When does an under-specified linear system have a unique solution?












2












$begingroup$


Consider the system of two equations with three real variables $x,y,z$:
$$
a_{11}cdot x + a_{12} cdot y + a_{13} cdot z = b_1
\
a_{21} cdot (1-x) + a_{22} cdot (1-y) + a_{23} cdot (1-z) = b_2
$$
where $x,y,zgeq 0$.



Since there are less equations than variables, this system may have infinitely many solutions. My question is: is there a simple condition on $b_1,b_2$ (as a function of the coefficients $a_{ij}$), that guarantees that this system has a unique solution?






linear-algebra systems-of-equations







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$endgroup$

















    2












    $begingroup$


    Consider the system of two equations with three real variables $x,y,z$:
    $$
    a_{11}cdot x + a_{12} cdot y + a_{13} cdot z = b_1
    \
    a_{21} cdot (1-x) + a_{22} cdot (1-y) + a_{23} cdot (1-z) = b_2
    $$
    where $x,y,zgeq 0$.



    Since there are less equations than variables, this system may have infinitely many solutions. My question is: is there a simple condition on $b_1,b_2$ (as a function of the coefficients $a_{ij}$), that guarantees that this system has a unique solution?






    linear-algebra systems-of-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Consider the system of two equations with three real variables $x,y,z$:
      $$
      a_{11}cdot x + a_{12} cdot y + a_{13} cdot z = b_1
      \
      a_{21} cdot (1-x) + a_{22} cdot (1-y) + a_{23} cdot (1-z) = b_2
      $$
      where $x,y,zgeq 0$.



      Since there are less equations than variables, this system may have infinitely many solutions. My question is: is there a simple condition on $b_1,b_2$ (as a function of the coefficients $a_{ij}$), that guarantees that this system has a unique solution?






      linear-algebra systems-of-equations







      share|cite|improve this question











      $endgroup$




      Consider the system of two equations with three real variables $x,y,z$:
      $$
      a_{11}cdot x + a_{12} cdot y + a_{13} cdot z = b_1
      \
      a_{21} cdot (1-x) + a_{22} cdot (1-y) + a_{23} cdot (1-z) = b_2
      $$
      where $x,y,zgeq 0$.



      Since there are less equations than variables, this system may have infinitely many solutions. My question is: is there a simple condition on $b_1,b_2$ (as a function of the coefficients $a_{ij}$), that guarantees that this system has a unique solution?






      linear-algebra systems-of-equations




      linear-algebra systems-of-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 13:56







      Erel Segal-Halevi

















      asked Jan 11 at 10:43









      Erel Segal-HaleviErel Segal-Halevi

      4,27011760




      4,27011760






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          For example if you have that
          $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if



          $b_2=a_{21}+a_{22}+a_{23}$



          So in the case in which
          begin{cases}
          b_1=0\
          b_2=a_{21}+a_{22}+a_{23}
          end{cases}



          your only solution is $x=y=z=0$



          The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,zgeq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,zgeq 0$ is only the origin.
          In this case your only solution will be the origin.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            In general the answer is no: you are looking for a solution $xgeq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $xgeq0$, or $xgeq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+zgeq0$, so you have at least two solutions.
            Since your $A$ is completely arbitrary, you have cases satisfying (1).






            share|cite|improve this answer











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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              2












              $begingroup$

              For example if you have that
              $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if



              $b_2=a_{21}+a_{22}+a_{23}$



              So in the case in which
              begin{cases}
              b_1=0\
              b_2=a_{21}+a_{22}+a_{23}
              end{cases}



              your only solution is $x=y=z=0$



              The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,zgeq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,zgeq 0$ is only the origin.
              In this case your only solution will be the origin.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                For example if you have that
                $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if



                $b_2=a_{21}+a_{22}+a_{23}$



                So in the case in which
                begin{cases}
                b_1=0\
                b_2=a_{21}+a_{22}+a_{23}
                end{cases}



                your only solution is $x=y=z=0$



                The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,zgeq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,zgeq 0$ is only the origin.
                In this case your only solution will be the origin.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For example if you have that
                  $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if



                  $b_2=a_{21}+a_{22}+a_{23}$



                  So in the case in which
                  begin{cases}
                  b_1=0\
                  b_2=a_{21}+a_{22}+a_{23}
                  end{cases}



                  your only solution is $x=y=z=0$



                  The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,zgeq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,zgeq 0$ is only the origin.
                  In this case your only solution will be the origin.






                  share|cite|improve this answer











                  $endgroup$



                  For example if you have that
                  $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if



                  $b_2=a_{21}+a_{22}+a_{23}$



                  So in the case in which
                  begin{cases}
                  b_1=0\
                  b_2=a_{21}+a_{22}+a_{23}
                  end{cases}



                  your only solution is $x=y=z=0$



                  The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,zgeq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,zgeq 0$ is only the origin.
                  In this case your only solution will be the origin.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 11 at 11:34

























                  answered Jan 11 at 11:24









                  Federico FalluccaFederico Fallucca

                  1,95919




                  1,95919























                      0












                      $begingroup$

                      In general the answer is no: you are looking for a solution $xgeq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $xgeq0$, or $xgeq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+zgeq0$, so you have at least two solutions.
                      Since your $A$ is completely arbitrary, you have cases satisfying (1).






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        In general the answer is no: you are looking for a solution $xgeq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $xgeq0$, or $xgeq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+zgeq0$, so you have at least two solutions.
                        Since your $A$ is completely arbitrary, you have cases satisfying (1).






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          In general the answer is no: you are looking for a solution $xgeq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $xgeq0$, or $xgeq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+zgeq0$, so you have at least two solutions.
                          Since your $A$ is completely arbitrary, you have cases satisfying (1).






                          share|cite|improve this answer











                          $endgroup$



                          In general the answer is no: you are looking for a solution $xgeq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $xgeq0$, or $xgeq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+zgeq0$, so you have at least two solutions.
                          Since your $A$ is completely arbitrary, you have cases satisfying (1).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 11 at 11:40

























                          answered Jan 11 at 11:29









                          Jose BroxJose Brox

                          3,13711028




                          3,13711028






























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