Constraints to make a matrix triangular [closed]












0












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I have a rank one matrix given below



$$A = bc^T$$



where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?










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closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    I have a rank one matrix given below



    $$A = bc^T$$



    where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?










    share|cite|improve this question









    $endgroup$



    closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$


      I have a rank one matrix given below



      $$A = bc^T$$



      where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?










      share|cite|improve this question









      $endgroup$




      I have a rank one matrix given below



      $$A = bc^T$$



      where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?







      linear-algebra matrices optimization






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      asked Nov 27 '18 at 4:46









      Dushyant SahooDushyant Sahoo

      528




      528




      closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






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          1












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          Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.



          Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?




          I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.







          share|cite|improve this answer









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            1












            $begingroup$

            Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.



            If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.






            share|cite|improve this answer









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              1












              $begingroup$

              We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.






              share|cite|improve this answer









              $endgroup$




















                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.



                Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?




                I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.







                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.



                  Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?




                  I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.







                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.



                    Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?




                    I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.







                    share|cite|improve this answer









                    $endgroup$



                    Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.



                    Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?




                    I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.








                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 '18 at 4:53









                    angryavianangryavian

                    40.5k23280




                    40.5k23280























                        1












                        $begingroup$

                        Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.



                        If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.



                          If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.



                            If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.






                            share|cite|improve this answer









                            $endgroup$



                            Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.



                            If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 27 '18 at 5:10









                            user1551user1551

                            72.4k566127




                            72.4k566127























                                1












                                $begingroup$

                                We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 29 '18 at 10:47









                                    Mostafa AyazMostafa Ayaz

                                    15.3k3939




                                    15.3k3939















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