Dec 5 '18 at 17:49












Well in that case, not so much. I'll expand my answer
– Federico
Dec 5 '18 at 17:51




Well in that case, not so much. I'll expand my answer
– Federico
Dec 5 '18 at 17:51












@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
– Federico
Dec 5 '18 at 17:59




@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
– Federico
Dec 5 '18 at 17:59




1




1




It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
– fleablood
Dec 5 '18 at 18:14




It's hardly "another" way of proving. It's clearly the exact attempt you were attempting but mangled to incomprehensibility but done correctly.
– fleablood
Dec 5 '18 at 18:14












alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
– Mathmeeeeen
Dec 5 '18 at 20:27




alright thank you both. That was exactly the reason for my question, I was doubting the rigidity of my "proof".
– Mathmeeeeen
Dec 5 '18 at 20:27


















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